To determine the internal resistance of a cell by using a potenti… - Vidyadip

MCQ

To determine the internal resistance of a cell by using a potentiometer the null point is at $1 m$ when the cell is shunted by $3 \Omega$ resistance and at a length $1.5 m$. When cell is shunted by $6 \Omega$ resistance, the internal resistance of the cell is

A
$4 \Omega$
B
$8 \Omega$
C
$3 \Omega$
✓
$6 \Omega$

✓

Answer

Correct option: D.

$6 \Omega$

(d) : When internal resistance is measured by potentiometer,
$
\frac{V_1}{V_2}=\frac{l_1}{l_2}=\frac{\varepsilon R_1 /\left(R_1+r\right)}{\varepsilon R_1 /\left(R_2+r\right)}=\frac{R_1\left(R_2+r\right)}{R_2\left(R_1+r\right)}
$
Here, $l_1=1 m , l_2=1.5 m ; R_1=3 \Omega ; R_2=6 \Omega$
$
\begin{aligned}
& \Rightarrow \quad \frac{1}{1.5}=\frac{3(6+r)}{6(3+r)} \\
& \Rightarrow 18+6 r=27+4.5 r \\
& \Rightarrow \quad 1.5 r=9 \\
& \Rightarrow r=9 / 1.5, r=6 \Omega
\end{aligned}
$

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