Current Electricity Ques 63
63. A cell can be balanced against $110 cm$ and 100 $cm$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 \Omega$. Its internal resistance is
[2008]
(a) $1.0 ohm$
(b) $0.5 ohm$
(c) $2.0 ohm$
(d) zero
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Answer:
Correct Answer: 63.(a)
Solution:
- (a) Here $E>\frac{ER}{R+r}$, hence the lengths $110 cm$ and $100 cm$ are interchanged.
Without being short-circuited through R, only the battery E is balanced.
$E=\frac{V}{L} \times l_1=\frac{V}{L} \times 110 \ldots . .(i)$
When $R$ is connected across E, $R i=\frac{V}{L} \times l_2$
or, $R(\frac{E}{R+r})=\frac{V}{L} \times 100$.
Dividing (i) by (ii), we get
$\frac{R+r}{R}=\frac{110}{100}$
or, $100 R+100 r=110 R$
or, $10 R=100 r$
$\therefore r=\frac{10 R}{100}=\frac{10 \times 10}{100}(\therefore R=10 \Omega)$
$\Rightarrow r=1 \Omega$.
BETA
