Question

Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer 1

The internal resistance of the cell is given as,

r=(l1−l2l2)R

Where, the internal resistance is r, the balance point of the cell in open circuit is l1, the new balance point in the presence of external resistance R is l2.

By substituting the given values in the above expression, we get

r=76.3−64.864.8×9.5=1.68 Ω

Thus, the internal resistance of the cell is 1.68 Ω.

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\orange{☆\:Solution\:☆}}}}}}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\hookrightarrow$ Internal resistance of the cell = r

$\hookrightarrow$ Balanced point of the cell in the open circuit, $\sf l_1\:=\:76.3\,cm$

$\hookrightarrow$ An external resistance (R) is connected to the circuit with $\sf R\:=\:9.5\Omega$

$\hookrightarrow$ New balanced point of the circuit, $\sf l_2\:=\:64.8\,cm$

$\hookrightarrow$ Current flowing through the circuit = I

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

The relation connection resistance and emf is

$\longrightarrow$ $\underline{\boxed{\sf{\red{r\:=\: \left(\dfrac{I_1-I_2}{I_2}\right)R}}}}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\leadsto$ $\sf r\:=\: \dfrac{76.3-64.8}{64.8}\times 9.5$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\leadsto$ $\sf r\:=\:1.68\Omega$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\longrightarrow$ Hence the internal resistance of the cell is $\underline{\sf{\orange{\:1.68\Omega}}}$