Question

For the arrangement of the potentiometer shown in the figure, the balance point is obtained at a distance $75cm$ from A when the key $k$ is open. The second balance point is obtained at $60cm$ from $A$ when the key $k$ is closed. Find the internal resistance $(in\Omega )$ of the battery ${E_1}$ .

Answer 1

Hint : Potentiometer working principle is that the potential drop in the wire which is carrying uniform current is directly proportional to its length. It is a device which is used to measure the internal resistance. The formula used is as follows:
$r = \left( {\dfrac{{{l_1} - {l_2}}}{{{l_2}}}} \right) \times R$
Where ${l_1}$ and ${l_2}$ are the balancing point when the key is open and the key is closed respectively. $R$ is the shunt resistance connected parallel to the battery.

Complete Step By Step Answer:
Let us consider the balance point obtained when the key is open as ${l_1}$
${l_1} = 75cm = 0.75m$
Now, let us consider the balance point obtained when the key is closed as ${l_2}$
When the key is closed, it produces a shunt and the balance point changes to $60$ . So,
${l_2} = 60cm = 0.60m$
Let the shunt resistance be $R$
$R = 6\Omega$
Now, we know that the formula to calculate the internal resistance is:
$r = \left( {\dfrac{{{l_1} - {l_2}}}{{{l_2}}}} \right) \times R$
Where, $r$ is the internal resistance.
So, substituting the values we will get,
r = \left( {\dfrac{{0.75 - 0.60}}{{0.60}}} \right) \times 6 \\\ r = \left( {\dfrac{{0.15}}{{0.60}}} \right) \times 6 \\\ r = \dfrac{6}{4} \\\ r = 1.5\Omega \\\
Hence, the internal resistance of the battery ${E_1}$ is $1.5\Omega$ .

Note :
It’s important to carry out necessary precautions while doing the experiment, to ensure that the calculated value is equal to the experimental value. The connections should be tight. The current should be uniform and it should not flow for a long time or else it heats up the wire. The cross sectional area of the wire should also be uniform.