ohm's law experiment with rheostat

Verification of Ohm’s Law experiment with data and graph

In the previous article, we discussed Ohm’s Law of current electricity. In this article, we’re going to perform an experiment for the verification of Ohm’s law. This practical verification of Ohm’s law is very important for the students of grades 10 and 12. This is a lab-based experiment to verify Ohm’s law or Ohm’s law practical.

Aim of the Experiment

Theory of the ohm’s law experiment.

From Ohm’s law , we know that the relation between electric current and potential difference is V = IR

or, \color{Blue}R=\frac{V}{I} ………….. (1)

or, resistivity, \color{Blue}\rho = \frac{RA}{L} ………. (2)

Where A is the cross-section area of the wire. A = πr 2 where r is the radius of the wire. L is the length of the wire.

Apparatus Used

The apparatus used for this experiment –

Circuit Diagram

Here, R is the resistance of the wire, A is the ammeter, V is the Voltmeter, Rh is the rheostat and K is the key. The arrow sign indicates the direction of the current flow in the circuit .

Formula used for the Ohm’s law lab experiment

\color{Blue}R = \frac{V}{I} ………….. (1) and \color{Blue}\rho = \frac{RA}{L} ………. (2)

Experimental data

The least count of Voltmeter = Smallest division of voltmeter = 0.05 Volt

ervation
1	0	0	–
2	0.50	0.50	1.00
3	0.65	0.65	1.00
4	0.80	0.80	1.00	1.02
5	1.00	1.05	1.05
6	1.15	1.20	1.04

We also need to plot I-V graph to confirm the experimental value of R.

Current versus Voltage graph (Ohm’s Law graph)

If we plot the Current as a function of voltage with the help of the above data then we will get a straight line passing through the origin.

Calculations

Calculation of resistance from the graph.

The inverse of the I-V graph gives the resistance of the wire. Now, from the graph, change in current, ∆I = AB = 0.5 amp corresponding change in voltage, ∆V = BC = 0.5 volt Thus, the Resistance from the graph, R = ∆V/∆I = 0.5/0.5 = 1.00 ohm

Calculation of resistivity of the wire

Length of the wire is, L = 50 cm = 0.5 m Radius of the wire. r = 0.25 mm = 0.25 × 10 -3 m So, the cross-section area of the wire, A = πr 2 = 3.14 × (0.25×10 -3 ) 2 = 0.196 × 10 -6 m 2 Thus from the equation-2 we get the resistivity of the material of the wire is, \rho = (1 × 0.196 ×10 -6 )/0.5 or, \rho = 0.392 × 10 -6 = 3.92 ×10 -7 ohm.m Thus the resistivity of the material of the wire is 3.92 ×10 -7 ohm.m

Final result

The resistance of the wire from the Current-Voltage graph is, R = 1.00 ohm The calculated value of the resistance of the wire is, R = 1.02 ohm. Resistivity of the material of the wire is 3.92 ×10 -7 ohm.m

Discussions

Current Electricity

Ohm’s law states the relationship between electric current and potential difference. The current that flows through most conductors is directly proportional to the voltage applied to it. Georg Simon Ohm, a German physicist was the first to verify Ohm’s law experimentally.

Ohm’s Law Explanation

One of the most basic and important laws of electric circuits is Ohm’s law.

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Mathematically, this current-voltage relationship is written as,

In the equation, the constant of proportionality, R, is called Resistance and has units of ohms, with the symbol Ω.

The same formula can be rewritten in order to calculate the current and resistance respectively as follows:

Ohm’s law only holds true if the provided temperature and the other physical factors remain constant. In certain components, increasing the current raises the temperature . An example of this is the filament of a light bulb, in which the temperature rises as the current is increased. In this case, Ohm’s law cannot be applied. The lightbulb filament violates Ohm’s Law .

: Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperature, remain constant.

: V = IR, where V is the voltage across the conductor, I is the current flowing through the conductor and R is the resistance provided by the conductor to the flow of current.

Relationship Between Voltage, Current and Resistance

Water Pipe Analogy for Ohm’s Law

Ohm’s Law describes the current flow through a resistance when different electric potentials (voltage) are applied at each end of the resistance . Since we can’t see electrons, the water-pipe analogy helps us understand the electric circuits better. Water flowing through pipes is a good mechanical system that is analogous to an electrical circuit.

Here, the voltage is analogous to water pressure, the current is the amount of water flowing through the pipe, and the resistance is the size of the pipe. More water will flow through the pipe (current) when more pressure is applied (voltage) and the bigger the pipe (lower the resistance).

The video below shows the physical demonstration of the Waterpipe analogy and explains to you the factors that affect the flow of current

Experimental Verification of Ohm’s Law

Ohm’s Law can be easily verified by the following experiment:

Apparatus Required:

Initially, the key K is closed and the rheostat is adjusted to get the minimum reading in ammeter A and voltmeter V.
The current in the circuit is increased gradually by moving the sliding terminal of the rheostat. During the process, the current flowing in the circuit and the corresponding value of potential difference across the resistance wire R are recorded.
This way different sets of values of voltage and current are obtained.
For each set of values of V and I, the ratio of V/I is calculated.
When you calculate the ratio V/I for each case, you will come to notice that it is almost the same. So V/I = R, which is a constant.
Plot a graph of the current against the potential difference, it will be a straight line. This shows that the current is proportional to the potential difference.

Ohm’s Law Magic Triangle

Ohm’s Law Solved Problems

Example 1: If the resistance of an electric iron is 50 Ω and a current of 3.2 A flows through the resistance. Find the voltage between two points.

If we are asked to calculate the value of voltage with the value of current and resistance, then cover V in the triangle. Now, we are left with I and R or more precisely I × R.

Therefore, we use the following formula to calculate the value of V:

Substituting the values in the equation, we get

V = 3.2 A × 50 Ω = 160 V

Example 2: An EMF source of 8.0 V is connected to a purely resistive electrical appliance (a light bulb). An electric current of 2.0 A flows through it. Consider the conducting wires to be resistance-free. Calculate the resistance offered by the electrical appliance.

When we are asked to determine the value of resistance when the values of voltage and current are given, we cover R in the triangle. This leaves us with only V and I, more precisely V ÷ I.

R = 8 V ÷ 2 A = 4 Ω

Calculating Electrical Power Using Ohm’s Law

The rate at which energy is converted from the electrical energy of the moving charges to some other form of energy like mechanical energy, heat energy, energy stored in magnetic fields or electric fields, is known as electric power. The unit of power is the watt. The electrical power can be calculated using Ohm’s law and by substituting the values of voltage, current and resistance.

Formula to find power

What is a Power Triangle?

The power triangle can be employed to determine the value of electric power, voltage and current when the values of the other two parameters are given to us. In the power triangle, the power (P) is on the top and current (I) and voltage (V) are at the bottom.

Ohm’s Law Pie Chart

Ohm’s Law Matrix Table

Ohm’s Law Applications

The main applications of Ohm’s law are:

To determine the voltage, resistance or current of an electric circuit.
Ohm’s law maintains the desired voltage drop across the electronic components.
Ohm’s law is also used in DC ammeter and other DC shunts to divert the current.

Limitations of Ohm’s Law

Following are the limitations of Ohm’s law:

Ohm’s law is not applicable for unilateral electrical elements like diodes and transistors as they allow the current to flow through in one direction only.
For non-linear electrical elements with parameters like capacitance, resistance etc the ratio ofvoltage and current won’t be constant with respect to time making it difficult to use Ohm’s law.

The video about conductance, resistance, and ohm’s law

Frequently Asked Questions – FAQs

What does ohm’s law state.

Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

What can Ohm’s law be used for?

Ohm’s law is used to validate the static values of circuit components such as current levels, voltage supplies, and voltage drops.

Is Ohm’s law Universal?

No. Ohm’s law is not a universal law. This is because Ohm’s law is only applicable to ohmic conductors such as iron and copper but is not applicable to non-ohmic conductors such as semiconductors.

Why is Ohm’s law not applicable to semiconductors?

Ohm’s law doesn’t apply to semiconducting devices because they are nonlinear devices. This means that the ratio of voltage to current doesn’t remain constant for variations in voltage.

When does Ohm’s law fail?

Ohm’s law fails to explain the behaviour of semiconductors and unilateral devices such as diodes. Ohm’s law may not give the desired results if the physical conditions such as temperature or pressure are not kept constant.

Watch the video and solve important questions in the chapter Electricity Class 10

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

Visit BYJU’S for all Physics related queries and study materials

Your result is as below

Request OTP on Voice Call

PHYSICS Related Links

Register with BYJU'S & Download Free PDFs

Network Sites:
Technical Articles
Market Insights

Or sign in with
iHeartRadio

Intro Lab - Ohm’s Law

Join our Engineering Community! Sign-in with:

DIY Electronics Projects

Basic Projects and Test Equipment

Intro Lab - How to Use a Voltmeter to Measure Voltage
Intro Lab - How to Use an Ohmmeter to Measure Resistance
Intro Lab - How to Use an Ammeter to Measure Current
Intro Lab - Resistor Power Dissipation
Intro Lab - A Simple Lighting Circuit
Intro Lab - Nonlinear Resistance
Intro Lab - Circuit With a Switch
Intro Lab - Build an Electromagnet
Intro Lab - Electromagnetic Induction

Project Overview

This project employs all of the skills explained earlier in this textbook for using an ohmmeter , voltmeter , and ammeter . You will build a simple circuit, as illustrated in Figure 1, to experimentally verify Ohm's law which defines the mathematical relationship between resistance , voltage, and current.

Figure 1. Test circuit for evaluating Ohm's law.

Parts and materials.

6 V battery
Assortment of resistors between 1 kΩ and 100 kΩ in value

For this experiment, I’m purposely restricting the resistance values between 1 kΩ and 100 kΩ for the sake of obtaining accurate voltage and current readings with your meter.

With very low resistance values, the internal resistance of the ammeter has a significant impact on measurement accuracy. Very high resistance values can cause problems for voltage measurement since the internal resistance of the voltmeter substantially changes circuit resistance when it is connected in parallel with a high-value resistor. At the recommended resistance values, there will still be a small amount of measurement error due to the impact of the meter, but not enough to cause serious disagreement with calculated values.

Learning Objectives

Application of Ohm’s law
Voltmeter use
Ammeter use
Ohmmeter use

Instructions

Step 1: Select a resistor from the assortment, and measure its resistance, R , with your multimeter set to the appropriate resistance range. Be sure not to hold the resistor terminals when measuring resistance, or else your hand-to-hand body resistance will influence the measurement! Record this resistance value for future use.

Step 2: Next, build the one-battery, one-resistor circuit, shown in the schematic diagram of Figure 2.

Figure 2. Test circuit schematic diagram for evaluating Ohm's law.

If you have two multimeters, you can construct exactly as shown in Figures 1 and 2 and measure both voltage and current simultaneously. However, in the following descriptions, we will assume you have only one multimeter and will perform the measurements in steps.

A terminal strip implementation of this circuit is demonstrated in Figure 1, but you can also use a breadboard or any other type of circuit construction method.

Step 3: Set your multimeter to the appropriate voltage range and measure the voltage, V , across the resistor as it is being powered by the battery. Record this voltage value.

Step 4: Set your multimeter to the highest current range available. Break the circuit and connect the ammeter within that break, so it becomes a part of the circuit, in series, with the battery and resistor.

Step 5: Select the best current range; whichever one gives the strongest meter indication without over-ranging the meter. If your multimeter is autoranging, of course, you need not bother with setting ranges. Record this current value along with the resistance and voltage values previously recorded.

Step 6: Taking the measured figures for voltage and resistance, use Ohm’s law equation to calculate circuit current.

$$I = \frac{V}{R}$$

I = current in amps (A)
V = voltage in volts (V)
R = resistance in ohms (Ω)

Compare this calculated figure for the current with the measured figure for the circuit current. They should be identical (or very nearly so).

Step 7: Taking the measured figures for voltage and current, use Ohm’s law equation to calculate circuit resistance. Compare this calculated figure with the measured figure for circuit resistance.

$$R = \frac{V}{I}$$

Step 8: Finally, taking the measured figures for resistance and current, use Ohm’s law equation to calculate circuit voltage. Again, compare this calculated figure with the measured figure for circuit voltage.

$$V = I \cdot R$$

There should be close agreement between all measured and all calculated figures. Any differences in respective quantities of voltage, current, or resistance are most likely due to meter inaccuracies. These differences should be rather small, no more than several percent. Some meters, of course, are more accurate than others!

Step 9: Select different resistors are repeat steps 1 through 8. Re-take all resistance, voltage, and current measurements. Then, re-calculate these figures and check for agreement with the experimental data (measured quantities).

Also, note the simple mathematical relationship between changes in resistor value and changes in circuit current. Voltage should remain approximately the same for any resistor size inserted into the circuit because it is the nature of a battery to maintain voltage at a constant level.

Lessons in Electric Circuits

Volumes ».

Direct Current (DC)
Alternating Current (AC)
Semiconductors
Digital Circuits
EE Reference

Chapters »

1 Introduction to Electronics Projects

Pages »

3 DC Circuit Projects
4 AC Circuit Projects
5 Discrete Semiconductor Circuit Projects
6 Analog IC Projects
7 Digital IC Projects
8 555 Timer Circuit Projects
9 Contributor List
Advanced Textbooks Practical Guide to Radio-Frequency Analysis and Design
Designing Analog Chips
Ohm’s Law Calculator
Silicon Labs Bluetooth Solutions
Ohm’s Law
Ohm’s Law, Kirchhoff’s Laws, and Power Equations
Innovative Bluetooth Technology with Silicon Labs
The Three Laws of Thermodynamics

Würth Elektronik at the PCIM Europe 2023 in Nuremberg

In Partnership with Würth Elektronik eiSos GmbH & Co. KG

Bluetooth Can Now Measure Distance, With Hardware Support En Route

by Duane Benson

Phison Carves Out New SSD Brand Along With Inaugural Series

by Aaron Carman

An Introduction to Using Logic Gates in LTspice

by Robert Keim

How to Analyze Transmission Line Transformers: The Easy Way and the Hard Way

by Dr. Steve Arar

Welcome Back

Don't have an AAC account? Create one now .

Forgot your password? Click here .

Learn all about Ohm's law

5 Error Sources in Ohm’s Law Experiment [How to avoid them]

The practical observations of Ohm’s law experiment never match the theoretical readings.

In fact, you can never match the theoretical calculations with practical values.

However, you can take some precautions to closely match the values.

Today’ you’ll learn the 5 error sources which are responsible for misleading readings. You’ll learn to keep you and your equipment safe by avoiding the blunders. You’ll also learn to obtain quite accurate readings. Let’s start off by understanding the types of errors.

Scientific measurement and instrumentation errors are often classified into three types:

Personal errors: Mistakes made by the user due to his inexperience.
Systematic: The faults in the instrument itself and the faults which may occur due to environmental conditions.
Random errors: An accidental error whose cause is unknown. (We’ll ignore it here).

What is a personal error [Don’t of Ohm’s law]

Generally, a personal error is an outright mistake which is made by the person himself. For example, you ignore a digit while taking observations. In case of Ohm’s law, you can commit a personal error by:

Wrong connecting the circuit

The ammeter is used to measure the current. It always connects in series with the circuit. Wrong connecting the ammeter will damage the instrument.

The voltmeter measures the potential difference between two points. It connects in parallel to the circuit. Wrong connecting the voltmeter will yield wrong readings.

Wrong taking the readings

Wrong measurements usually happen due to careless handling behavior. Carefully take the readings to avoid the errors.

Systematic errors

Tolerance values of resistors.

Carbon and metal film resistors are the most popular class of resistors which are employed in our labs. Such resistors have a tolerance value which ranges between 0.05-20%. The leftmost band of carbon resistors indicates the possible tolerance of resistance. A silver band indicates a tolerance of 10%, the golden band indicates 5% and brown band indicates 1%. More tolerance means your resistance, and thus the voltage/current will fluctuate away from the theoretical value.

You have two choices to bypass this error.

Use a brown [1%] or grey [0.05%] band resistor which has low tolerance value and thus will provide a lower error.

Measure the resistance first and base your theoretical formula calculations on this value.

Quality of Multimeter

Your multimeter is the actual tool which measures the electrical quantities. While low-quality multimeters yield wrong observations, they are equally dangerous. Again you have two choices.

Variable DC Power Supply

A variable power supply displays the output voltages on its main screen. For the time being, the accuracy of components decreases and your supply might display wrong results. Such cases are common in general labs where supplies are used thousands of times.

Use your multimeter to confirm the actual volts coming out of power supply.

Let’s summarize our results:

← Ohm’s Law in Series Circuits
Theory VS Experimental Verification of Ohm’s Law →

Labkafe Blog

Get in Touch

Get A Free Lab Consultation

How to perform ohm’s law experiment for class 10 | labkafe, how to perform ohm’s law experiment.

Study the dependence of potential difference (V) across a resistor on the current (I) passing through it and determine its resistance. (Ohm’s Law)

One of the most common tasks you can face in the 10th-grade physics or science laboratory is to perform the Ohm’s Law Experiment. It is common in class IX or class X in most boards like CBSE, ICSE, state boards, and even in IGCSE or IB curriculums. This experiment is also there for certain class 12 syllabus.

We have covered the theory of Ohm’s law experiment before, now let’s move on to the practical session of verifying Ohm’s law. But first, let’s state Ohm’s law for convenience’s sake once.

The current through a conductor between two points is directly proportional to the voltage across the two points.

Ohm’s Law Mathematical Expression

V = voltage across a given conductor
I = current through the circuit
R = resistance of the conductor

Base Idea of the Experiment

Here V would be measured across the conductor, and I would be measured in series to it. We will change the voltage, and record the current as a result. When we have a bunch of readings, we will plot the readings in a graph.

Our intention is to see if the graph would come out as close to a straight line as possible. That would mean that the resistance remained the same, denoting V was proportional to I . When we see that V and I are proportional, we will get the value of the resistance from the above formula.

The Experiment

To perform the Ohm’s law experiment, you will need the following equipment:

1 battery eliminator (0-12 volts, 2 Amps)
1 voltmeter
1 resistance box (or an unknown resistor)
Connecting wires

You can find all of those in Labkafe’s Composite Lab Package or Physics Lab Package .

We will need to build an electronic circuit with the battery and the resistor, through which the current is supposed to pass. To measure that current, you will need to connect the ammeter in between the battery and the resistor in series . And to measure the voltage, you will have to connect the voltmeter in parallel to the resistor.

Since the ammeter measures the current passing through the circuit, that is why it has to be a part of the circuit itself. That means, you will have to connect the positive (red) pole of the battery to the negative (black) pole of the ammeter, and connect the positive pole of the ammeter to the rest of the circuit. Since an ideal ammeter has ZERO resistance, we can assume that there will be no loss of current across the ammeter itself.

How to perform the experiment

Connect the equipment as shown in the circuit diagram. Keep the switch off for now and the rheostat at the least position.
Take a notebook and make two columns for voltage ( V ) and current ( I ).
Turn on the switch.
Observe the voltmeter and ammeter carefully. In the notebook, note down the values shown in the meters in the respective columns for V and I .
Turn off the power and move the rheostat to change its resistance a little. This will change the reading in the voltmeter.
Repeat steps 3, 4 and 5 a few more times till you get at least half a dozen different readings.
Plot the V vs I graph in a fresh graph paper. What does the graph look like?

Interpreting the Results

We have performed the Ohm’s law experiment with our electronics lab package, and here is the result data as chart and graph:

As you can see, the above data shows that the current rose steadily as the voltage increased and we had a pretty much straight lined graph. That means I was proportional to V . Proved!

Finding the resistance of the conductor

Since V = IR ,

So, R = V/I

From the above results, we can figure the average voltage as, V = 3.96v

And the average current was, I = 0.203A

Therefore, the effective value of the resistor in the experiment would be:

R = V/I = 3.96/0.203 = 19.53Ω .

How accurate are the results of Ohm’s Law experiment?

In reality, we used a 19 ohms resistor for the experiment. The error (0.53 ohms) probably came due to:

Heating issues during the Ohm’s law experiment
Internal resistances of the voltmeter, ammeter, or other components
The equipment being used not precisely perfect to theory

Do keep in mind that due to various natural issues it would be hard for you to get an accurate value of the resistor ‒ there can probably be a ~10% error margin. The older your equipment is, the more error you can expect.

All the equipment used in this experiment came from Labkafe’s standard composite lab package.

11.2 Ohm's Law

11.2 Ohm's Law (ESBQ6)

Three quantities which are fundamental to electric circuits are current, voltage (potential difference) and resistance . To recap:

Electrical current, $I$, is defined as the rate of flow of charge through a circuit.

Potential difference or voltage, $V$, is the amount of energy per unit charge needed to move that charge between two points in a circuit.

Resistance, $R$, is a measure of how `hard' it is to push current through a circuit element.

We will now look at how these three quantities are related to each other in electric circuits.

An important relationship between the current, voltage and resistance in a circuit was discovered by Georg Simon Ohm and it is called Ohm's Law .

The amount of electric current through a metal conductor, at a constant temperature, in a circuit is proportional to the voltage across the conductor and can be described by

where $I$ is the current through the conductor, $V$ is the voltage across the conductor and $R$ is the resistance of the conductor. In other words, at constant temperature, the resistance of the conductor is constant, independent of the voltage applied across it or current passed through it.

Ohm's Law tells us that if a conductor is at a constant temperature, the current flowing through the conductor is directly proportional to the voltage across it. This means that if we plot voltage on the x-axis of a graph and current on the y-axis of the graph, we will get a straight-line.

The gradient of the straight-line graph is related to the resistance of the conductor as \[\frac{I}{V} = \frac{1}{R}\] This can be rearranged in terms of the constant resistance as: \[R = \frac{V}{I}\]

To determine the relationship between the current going through a resistor and the potential difference (voltage) across the same resistor.

4 cells, 4 resistors, an ammeter, a voltmeter, connecting wires

This experiment has two parts. In the first part we will vary the applied voltage across the resistor and measure the resulting current through the circuit. In the second part we will vary the current in the circuit and measure the resulting voltage across the resistor. After obtaining both sets of measurements, we will examine the relationship between the current and the voltage across the resistor.

Varying the voltage:

Set up the circuit according to circuit diagram 1), starting with just one cell.

Draw the following table in your lab book.

Number of cells	Voltage, V ($\text{V}$)	Current, I ($\text{A}$)
$\text{1}$
$\text{2}$
$\text{3}$
$\text{4}$

Get your teacher to check the circuit before turning the power on.

Measure the voltage across the resistor using the voltmeter, and the current in the circuit using the ammeter.

Add one more $\text{1,5}$ $\text{V}$ cell to the circuit and repeat your measurements.

Repeat until you have four cells and you have completed your table.

Varying the current:

Set up the circuit according to circuit diagram 2), starting with only 1 resistor in the circuit.

Voltage, V ($\text{V}$)	Current, I ($\text{A}$)

Get your teacher to check your circuit before turning the power on.

Measure the current and measure the voltage across the single resistor.

Now add another resistor in series in the circuit and measure the current and the voltage across only the original resistor again. Continue adding resistors until you have four in series, but remember to only measure the voltage across the original resistor each time. Enter the values you measure into the table.

Analysis and results

Using the data you recorded in the first table, draw a graph of current versus voltage. Since the voltage is the variable which we are directly varying, it is the independent variable and will be plotted on the $x$-axis. The current is the dependent variable and must be plotted on the $y$-axis.

Using the data you recorded in the second table, draw a graph of voltage vs. current. In this case the independent variable is the current which must be plotted on the $x$-axis, and the voltage is the dependent variable and must be plotted on the $y$-axis.

Conclusions

Examine the graph you made from the first table. What happens to the current through the resistor when the voltage across it is increased? i.e. Does it increase or decrease?

Examine the graph you made from the second table. What happens to the voltage across the resistor when the current increases through the resistor? i.e. Does it increase or decrease?

Do your experimental results verify Ohm's Law? Explain.

Questions and discussion

For each of your graphs, calculate the gradient and from this determine the resistance of the original resistor. Do you get the same value when you calculate it for each of your graphs?

How would you go about finding the resistance of an unknown resistor using only a power supply, a voltmeter and a known resistor $R_0$?

Simulation: 242K

Use the data in the table below to answer the following questions.


$\text{3,0}$	$\text{0,4}$
$\text{6,0}$	$\text{0,8}$
$\text{9,0}$	$\text{1,2}$
$\text{12,0}$	$\text{1,6}$

Plot a graph of voltage (on the x-axis) and current (on the y-axis).

What type of graph do you obtain (straight-line, parabola, other curve)

Calculate the gradient of the graph.

The gradient of the graph ($m$) is the change in the current divided by the change in the voltage:

Yes. A straight line graph is obtained when we plot a graph of voltage vs. current.

How would you go about finding the resistance of an unknown resistor using only a power supply, a voltmeter and a known resistor $R_{0}$?

You start by connecting the known resistor in a circuit with the power supply. Now you read the voltage of the power supply and note this down.

Next you connect the two resistors in series. You can now take the voltage measurements for each of resistors.

So we can find the voltages for the two resistors. Now we note that:

So using this and the fact that for resistors in series, the current is the same everywhere in the circuit we can find the unknown resistance.

Ohmic and non-ohmic conductors (ESBQ7)

Conductors which obey Ohm's Law have a constant resistance when the voltage is varied across them or the current through them is increased. These conductors are called ohmic conductors. A graph of the current vs. the voltage across these conductors will be a straight-line. Some examples of ohmic conductors are circuit resistors and nichrome wire.

As you have seen, there is a mention of constant temperature when we talk about Ohm's Law. This is because the resistance of some conductors changes as their temperature changes. These types of conductors are called non-ohmic conductors, because they do not obey Ohm's Law. A light bulb is a common example of a non-ohmic conductor. Other examples of non-ohmic conductors are diodes and transistors.

In a light bulb, the resistance of the filament wire will increase dramatically as it warms from room temperature to operating temperature. If we increase the supply voltage in a real lamp circuit, the resulting increase in current causes the filament to increase in temperature, which increases its resistance. This effectively limits the increase in current. In this case, voltage and current do not obey Ohm's Law.

The phenomenon of resistance changing with variations in temperature is one shared by almost all metals, of which most wires are made. For most applications, these changes in resistance are small enough to be ignored. In the application of metal lamp filaments, which increase a lot in temperature (up to about $\text{1 000}$ $\text{℃}$, and starting from room temperature) the change is quite large.

In general, for non-ohmic conductors, a graph of voltage against current will not be a straight-line, indicating that the resistance is not constant over all values of voltage and current.

A recommended experiment for informal assessment is included. In this experiment learners will obtain current and voltage data for a resistor and light bulb and determine which obeys Ohm's law. You will need light bulbs, resistors, connecting wires, power source, ammeter and voltmeter. Learners should find that the resistor obeys Ohm's law, while the light bulb does not.

Ohmic and non-ohmic conductors

To determine whether two circuit elements (a resistor and a lightbulb) obey Ohm's Law

4 cells, a resistor, a lightbulb, connecting wires, a voltmeter, an ammeter

The two circuits shown in the diagrams above are the same, except in the first there is a resistor and in the second there is a lightbulb. Set up both the circuits above, starting with 1 cell. For each circuit:

Measure the voltage across the circuit element (either the resistor or lightbulb) using the voltmeter.

Measure the current in the circuit using the ammeter.

Add another cell and repeat your measurements until you have 4 cells in your circuit.

Draw two tables which look like the following in your book. You should have one table for the first circuit measurements with the resistor and another table for the second circuit measurements with the lightbulb.

Using the data in your tables, draw two graphs of $I$ ($y$-axis) vs. $V$ ($x$-axis), one for the resistor and one for the lightbulb.

Questions and Discussion

Examine your graphs closely and answer the following questions:

What should the graph of $I$ vs. $V$ look like for a conductor which obeys Ohm's Law?

Do either or both your graphs look like this?

What can you conclude about whether or not the resistor and/or the lightbulb obey Ohm's Law?

Is the lightbulb an ohmic or non-ohmic conductor?

Using Ohm's Law (ESBQ8)

We are now ready to see how Ohm's Law is used to analyse circuits.

Consider a circuit with a cell and an ohmic resistor, R. If the resistor has a resistance of $\text{5}$ $\text{Ω}$ and voltage across the resistor is $\text{5}$ $\text{V}$, then we can use Ohm's Law to calculate the current flowing through the resistor. Our first task is to draw the circuit diagram. When solving any problem with electric circuits it is very important to make a diagram of the circuit before doing any calculations. The circuit diagram for this problem looks like the following:

The equation for Ohm's Law is: \[R = \frac{V}{I}\]

which can be rearranged to: \[I = \frac{V}{R}\]

The current flowing through the resistor is:

\begin{align*} I &= \frac{V}{R} \\ &= \frac{\text{5}\text{ V}}{\text{5 }\Omega} \\ &= \text{1}\text{ A} \end{align*}

Worked example 1: Ohm's Law

Study the circuit diagram below:

The resistance of the resistor is $\text{10}$ $\text{Ω}$ and the current going through the resistor is $\text{4}$ $\text{A}$. What is the potential difference (voltage) across the resistor?

Determine how to approach the problem

We are given the resistance of the resistor and the current passing through it and are asked to calculate the voltage across it. We can apply Ohm's Law to this problem using: \[R = \frac{V}{I}.\]

Solve the problem

Rearrange the equation above and substitute the known values for $R$ and $I$ to solve for $V$. \begin{align*} R &= \frac{V}{I} \\ R \times I&= \frac{V}{I} \times I\\ V &= I \times R \\ &= \text{10} \times \text{4} \\ &= \text{40}\text{ V} \end{align*}

Write the final answer

The voltage across the resistor is $\text{40}$ $\text{V}$.

Calculate the resistance of a resistor that has a potential difference of $\text{8}$ $\text{V}$ across it when a current of $\text{2}$ $\text{A}$ flows through it. Draw the circuit diagram before doing the calculation.

The resistance of the unknown resistor is:

What current will flow through a resistor of $\text{6}$ $\text{Ω}$ when there is a potential difference of $\text{18}$ $\text{V}$ across its ends? Draw the circuit diagram before doing the calculation.

What is the voltage across a $\text{10}$ $\text{Ω}$ resistor when a current of $\text{1,5}$ $\text{A}$ flows though it? Draw the circuit diagram before doing the calculation.

Recap of resistors in series and parallel (ESBQ9)

In Grade 10, you learnt about resistors and were introduced to circuits where resistors were connected in series and in parallel. In a series circuit there is one path along which current flows. In a parallel circuit there are multiple paths along which current flows.

When there is more than one resistor in a circuit, we are usually able to calculate the total combined resistance of all the resistors. This is known as the equivalent resistance .

Equivalent series resistance

In a circuit where the resistors are connected in series, the equivalent resistance is just the sum of the resistances of all the resistors.

For n resistors in series the equivalent resistance is:

Let us apply this to the following circuit.

The resistors are in series, therefore:

Simulation: 242R

Video: 242S

Equivalent parallel resistance

In a circuit where the resistors are connected in parallel, the equivalent resistance is given by the following definition.

For $n$ resistors in parallel, the equivalent resistance is:

Let us apply this formula to the following circuit.

What is the total (equivalent) resistance in the circuit?

Video: 242T

Video: 242V

Series and parallel resistance

Two $\text{10}$ $\text{kΩ}$ resistors are connected in series. Calculate the equivalent resistance.

Since the resistors are in series we can use:

The equivalent resistance is:

Two resistors are connected in series. The equivalent resistance is $\text{100}$ $\text{Ω}$. If one resistor is $\text{10}$ $\text{Ω}$, calculate the value of the second resistor.

Two $\text{10}$ $\text{kΩ}$ resistors are connected in parallel. Calculate the equivalent resistance.

Since the resistors are in parallel we can use:

Two resistors are connected in parallel. The equivalent resistance is $\text{3,75}$ $\text{Ω}$. If one resistor has a resistance of $\text{10}$ $\text{Ω}$, what is the resistance of the second resistor?

Calculate the equivalent resistance in each of the following circuits:

a) The resistors are in parallel and so we use:

b) The resistors are in parallel and so we use:

c) The resistors are in series and so we use:

d) The resistors are in series and so we use:

Use of Ohm's Law in series and parallel circuits (ESBQB)

Using the definitions for equivalent resistance for resistors in series or in parallel, we can analyse some circuits with these setups.

Series circuits

Consider a circuit consisting of three resistors and a single cell connected in series.

The first principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to flow in a series circuit. From the way that the battery is connected, we can tell in which direction the current will flow. We know that current flows from positive to negative by convention. Conventional current in this circuit will flow in a clockwise direction, from point A to B to C to D and back to A.

We know that in a series circuit the current has to be the same in all components. So we can write:

We also know that total voltage of the circuit has to be equal to the sum of the voltages over all three resistors. So we can write:

Using this information and what we know about calculating the equivalent resistance of resistors in series, we can approach some circuit problems.

Worked example 2: Ohm's Law, series circuit

Calculate the current (I) in this circuit if the resistors are both ohmic in nature.

Determine what is required

We are required to calculate the current flowing in the circuit.

Since the resistors are ohmic in nature, we can use Ohm's Law. There are however two resistors in the circuit and we need to find the total resistance.

Find total resistance in circuit

Since the resistors are connected in series, the total (equivalent) resistance R is:

Apply Ohm's Law

\begin{align*} R & = \frac{V}{I} \\ R \times \frac{I}{R} & = \frac{V}{I} \times \frac{I}{R} \\ I & = \frac{V}{R} \\ & = \frac{12}{6} \\ & = \text{2}\text{ A} \end{align*}

A current of $\text{2}$ $\text{A}$ is flowing in the circuit.

Worked example 3: Ohm's Law, series circuit

Two ohmic resistors ($R_{1}$ and $R_{2}$) are connected in series with a cell. Find the resistance of $R_{2}$, given that the current flowing through $R_{1}$ and $R_{2}$ is $\text{0,25}$ $\text{A}$ and that the voltage across the cell is $\text{1,5}$ $\text{V}$. $R_{1}$ =$\text{1}$ $\text{Ω}$.

Draw the circuit and fill in all known values.

Determine how to approach the problem.

We can use Ohm's Law to find the total resistance R in the circuit, and then calculate the unknown resistance using:

because it is in a series circuit.

Find the total resistance

Find the unknown resistance.

We know that:

Worked example 4: Ohm's Law, series circuit

For the following circuit, calculate:

the voltage drops $V_1$, $V_2$ and $V_3$ across the resistors $R_1$, $R_2$, and $R_3$

the resistance of $R_3$.

We are given the voltage across the cell and the current in the circuit, as well as the resistances of two of the three resistors. We can use Ohm's Law to calculate the voltage drop across the known resistors. Since the resistors are in a series circuit the voltage is $V = V_1 + V_2 + V_3$ and we can calculate $V_3$. Now we can use this information to find the voltage across the unknown resistor $R_3$.

Calculate voltage drop across $R_1$

Using Ohm's Law: \begin{align*} R_1 &= \frac{V_1}{I} \\ I \cdot R_1 &= I \cdot \frac{V_1}{I} \\ V_1 &= {I}\cdot{R_1}\\ &= 2 \cdot 1 \\ V_1 &= \text{2}\text{ V} \end{align*}

Calculate voltage drop across $R_2$

Again using Ohm's Law: \begin{align*} R_2 &= \frac{V_2}{I} \\ I \cdot R_2 &= I \cdot \frac{V_2}{I} \\ V_2 &= {I}\cdot{R_2}\\ &= 2 \cdot 3 \\ V_2 &= \text{6}\text{ V} \end{align*}

Calculate voltage drop across $R_3$

Since the voltage drop across all the resistors combined must be the same as the voltage drop across the cell in a series circuit, we can find $V_3$ using: \begin{align*} V &= V_1 + V_2 + V_3\\ V_3 &= V - V_1 - V_2 \\ &= 18 - 2 - 6 \\ V_3&= \text{10}\text{ V} \end{align*}

Find the resistance of $R_3$

We know the voltage across $R_3$ and the current through it, so we can use Ohm's Law to calculate the value for the resistance: \begin{align*} R_3 &= \frac{V_3}{I}\\ &= \frac{10}{2} \\ R_3&= \text{5 } \Omega \end{align*}

$V_1 = \text{2}\text{ V}$

$V_2 = \text{6}\text{ V}$

$V_3 = \text{10}\text{ V}$

$R_1 = \text{5 } \Omega$

Parallel circuits

Consider a circuit consisting of a single cell and three resistors that are connected in parallel.

The first principle to understand about parallel circuits is that the voltage is equal across all components in the circuit. This is because there are only two sets of electrically common points in a parallel circuit, and voltage measured between sets of common points must always be the same at any given time. So, for the circuit shown, the following is true:

The second principle for a parallel circuit is that all the currents through each resistor must add up to the total current in the circuit:

Using these principles and our knowledge of how to calculate the equivalent resistance of parallel resistors, we can now approach some circuit problems involving parallel resistors.

Worked example 5: Ohm's Law, parallel circuit

Find the equivalent resistance in circuit

Since the resistors are connected in parallel, the total (equivalent) resistance R is:

The current flowing in the circuit is $\text{9}$ $\text{A}$.

Worked example 6: Ohm's Law, parallel circuit

Two ohmic resistors ($R_1$ and $R_2$) are connected in parallel with a cell. Find the resistance of $R_2$, given that the current flowing through the cell is $\text{4,8}$ $\text{A}$ and that the voltage across the cell is $\text{9}$ $\text{V}$.

We need to calculate the resistance $R_2$.

Since the resistors are ohmic and we are given the voltage across the cell and the current through the cell, we can use Ohm's Law to find the equivalent resistance in the circuit. \begin{align*} R & = \frac{V}{I} \\ & = \frac{9}{\text{4,8}} \\ & = \text{1,875} \ \Omega \end{align*}

Calculate the value for $R_2$

Since we know the equivalent resistance and the resistance of $R_1$, we can use the formula for resistors in parallel to find the resistance of $R_2$. \begin{align*} \frac{1}{R} & = \frac{1}{R_1} + \frac{1}{R_2} \end{align*} Rearranging to solve for $R_2$: \begin{align*} \frac{1}{R_2} & = \frac{1}{R} - \frac{1}{R_1} \\ & = \frac{1}{\text{1,875}} - \frac{1}{3}\\ & = \text{0,2} \\ R_2 & = \frac{1}{\text{0,2}} \\ & = \text{5} \ \Omega \end{align*}

The resistance $R_2$ is $\text{5}$ $\Omega$

Worked example 7: Ohm's Law, parallel circuit

An 18 volt cell is connected to two parallel resistors of $\text{4}$ $\Omega$ and $\text{12}$ $\Omega$ respectively. Calculate the current through the cell and through each of the resistors.

First draw the circuit before doing any calculations

We need to determine the current through the cell and each of the parallel resistors. We have been given the potential difference across the cell and the resistances of the resistors, so we can use Ohm's Law to calculate the current.

Calculate the current through the cell

To calculate the current through the cell we first need to determine the equivalent resistance of the rest of the circuit. The resistors are in parallel and therefore: \begin{align*} \frac{1}{R} &= \frac{1}{R_1} + \frac{1}{R_2} \\ &= \frac{1}{4} + \frac{1}{12} \\ &= \frac{3+1}{12} \\ &= \frac{4}{12} \\ R &= \frac{12}{4} = \text{3} \ \Omega \end{align*} Now using Ohm's Law to find the current through the cell: \begin{align*} R &= \frac{V}{I} \\ I &= \frac{V}{R} \\ &= \frac{18}{3} \\ I &= \text{6}\text{ A} \end{align*}

Now determine the current through one of the parallel resistors

We know that for a purely parallel circuit, the voltage across the cell is the same as the voltage across each of the parallel resistors. For this circuit: \begin{align*} V &= V_1 = V_2 = \text{18}\text{ V} \end{align*} Let's start with calculating the current through $R_1$ using Ohm's Law: \begin{align*} R_1 &= \frac{V_1}{I_1} \\ I_1 &= \frac{V_1}{R_1} \\ &= \frac{18}{4} \\ I_1 &= \text{4,5}\text{ A} \end{align*}

Calculate the current through the other parallel resistor

We can use Ohm's Law again to find the current in $R_2$: \begin{align*} R_2 &= \frac{V_2}{I_2} \\ I_2 &= \frac{V_2}{R_2} \\ &= \frac{18}{12} \\ I_2 &= \text{1,5}\text{ A} \end{align*} An alternative method of calculating $I_2$ would have been to use the fact that the currents through each of the parallel resistors must add up to the total current through the cell: \begin{align*} I &= I_1 + I_2 \\ I_2 &= I - I_1 \\ &= 6 - 4.5 \\ I_2 &= \text{1,5}\text{ A} \end{align*}

The current through the cell is $\text{6}$ $\text{A}$.

The current through the $\text{4}$ $\Omega$ resistor is $\text{4,5}$ $\text{A}$.

The current through the $\text{12}$ $\Omega$ resistor is $\text{1,5}$ $\text{A}$.

Ohm's Law in series and parallel circuits

Calculate the value of the unknown resistor in the circuit:

We first use Ohm's law to calculate the total series resistance:

Now we can find the unknown resistance:

Calculate the value of the current in the following circuit:

We first find the total resistance:

Now we can calculate the current:

Three resistors with resistance $\text{1}$ $\text{Ω}$, $\text{5}$ $\text{Ω}$ and $\text{10}$ $\text{Ω}$ respectively, are connected in series with a $\text{12}$ $\text{V}$ battery. Calculate the value of the current in the circuit.

We draw the circuit diagram:

We now find the total resistance:

Calculate the current through the cell if the resistors are both ohmic in nature.

Calculate the value of the unknown resistor $R_{4}$ in the circuit:

Now we can calculate the unknown resistance:

Three resistors with resistance $\text{1}$ $\text{Ω}$, $\text{5}$ $\text{Ω}$ and $\text{10}$ $\text{Ω}$ respectively, are connected in parallel with a $\text{20}$ $\text{V}$ battery. All the resistors are ohmic in nature. Calculate:

the value of the current through the battery

We draw a circuit diagram:

To calculate the value of the current through the battery we first need to calculate the equivalent resistance:

Now we can calculate the current through the battery:

the value of the current in each of the three resistors.

For a parallel circuit the voltage across cell is the same as the voltage across each of the resistors. For this circuit:

Now we can calculate the current through each resistor. We will start with $R_{1}$:

Next we calculate the current through $R_{2}$:

And finally we calculate the current through $R_{3}$:

You can check that these add up to the total current.

Series and parallel networks of resistors (ESBQC)

Now that you know how to handle simple series and parallel circuits, you are ready to tackle circuits which combine these two setups such as the following circuit:

It is relatively easy to work out these kind of circuits because you use everything you have already learnt about series and parallel circuits. The only difference is that you do it in stages. In Figure 11.1 , the circuit consists of 2 parallel portions that are then in series with a cell. To work out the equivalent resistance for the circuit, you start by calculating the total resistance of each of the parallel portions and then add up these resistances in series. If all the resistors in Figure 11.1 had resistances of $\text{10}$ $\text{Ω}$, we can calculate the equivalent resistance of the entire circuit.

We start by calculating the total resistance of Parallel Circuit 1 .

The value of $R_{p1}$ is: \begin{align*} \frac{1}{R_{p1}} &= \frac{1}{R_1} + \frac{1}{R_2} \\ R_{p1}&= \left(\frac{1}{10} + \frac{1}{10} \right)^{-1} \\ &= \left(\frac{1+1}{10} \right)^{-1} \\ &= \left(\frac{2}{10} \right)^{-1} \\ &= \text{5} \, \Omega \end{align*}

We can similarly calculate the total resistance of Parallel Circuit 2 : \begin{align*} \frac{1}{R_{p2}} &= \frac{1}{R_3} + \frac{1}{R_4} \\ R_{p2}&= \left(\frac{1}{10} + \frac{1}{10} \right)^{-1} \\ &= \left(\frac{1+1}{10} \right)^{-1} \\ &= \left(\frac{2}{10} \right)^{-1} \\ &= \text{5} \, \Omega \end{align*}

You can now treat the circuit like a simple series circuit as follows:

Therefore the equivalent resistance is: \begin{align*} R &= R_{p1} + R_{p2} \\ &= 5 + 5 \\ &= 10 \, \Omega \end{align*}

The equivalent resistance of the circuit in Figure 11.1 is $\text{10}$ $\text{Ω}$.

Series and parallel networks

Determine the equivalent resistance of the following circuits:

We start by determining the equivalent resistance of the parallel combination:

Now we have a circuit with two resistors in series so we can calculate the equivalent resistance:

Now we have a circuit with three resistors in series so we can calculate the equivalent resistance:

Examine the circuit below:

If the potential difference across the cell is $\text{12}$ $\text{V}$, calculate:

the current $I$ through the cell.

To find the current $I$ we first need to find the equivalent resistance. We start by calculating the equivalent resistance of the parallel combination:

So the current through the cell is:

the current through the $\text{5}$ $\text{Ω}$ resistor.

The current through the parallel combination of resistors is $\text{4,52}$ $\text{A}$. (The current is the same through series combinations of resistors and we can consider the entire parallel set of resistors as one series resistor.)

Using this we can find the voltage through the parallel combination of resistors (remember to use the equivalent parallel resistance and not the equivalent resistance of the circuit):

Since the voltage across each resistor in the parallel combination is the same, this is also the voltage across the $\text{5}$ $\text{Ω}$ resistor.

So now we can calculate the current through the resistor:

If current flowing through the cell is $\text{2}$ $\text{A}$, and all the resistors are ohmic, calculate the voltage across the cell and each of the resistors, $R_1$, $R_2$, and $R_3$ respectively.

To find the voltage we first need to find the equivalent resistance. We start by calculating the equivalent resistance of the parallel combination:

So the voltage across the cell is:

The current through the parallel combination of resistors is $\text{2}$ $\text{A}$. (The current is the same through series combinations of resistors and we can consider the entire parallel set of resistors as one series resistor.)

Using this we can find the voltage through the each of the resistors. We start by finding the voltage across $R_{1}$:

Now we find the voltage across the parallel combination:

Since the voltage across each resistor in the parallel combination is the same, this is also the voltage across resistors $R_{2}$ and $R_{3}$.

the current through the cell

To find the current we first need to find the equivalent resistance. We start by calculating the equivalent resistance of the parallel combination:

the voltage drop across $R_4$

The current through all the resistors is $\text{2,5}$ $\text{A}$. (The current is the same through series combinations of resistors and we can consider the entire parallel set of resistors as one series resistor.)

Using this we can find the voltage through $R_{4}$:

the current through $R_2$

Using this we can find the current through $R_{2}$.

We first need to find the voltage across the parallel combination:

Now we can find the current through $R_{2}$ using the fact that the voltage is the same across each resistor in the parallel combination:

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Ohm's law experiment

I was going through my physics laboratory manual. In the Ohm's Law Experiment, the book states a few precautions without any reasoning.

1.The wire whose resistance has to be determined should ideally be made up of alloys such as Manganin and not a metal . 2. A low resistance rheostat should be used.

Any reason why this is so?

Also the book says that the resistance wire must be wound on its self before it is wound on bobbin or reel to avoid induction effect.I did not understand this.

PS:I am connecting my sample resistance wire(with its length constant) and taking different values of V and I by altering the rheostat resistance.

electromagnetism
electricity
electric-circuits
electric-current

$\begingroup$ Please describe the experiment. It sounds like you are connecting a rheostat in series with a voltage source, connecting your sample across that, and measuring the current through the sample as a function of the rheostat resistance. Are you using different lengths of sample? Are you making a measurement very quickly? It would appear so as the induction effect will die away with time. $\endgroup$ – Ross Millikan Commented Feb 7, 2016 at 15:37
$\begingroup$ resistance of manganin and constantan do not vary much with temperature. Also for higher resistance. $\endgroup$ – Anubhav Goel Commented Feb 7, 2016 at 15:49

2 Answers 2

The reason to use the alloy is because it has a much higher resistivity than copper and so the alloy wire will have a higher resistance which with standard laboratory apparatus can be measured more accurately. I also seem to remember that the temperature of coefficient is lower for some of these alloys ie for a given increase in temperature the resistance of the alloy will change less.

The rheostat having a low resistance possibly means that it is used as a potential divider. If that is the case then you would have more control (use a greater range of the slider) on setting the voltage.

The only reason to worry about self inductance is if you are using alternating voltages and currents.

$\begingroup$ but the rheostat wire is made up of constantan which has a pretty high resistivity.So how can we possibly have a low resistance rheostat by using such a wire? $\endgroup$ – Karan Singh Commented Feb 7, 2016 at 15:58
$\begingroup$ If you look at the windings you would probably find that they are of fairly large diameter for a low resistance rheostat. $\endgroup$ – Farcher Commented Feb 7, 2016 at 16:00

The choice of a non-metallic sample is to get the resistance high enough to measure. The resistance of the rheostat, the voltage supply, and the hookup wires can make the measurement inaccurate if you do not account for them. If they are much smaller than the resistance of the sample they will not matter much. You also will not try to draw too much current for your voltage supply to source.

If you make a coil of wire, it will have a certain amount of inductance. This will resist changes in the current by making a back emf of $L\frac {di}{dt}$. If you allow the transient to die away before you measure the current this will not matter. If you measure quickly after establishing the circuit, you want to keep $L$ small to minimize the effect.

$\begingroup$ so does wounding the resistance wire on itself help me to keep L small?How? $\endgroup$ – Karan Singh Commented Feb 7, 2016 at 15:54
$\begingroup$ Depending on how you wind it, you can cancel out the magnetic field it makes, reducing $L$. You did not provide any details of the winding approach, so it is hard to tell. $\endgroup$ – Ross Millikan Commented Feb 7, 2016 at 16:00
1 $\begingroup$ The standard way is to fold the wire in half like a squashed U. Then the currents in adjacent bits of the wire flow in opposite directions and so the magnetic field due to wires cancel out. $\endgroup$ – Farcher Commented Feb 7, 2016 at 16:04

Your Answer

Required, but never shown

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged electromagnetism electricity electric-circuits electric-current or ask your own question .

Featured on Meta
User activation: Learnings and opportunities
Preventing unauthorized automated access to the network

Hot Network Questions

If I distribute GPLv2-licensed software as a bundle that uses a virtual machine to run it, do I need to open-source the virtual machine too?
Could you suffocate someone to death with a big enough standing wave?
Should punctuation (comma, period, etc.) be placed before or after the inches symbol when listing heights?
cURL in bash script reads $HOME as /root/
The meaning(s) of 'She couldn't buy a car yesterday.'
Does a ball fit in a pipe if they are exactly the same diameter?
How do I link a heading containing spaces in Markdown?
If a professor wants to hire a student themselves, how can they write a letter of recommendation for other universities?
What evidence exists for the historical name of Kuwohi Mountain (formerly Clingmans Dome)?
The answer is not single
Why is China not mentioned in the Fallout TV series despite its significant role in the games' lore?
Classification of countable subsets of the real line
2000s creepy independant film with a voice-over of a radio host giving bad self-help advice
Asymptotics of A000613
Do "always prepared" Spells cost spell slots?
How can moving observer explain non-simultaneity?
Day Convolution of Sheaves
Complexity of computing minimum unsatisfiable core
Do all languages distinguish between persons and non-persons?
Tic-tac-toe encode them all
Did Gauss really call Archimedes an idiot?
I need a temporary hoist and track system to lift a dead chest freezer up through a tight stairwell
Musicians wearing Headphones
How does a Trinitarian explain the role each "person" of the Trinity plays, in the creation of Adam or an animal?

9.4 Ohm's Law

Learning objectives.

By the end of this section, you will be able to:

Describe Ohm’s law
Recognize when Ohm’s law applies and when it does not

We have been discussing three electrical properties so far in this chapter: current, voltage, and resistance. It turns out that many materials exhibit a simple relationship among the values for these properties, known as Ohm’s law. Many other materials do not show this relationship, so despite being called Ohm’s law, it is not considered a law of nature, like Newton’s laws or the laws of thermodynamics. But it is very useful for calculations involving materials that do obey Ohm’s law.

Description of Ohm’s Law

The current that flows through most substances is directly proportional to the voltage V applied to it. The German physicist Georg Simon Ohm (1787–1854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied :

This important relationship is the basis for Ohm’s law . It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law, which is to say that it is an experimentally observed phenomenon, like friction. Such a linear relationship doesn’t always occur. Any material, component, or device that obeys Ohm’s law, where the current through the device is proportional to the voltage applied, is known as an ohmic material or ohmic component. Any material or component that does not obey Ohm’s law is known as a nonohmic material or nonohmic component.

Ohm’s Experiment

In a paper published in 1827, Georg Ohm described an experiment in which he measured voltage across and current through various simple electrical circuits containing various lengths of wire. A similar experiment is shown in Figure 9.19 . This experiment is used to observe the current through a resistor that results from an applied voltage. In this simple circuit, a resistor is connected in series with a battery. The voltage is measured with a voltmeter, which must be placed across the resistor (in parallel with the resistor). The current is measured with an ammeter, which must be in line with the resistor (in series with the resistor).

In this updated version of Ohm’s original experiment, several measurements of the current were made for several different voltages. When the battery was hooked up as in Figure 9.19 (a), the current flowed in the clockwise direction and the readings of the voltmeter and ammeter were positive. Does the behavior of the current change if the current flowed in the opposite direction? To get the current to flow in the opposite direction, the leads of the battery can be switched. When the leads of the battery were switched, the readings of the voltmeter and ammeter readings were negative because the current flowed in the opposite direction, in this case, counterclockwise. Results of a similar experiment are shown in Figure 9.20 .

In this experiment, the voltage applied across the resistor varies from −10.00 to +10.00 V, by increments of 1.00 V. The current through the resistor and the voltage across the resistor are measured. A plot is made of the voltage versus the current, and the result is approximately linear. The slope of the line is the resistance, or the voltage divided by the current. This result is known as Ohm’s law :

where V is the voltage measured in volts across the object in question, I is the current measured through the object in amps, and R is the resistance in units of ohms. As stated previously, any device that shows a linear relationship between the voltage and the current is known as an ohmic device. A resistor is therefore an ohmic device.

Example 9.8

Measuring resistance.

(b) First, the resistance is temperature dependent so the new resistance after the resistor has been heated can be found using R = R 0 ( 1 + α Δ T ) R = R 0 ( 1 + α Δ T ) . The current can be found using Ohm’s law in the form I = V / R I = V / R .

Using Ohm’s law and solving for the resistance yields the resistance at room temperature: R = V I = 9.00 V 3.00 × 10 −3 A = 3.00 × 10 3 Ω = 3.00 k Ω . R = V I = 9.00 V 3.00 × 10 −3 A = 3.00 × 10 3 Ω = 3.00 k Ω .
The resistance at 60 ° C 60 ° C can be found using R = R 0 ( 1 + α Δ T ) R = R 0 ( 1 + α Δ T ) where the temperature coefficient for carbon is α = −0.0005 α = −0.0005 . R = R 0 ( 1 + α Δ T ) = 3.00 × 10 3 ( 1 − 0.0005 ( 60 ° C − 20 ° C ) ) = 2.94 k Ω R = R 0 ( 1 + α Δ T ) = 3.00 × 10 3 ( 1 − 0.0005 ( 60 ° C − 20 ° C ) ) = 2.94 k Ω . The current through the heated resistor is I = V R = 9.00 V 2.94 × 10 3 Ω = 3.06 × 10 −3 A = 3.06 mA . I = V R = 9.00 V 2.94 × 10 3 Ω = 3.06 × 10 −3 A = 3.06 mA .

Significance

Check your understanding 9.8.

The voltage supplied to your house varies as V ( t ) = V max sin ( 2 π f t ) V ( t ) = V max sin ( 2 π f t ) . If a resistor is connected across this voltage, will Ohm’s law V = I R V = I R still be valid?

Interactive

See how the equation form of Ohm’s law relates to a simple circuit by engaging the simulation below. Adjust the voltage and resistance, and see the current change according to Ohm’s law. The sizes of the symbols in the equation change to match the circuit diagram.

Nonohmic devices do not exhibit a linear relationship between the voltage and the current. One such device is the semiconducting circuit element known as a diode. A diode is a circuit device that allows current flow in only one direction. A diagram of a simple circuit consisting of a battery, a diode, and a resistor is shown in Figure 9.21 . Although we do not cover the theory of the diode in this section, the diode can be tested to see if it is an ohmic or a nonohmic device.

A plot of current versus voltage is shown in Figure 9.22 . Note that the behavior of the diode is shown as current versus voltage, whereas the resistor operation was shown as voltage versus current. A diode consists of an anode and a cathode. When the anode is at a negative potential and the cathode is at a positive potential, as shown in part (a), the diode is said to have reverse bias. With reverse bias, the diode has an extremely large resistance and there is very little current flow—essentially zero current—through the diode and the resistor. As the voltage applied to the circuit increases, the current remains essentially zero, until the voltage reaches the breakdown voltage and the diode conducts current, as shown in Figure 9.22 . When the battery and the potential across the diode are reversed, making the anode positive and the cathode negative, the diode conducts and current flows through the diode if the voltage is greater than 0.7 V. The resistance of the diode is close to zero. (This is the reason for the resistor in the circuit; if it were not there, the current would become very large.) You can see from the graph in Figure 9.22 that the voltage and the current do not have a linear relationship. Thus, the diode is an example of a nonohmic device.

Ohm’s law is commonly stated as V = I R V = I R , but originally it was stated as a microscopic view, in terms of the current density, the conductivity, and the electrical field. This microscopic view suggests the proportionality V ∝ I V ∝ I comes from the drift velocity of the free electrons in the metal that results from an applied electrical field. As stated earlier, the current density is proportional to the applied electrical field. The reformulation of Ohm’s law is credited to Gustav Kirchhoff, whose name we will see again in the next chapter.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction

Authors: Samuel J. Ling, William Moebs, Jeff Sanny
Publisher/website: OpenStax
Book title: University Physics Volume 2
Publication date: Oct 6, 2016
Location: Houston, Texas
Book URL: https://openstax.org/books/university-physics-volume-2/pages/1-introduction
Section URL: https://openstax.org/books/university-physics-volume-2/pages/9-4-ohms-law

© Jul 23, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Remember Me

Shop Experiment Ohm’s Law Experiments

Ohm’s law.

Experiment #22 from Physics with Vernier

Video Overview

Introduction

The fundamental relationship among the three important electrical quantities current , voltage , and resistance was discovered by Georg Simon Ohm. The relationship and the unit of electrical resistance were both named for him to commemorate this contribution to physics. One statement of Ohm’s law is that the current through a resistor is proportional to the potential difference, in volts, across the resistor. In this experiment, you will see if Ohm’s law is applicable to several different circuits using a Current Probe and a Differential Voltage Probe.

Current and potential difference, in volts, can be difficult to understand, because they cannot be observed directly. To clarify these terms, some people make the comparison between electrical circuits and water flowing in pipes. Here is a chart of the three electrical units we will study in this experiment.

Electrical Quantity	Description	Unit	Water Analogy
Voltage or Potential Difference	A measure of the energy difference per unit charge between two points in a circuit.	volt (V)	Water pressure
Current	A measure of the flow of charge in a circuit.	ampere (A)	Amount of water flowing
Resistance	A measure of how difficult it is for current to flow in a circuit.	ohm (*)	A measure of how difficult it is for water to flow through a pipe.

Determine the mathematical relationship between current, potential difference, and resistance in a simple circuit.
Compare the potential vs. current behavior of a resistor to that of a light bulb.

Sensors and Equipment

This experiment features the following sensors and equipment. Additional equipment may be required.

Correlations

Teaching to an educational standard? This experiment supports the standards below.

Ready to Experiment?

Ask an expert.

Get answers to your questions about how to teach this experiment with our support team.

Call toll-free: 888-837-6437
Chat with Us
Email [email protected]

Purchase the Lab Book

This experiment is #22 of Physics with Vernier . The experiment in the book includes student instructions as well as instructor information for set up, helpful hints, and sample graphs and data.

New Products
Top Sellers
Gift Certificates
Raspberry Pi
Single Board Comp.
Microcontrollers
Machine Learning
Prototyping Boards
all development
Environment
all sensors
3D Printing
Instruments
Arts/Crafts Supplies
LED & Illumination
Buttons & Switches
LCDs & OLEDs
Cables & Wire
all components
Sewable Electronics
E-Textile Power
E-Textile Kits
all e-textiles
Motors & Drivers
Robotics Kits
all robotics
GPS & GNSS
Wireless Kits
all wireless/IoT
Audio Boards
Audio Cables
Audio Chips

Voltage, Current, Resistance, and Ohm's Law

Electricity Basics

When beginning to explore the world of electricity and electronics, it is vital to start by understanding the basics of voltage, current, and resistance. These are the three basic building blocks required to manipulate and utilize electricity. At first, these concepts can be difficult to understand because we cannot "see" them. One cannot see with the naked eye the energy flowing through a wire or the voltage of a battery sitting on a table. Even the lightning in the sky, while visible, is not truly the energy exchange happening from the clouds to the earth, but a reaction in the air to the energy passing through it. In order to detect this energy transfer, we must use measurement tools such as multimeters, spectrum analyzers, and oscilloscopes to visualize what is happening with the charge in a system. Fear not, however, this tutorial will give you the basic understanding of voltage, current, and resistance and how the three relate to each other.

Covered in this Tutorial

How electrical charge relates to voltage, current, and resistance.
What voltage, current, and resistance are.
What Ohm's Law is and how to use it to understand electricity.
A simple experiment to demonstrate these concepts.

Electrical Charge

Electricity is the movement of electrons. Electrons create charge, which we can harness to do work. Your lightbulb, your stereo, your phone, etc., are all harnessing the movement of the electrons in order to do work. They all operate using the same basic power source: the movement of electrons.

The three basic principles for this tutorial can be explained using electrons, or more specifically, the charge they create:

Voltage is the difference in charge between two points.
Current is the rate at which charge is flowing.
Resistance is a material's tendency to resist the flow of charge (current).

So, when we talk about these values, we're really describing the movement of charge, and thus, the behavior of electrons. A circuit is a closed loop that allows charge to move from one place to another. Components in the circuit allow us to control this charge and use it to do work.

Georg Ohm was a Bavarian scientist who studied electricity. Ohm starts by describing a unit of resistance that is defined by current and voltage. So, let's start with voltage and go from there.

We define voltage as the amount of potential energy between two points on a circuit. One point has more charge than another. This difference in charge between the two points is called voltage. It is measured in volts, which, technically, is the potential energy difference between two points that will impart one joule of energy per coulomb of charge that passes through it (don't panic if this makes no sense, all will be explained). The unit "volt" is named after the Italian physicist Alessandro Volta who invented what is considered the first chemical battery. Voltage is represented in equations and schematics by the letter "V".

When describing voltage, current, and resistance, a common analogy is a water tank. In this analogy, charge is represented by the water amount , voltage is represented by the water pressure , and current is represented by the water flow . So for this analogy, remember:

Water = Charge
Pressure = Voltage
Flow = Current

Consider a water tank at a certain height above the ground. At the bottom of this tank there is a hose.

Voltage is like the pressure created by the water.

The pressure at the end of the hose can represent voltage. The water in the tank represents charge. The more water in the tank, the higher the charge, the more pressure is measured at the end of the hose.

We can think of this tank as a battery, a place where we store a certain amount of energy and then release it. If we drain our tank a certain amount, the pressure created at the end of the hose goes down. We can think of this as decreasing voltage, like when a flashlight gets dimmer as the batteries run down. There is also a decrease in the amount of water that will flow through the hose. Less pressure means less water is flowing, which brings us to current.

We can think of the amount of water flowing through the hose from the tank as current. The higher the pressure, the higher the flow, and vice-versa. With water, we would measure the volume of the water flowing through the hose over a certain period of time. With electricity, we measure the amount of charge flowing through the circuit over a period of time. Current is measured in Amperes (usually just referred to as "Amps"). An ampere is defined as 6.241*10^18 electrons (1 Coulomb) per second passing through a point in a circuit. Amps are represented in equations by the letter "I".

Let's say now that we have two tanks, each with a hose coming from the bottom. Each tank has the exact same amount of water, but the hose on one tank is narrower than the hose on the other.

These two tanks create different pressures.

We measure the same amount of pressure at the end of either hose, but when the water begins to flow, the flow rate of the water in the tank with the narrower hose will be less than the flow rate of the water in the tank with the wider hose. In electrical terms, the current through the narrower hose is less than the current through the wider hose. If we want the flow to be the same through both hoses, we have to increase the amount of water (charge) in the tank with the narrower hose.

These two tanks create the same pressure.

This increases the pressure (voltage) at the end of the narrower hose, pushing more water through the tank. This is analogous to an increase in voltage that causes an increase in current.

Now we're starting to see the relationship between voltage and current. But there is a third factor to be considered here: the width of the hose. In this analogy, the width of the hose is the resistance. This means we need to add another term to our model:

Water = Charge (measured in Coulombs)
Pressure = Voltage (measured in Volts)
Flow = Current (measured in Amperes, or "Amps" for short)
Hose Width = Resistance

Consider again our two water tanks, one with a narrow pipe and one with a wide pipe.

The tank with the narrow pipe creates a higher resistance.

It stands to reason that we can't fit as much volume through a narrow pipe than a wider one at the same pressure. This is resistance. The narrow pipe "resists" the flow of water through it even though the water is at the same pressure as the tank with the wider pipe.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

This brings us back to Georg Ohm. Ohm defines the unit of resistance of "1 Ohm" as the resistance between two points in a conductor where the application of 1 volt will push 1 ampere, or 6.241×10^18 electrons. This value is usually represented in schematics with the greek letter "Ω", which is called omega, and pronounced "ohm".

Combining the elements of voltage, current, and resistance, Ohm developed the formula:

V = Voltage in volts
I = Current in amps
R = Resistance in ohms

This is called Ohm's law. Let's say, for example, that we have a circuit with the potential of 1 volt, a current of 1 amp, and resistance of 1 ohm. Using Ohm's Law we can say:

Let's say this represents our tank with a wide hose. The amount of water in the tank is defined as 1 volt and the "narrowness" (resistance to flow) of the hose is defined as 1 ohm. Using Ohms Law, this gives us a flow (current) of 1 amp.

Using this analogy, let's now look at the tank with the narrow hose. Because the hose is narrower, its resistance to flow is higher. Let's define this resistance as 2 ohms. The amount of water in the tank is the same as the other tank, so, using Ohm's Law, our equation for the tank with the narrow hose is

But what is the current? Because the resistance is greater, and the voltage is the same, this gives us a current value of 0.5 amps:

Tanks with their equivalent electrical meanings.

So, the current is lower in the tank with higher resistance. Now we can see that if we know two of the values for Ohm's law, we can solve for the third. Let's demonstrate this with an experiment.

An Ohm's Law Experiment

For this experiment, we want to use a 9 volt battery to power an LED. LEDs are fragile and can only have a certain amount of current flowing through them before they burn out. In the documentation for an LED, there will always be a "current rating". This is the maximum amount of current that can flow through the particular LED before it burns out.

Materials Required

In order to perform the experiments listed at the end of the tutorial, you will need:

A multimeter
A 9-Volt battery
A 560-Ohm resistor(or the next closest value)

NOTE: LEDs are what's known as a "non-ohmic" devices. This means that the equation for the current flowing through the LED itself is not as simple as V=IR. The LED introduces something called a "voltage drop" into the circuit, thus changing the amount of current running through it. However, in this experiment we are simply trying to protect the LED from over-current, so we will neglect the current characteristics of the LED and choose the resistor value using Ohm's Law in order to be sure that the current through the LED is safely under 20mA.

For this example, we have a 9 volt battery and a red LED with a current rating of 20 milliamps, or 0.020 amps. To be safe, we'd rather not drive the LED at its maximum current but rather its suggested current, which is listed on its datasheet as 18mA, or 0.018 amps. If we simply connect the LED directly to the battery, the values for Ohm's law look like this:

and since we have no resistance yet:

Dividing by zero gives us infinite current! Well, not infinite in practice, but as much current as the battery can deliver. Since we do NOT want that much current flowing through our LED, we're going to need a resistor. Our circuit should look like this:

We can use Ohm's Law in the exact same way to determine the reistor value that will give us the desired current value:

plugging in our values:

solving for resistance:

So, we need a resistor value of around 500 ohms to keep the current through the LED under the maximum current rating.

500 ohms is not a common value for off-the-shelf resistors, so this device uses a 560 ohm resistor in its place. Here's what our device looks like all put together.

Success! We've chosen a resistor value that is high enough to keep the current through the LED below its maximum rating, but low enough that the current is sufficient to keep the LED nice and bright.

This LED/current-limiting resistor example is a common occurrence in hobby electronics. You'll often need to use Ohm's Law to change the amount of current flowing through the circuit. Another example of this implementation is seen in the LilyPad LED boards.

With this setup, instead of having to choose the resistor for the LED, the resistor is already on-board with the LED so the current-limiting is accomplished without having to add a resistor by hand.

Current Limiting Before or After the LED?

To make things a little more complicated, you can place the current limiting resistor on either side of the LED, and it will work just the same!

Many folks learning electronics for the first time struggle with the idea that a current limiting resistor can live on either side of the LED and the circuit will still function as usual.

Imagine a river in a continuous loop, an infinite, circular, flowing river. If we were to place a dam in it, the entire river would stop flowing, not just one side. Now imagine we place a water wheel in the river which slows the flow of the river. It wouldn't matter where in the circle the water wheel is placed, it will still slow the flow on the entire river .

This is an oversimplification, as the current limiting resistor cannot be placed anywhere in the circuit ; it can be placed on either side of the LED to perform its function.

For a more scientific answer, we turn to Kirchoff's Voltage Law . It is because of this law that the current limiting resistor can go on either side of the LED and still have the same effect. For more info and some practice problems using KVL, visit this website .

Resources and Going Further

Now you should understand the concepts of voltage, current, resistance, and how the three are related. Congratulations! The majority of equations and laws for analyzing circuits can be derived directly from Ohm's Law. By knowing this simple law, you understand the concept that is the basis for the analysis of any electrical circuit!

Interested in learning more foundational topics?

See our Engineering Essentials page for a full list of cornerstone topics surrounding electrical engineering.

Take me there!

These concepts are just the tip of the iceberg. If you're looking to study further into more complex applications of Ohm's Law and the design of electrical circuits, be sure to check out the following tutorials.

Series vs. Parallel Circuits
Electric Power
Analog vs. Digital Circuits
How to Use a Multimeter
Electrical Engineering
Start a Project
Creative Commons tutorials are CC BY-SA 4.0
Your Account

Pardon Our Interruption

As you were browsing something about your browser made us think you were a bot. There are a few reasons this might happen:

You've disabled JavaScript in your web browser.
You're a power user moving through this website with super-human speed.
You've disabled cookies in your web browser.
A third-party browser plugin, such as Ghostery or NoScript, is preventing JavaScript from running. Additional information is available in this support article .

To regain access, please make sure that cookies and JavaScript are enabled before reloading the page.

IMAGES

Voltmeter Ammeter Rheostat Battery Key Resistance
Ohm S Law Circuit Diagram With Rheostat
Draw a labelled diagram of the experiment explaining Ohm’s Law
How To Adjust Voltage Using A Rheostat
State Ohm's law and draw a neat labelled circuit diagram containing a
Ohm's Law Experiment Class 12|Practical|Resistance Rheostat

VIDEO

Ohm's Law Experiment
Ohm's Law Experiment Connection
Rheostat #physics #physicslab
ohm's law experiment
Hindi
ohm's law experiment #physics #science #shorts

COMMENTS

Verification of Ohm's Law experiment with data and graph
Learn how to verify Ohm's law by measuring current, voltage and resistance of a wire using a power supply, ammeter, voltmeter and rheostat. See the circuit diagram, experimental data, graph and calculations for the lab-based experiment.
Ohm's law experiment
Learn how to perform the Ohm's law experiment with a step by step guide, circuit diagram, and color coding chart. Find out how to use voltmeter, ammeter, resistor, and variable dc power supply in series and parallel circuits.
Ohm's Law
Explore the equation form of Ohm's law in a simple circuit with this online tool. Adjust the voltage and resistance, and see the current change accordingly.
Ohm's Law
Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends. ... Ohm's Law can be easily verified by the following experiment: Apparatus Required: Resistor; Ammeter; Voltmeter; Battery; Plug Key; Rheostat; Circuit Diagram: Procedure: Initially, the key K is ...
Ohm's Law Lab Report [With Graph, Observations and ...
Learn how to verify Ohm's law using a 1kΩ resistor, a variable DC supply, and an ammeter. See the graph, observations, and calculations for the experiment, and find answers to basic questions about resistors and ammeters.
PDF Experiment 15: Ohm's Law
80 Experiment 15: Ohm's Law Advance Reading Text: Ohm's Law, voltage, resistance, current. Lab Manual: Appendix B, Appendix C -DMM Objective The objective of this lab is to determine the resistance of several resistors by applying Ohm's Law. Students will also be introduced to the resistor color code and refresh their graphing skills. Theory
Intro Lab
Learn how to use an ohmmeter, voltmeter, and ammeter to verify Ohm's law, which defines the relationship between resistance, voltage, and current. Build a simple circuit with a battery and a resistor and measure the values of each component.
5 Error Sources in Ohm's Law Experiment [How to avoid them]
The practical observations of Ohm's law experiment never match the theoretical readings. In fact, you can never match the theoretical calculations with. Ohm Law. Learn all about Ohm's law. ... Ohm performed repeated experiments on a resistor, applied different voltages, measured current and found relationship between these quantities. ...
How to Perform Ohm's Law Experiment for Class 10
The Experiment Equipment. To perform the Ohm's law experiment, you will need the following equipment: 1 battery eliminator (0-12 volts, 2 Amps) 1 voltmeter 1 ammeter; 1 rheostat; 1 resistance box (or an unknown resistor) Connecting wires; Ohm's Law Experiment Equipment
PDF Experiment 5: Simple Resistor Circuits
current (I) passing through a resistor, the potential di erence (V) across the resistor, and the resistance (R) of the resistor is described by Ohm's Law: V = IR (1) In Part 1 of the experiment, we will verify this law by measuring the potential di erence (i.e. voltage drop) across resistors in series and parallel.
Resistors Obey Ohm's Law (Experiment)
There are various ways to prove that resistors obey Ohm's law (V=IR). Ohm's law says that voltage across a component is proportional to the current going thr...
11.2 Ohm's Law
Learn how to use Ohm's Law to calculate current, voltage and resistance in electric circuits. The web page explains the experiment, the formula and the graph of Ohm's Law, and provides interactive exercises and textbook questions.
NCERT Class 10 Science Lab Manual
Question 38: The best graph plotted by a student for Ohm's experiment is: Question 39: The given graph, is plotted for V-I to verify Ohm's law. The resistance of the conductor used in the experiment is: (a) 1 Ω (b) 1.5 Ω (c) 3 Ω (d) 2 Ω. Question 40: A student wanted to make a battery of 6 V of cells with e.m.f 1.5 V each. The correct ...
PDF General Physics II Lab (PHYS-2021) Experiment ELEC-2: Ohm's Law
Learn how to verify Ohm's Law using a resistive network and a signal generator. Follow the steps to set up, calibrate, and analyze the data, and write a formal lab report with conclusions and uncertainties.
Ohm's law
Ohm's Law assumed a position of great importance in the nineteenth century when telegraph lines were designed and electrical engineering was developing. How Science Works extension: This experiment provides an excellent opportunity to focus on the range and number of results, as well as the analysis of them. Typically it yields an accurate set.
PDF Experiment 6: Ohm's Law, RC and RL Circuits
Experiment 6: Ohm's Law, RC and RL Circuits OBJECTIVES 1. To explore the measurement of voltage & current in circuits 2. To see Ohm's law in action for resistors ... Imagine you wish to measure the voltage drop across and current through a resistor in a circuit. To do so, you would use a voltmeter and an ammeter - similar devices that ...
Ohm's law experiment
In the Ohm's Law Experiment, the book states a few precautions without any reasoning. ... The rheostat having a low resistance possibly means that it is used as a potential divider. If that is the case then you would have more control (use a greater range of the slider) on setting the voltage. ...
9.4 Ohm's Law
Learn about Ohm's law, which relates the current, voltage, and resistance in a circuit. This web page is part of an open-source textbook on university physics, but it ...
Ohm's Law > Experiment 22 from Physics with Vernier
The fundamental relationship among the three important electrical quantities current, voltage, and resistance was discovered by Georg Simon Ohm. The relationship and the unit of electrical resistance were both named for him to commemorate this contribution to physics. One statement of Ohm's law is that the current through a resistor is proportional to the potential difference, in volts, across ...
Voltage, Current, Resistance, and Ohm's Law
Learn the basics of electricity and electronics with this tutorial on voltage, current, resistance, and Ohm's Law. Explore the concepts with analogies, examples, and a simple experiment.
‪Ohm's Law‬
‪Ohm's Law‬ - PhET Interactive Simulations
Ohms Law Worksheet Week5 V4 (docx)
Ohm's Law Worksheet Name: Jacob Mitchell Course: STEM100 Instructor: Dr. Jason Locklin Date: 06/09/2024 Before you begin the lab simulation, answer the first two questions. No 'cheating'! Try to answer based off your understanding of the reading. Include the Ohm's Law equation in your answers and be specific. As you change the value of the battery voltage, what does this do to magnitude of the ...

Number of cells	Voltage, V (\(\text{V}\))	Current, I (\(\text{A}\))
\(\text{1}\)
\(\text{2}\)
\(\text{3}\)
\(\text{4}\)


\(\text{3,0}\)	\(\text{0,4}\)
\(\text{6,0}\)	\(\text{0,8}\)
\(\text{9,0}\)	\(\text{1,2}\)
\(\text{12,0}\)	\(\text{1,6}\)