Question 112: Current Electricity

Question

In the potentiometer, when the cell in the secondary circuit is shunted with $4 \Omega$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with $12 \Omega$ resistance, the balance is shifted to a length of 180 cm . The internal resistance of cell is $\_\_\_\_$ $\Omega$

Accepted Answer

Let the current in the primary circuit is $i_p$. If the resistance of potentiometer wire of length $L$ is $R$, then the resistance of the balancing length $l$ will be $R_l=\left(\frac{R}{L}ight) l$
The voltage drop at point $A$ is. $V=i_p R_l=\left(\frac{{ }^i p R}{L}ight) l=k \cdot l$, where $k$ is constant.
the balancing length l is proportional to the terminal potential difference V :
$$ \mathrm{V}=\mathrm{k} \cdot \mathrm{l} \ldots 	ext { (i) } $$
When a cell of EMF E and internal resistance $r$ is shunted (connected in parallel) with an external resistance $R$,
The current in the secondary circuit is, $\mathrm{i}_{\mathrm{s}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}$
Then terminal voltage is, $\mathrm{V}=\mathrm{iR}=\frac{\mathrm{ER}}{\mathrm{R}+\mathrm{r}} \ldots$ (ii)
From (i) and (ii),
$\mathrm{k} \cdot \mathrm{l}=\frac{\mathrm{E} \cdot \mathrm{R}}{\mathrm{R}+\mathrm{r}}$
$$\Rightarrow $$ $$ l=\frac{1}{k}\left(\frac{R}{R+r}ight) $$
Case 1 when the external resistance is $\mathrm{R}_1=4 \Omega$ and the balancing length $\mathrm{l}_1=120 \mathrm{~cm}$
$$ 120=\frac{1}{\mathrm{k}}\left[\frac{4}{4+\mathrm{r}}ight] \ldots $$
Case 2 when the external resistance is $\mathrm{R}_2=12 \Omega$ and the balancing length $\mathrm{l}_2=180 \mathrm{~cm}$
$$ 180=\frac{1}{\mathrm{k}}\left[\frac{12}{12+\mathrm{r}}ight] . . $$
Dividing equation (iv) by equation (iii) :
$$ \frac{180}{120}=\frac{\frac{12}{12+\mathrm{r}}}{\frac{4}{4+\mathrm{r}}} $$
$$\Rightarrow $$ $\frac{3}{2}=\frac{3 \cdot(4+r)}{12+r}$
$$\Rightarrow $$ $12+r=2(4+r)$
$$\Rightarrow $$ $12+r=2(4+r)$
$$\Rightarrow $$ $$ \mathrm{r}=4 \Omega $$
Therefore, the internal resistance of the cell is $4 \Omega$. Hence, the correct option is (A).

Answer

4

Answer

12

Answer

6

Answer

3

Current Electricity

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