Verification of Ohm's Law using Voltmeter and Ammeter

Ohm's law states that the current $I$ through a conductor is proportional to the voltage $V$ across its ends. It is written as $V=IR$, where $R$ is the resistance of the conductor.

Experiment to Verify Ohm's Law

Ohm's law can be easily verified in the lab or at home. You need a voltmeter, an ammeter, power supply (dry cells), resistors, and connecting wires. A simple procedure to verify Ohm's law is given below:

Take four or five dry cells, a thin wire (AB), a voltmeter, an ammeter, a plug key and some thick connecting wires. Connect the circuit as shown in figure, using one cell. The plug key allows you to switch off the current when not required. The wire becomes quite hot when current passes through it for some time. This drains the cell as well. Therefore, insert the key into the plug to switch on the current only when taking measurements.

The ammeter measures the current $I$ through the circuit, and the voltmeter measures the potential difference $V$ between the ends A and B of the wire. Note these values. Now, connect two cells in series in the circuit. You will find that the reading of the voltmeter increases, indicating the fact that a larger potential difference has been applied across the wire AB. You will also find that the reading of the ammeter increases as well. Note down the new values of $V$ and $I$. Repeat the experiment by connecting in series three cells, four cells, and so on. In each case measure the potential difference and the current. If you calculate $V/I$ for each case, you will find that it is almost the same. So, $V/I=R$ is a constant, which is another way of stating Ohm's law. Here, $R$ is resistance of the wire AB. If you plot a graph of the current of the current $I$ against the potential difference $V$, it will be a straight line. This shows that the current is proportional to the potential difference.

Ohm's law verification using Rheostat — A circuit with Rheostat to verify Ohm's law

The experimental setup used in the laboratory makes use of a Rheostat to vary the potential difference $V$ across a standard resistor $R$.

Solved Problems on Verification of Ohm's Law

Problem on Least Count and Zero Error

The needles in an ammeter and voltmeter when not connected in a circuit are as shown in the figure. The least count and zero error of these two instruments are

Least count and zero error of ammeter and voltmeter

(2 mA, 0.1 V) and (-2 mA, 0.2 V)
(2 A, 0.1 V) and (-2 A, 0.2 V)
(1 mA, 0.1 V) and (-1 mA, 0.2 V)
(2 mA, 0.1 V) and (-2 mA, 0.1 V)

Solution: The least count of an instrument is the minimum value it can measure. It is the value of 1 division on the scale. For ammeter, 5 divisions are equal to 10 mA. Thus, the least count of the ammeter is 10/5=2 mA. For voltmeter, 10 divisions are equal to 1 V. Thus, the least count of the voltmeter is 1/10=0.1 V.

The zero error for the given ammeter is 1 division on the negative side which is equal to $-2$ mA. For voltmeter, zero error is 2 divisions on the positive side which is equal to 0.2 V.

Problem on Relative Error and Significant Figures

In an experiment to verify Ohm's law, a student measured the potential drop across the resistor as 4.20 V and the current through the resistor as 1.4 A. The most suitable way to report the resistance is

$3.0 \Omega$
$(3.0 \pm 0.2) \Omega$
$(3 \pm 0.2) \Omega$
$(3.0 \pm 0.1) \Omega$

Solution: From given data, the least count of the voltmeter is $\Delta V=0.01$ V and that of the ammeter is $\Delta I=0.1$ A. Ohm's law gives resistance as $R=V/I=4.20/1.4=3.0\;\mathrm{\Omega}$ (recall the rules of significant figures).

The relative error in resistance is \begin{align} \frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}=0.07,\nonumber \end{align} which gives $\Delta R=0.2\,\mathrm{\Omega}$. Thus, the resistance should be reported as $(3.0 \pm 0.2) \Omega$.

Problem from IIT JEE 2003

Which of the following set-up can be used to verify Ohm's law?

Solution: The verification of Ohm's law ($V=IR$) requires the measurements of current through and voltage across the variable resistance. Hence answer is B.

Problem from IIT JEE 2004

Draw the circuit for experimental verification of Ohm's law using a source of variable DC voltage, a main resistance of $100\; \Omega$, two galvanometers and two resistances of values $10^6\; \Omega$ and $10^{-3}\;\Omega$, respectively. Clearly show the positions of the voltmeter and the ammeter.

Solution: Ohm's law verification requires measurement of the voltage and current. The galvanometer can be converted to a voltmeter by connecting a very high resistance (${10}^{6}\;\Omega$) in series and to an ammeter by connecting very low resistance ${10}^{-3}\; \Omega$) in parallel, as shown.

Solved problem on Ohm's law verification from IIT JEE 2004 — A galvanometer is converted to a voltmeter by connecting very high resistance in series. It is converted to an ammeter by connecting very low resistance in parallel.

Problem from IIT JEE 2010

To verify Ohm's law, a student is provided with a test resister $R_T$, a high resistance $R_1$, and a small resistance $R_2$, two identical galvanometers $G_1$ and $G_2$, and a variable voltage source $V$. The correct circuit to carry out the experiment is,

Solution: To verify Ohm's law, we need to measure voltage across the test resistance $R_T$ and current passing through it. The voltage can be measured by connecting high resistance $R_1$ in series with galvanometer. This combination becomes a voltmeter and shall be connected in parallel to $R_T$. The current can be measured by connecting low resistance $R_2$ in parallel with galvanometer. This combination becomes an ammeter and shall be connected in series to measure current through $R_T$. Hence answer is C.

Questions on Verification of Ohm's Law

Question 1: The internal resistance of the voltmeter is very high (ideal voltmeter has infinite resistance), whereas that of an ammeter is very low (ideal ammeter has zero resistance). The primary reason for this is

Question 2: In an experiment to verify Ohm’s law, a student used a torch bulb as a resistor. When he plotted the voltage versus the current graph, he obtained a slightly curved line instead of the expected straight line. This may be due to