Current Electricity Ques 54
54. A current of 2 A flows through a $2 \Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5 A$ when connected across a $9 \Omega$ resistor. The internal resistance of the battery is
(a) $0.5 \Omega$
(b) $1 / 3 \Omega$
(c) $1 / 4 \Omega$
(d) $1 \Omega$
[2011]
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Answer:
Correct Answer: 54.(b)
Solution:
- (b) Let the internal resistance of the battery be $r$. Then the current flowing through the circuit is given by
$i=\frac{E}{R+r}$
In first case,
$2=\frac{E}{2+r}$
In second case,
$0.5=\frac{E}{9+r}$
From (1) & (2),
$4+2 r=4.5+0.5 r$
$\Rightarrow 1.5 r=0.5 \Rightarrow r=\frac{1}{3} \Omega$.
If $I_1$ be the current in a circuit with an external resistance $R_1$ and $I_2$ be the current in the circuit with external resistance $R$, then internal resistance of cell can be find using
$r=\frac{I_2 R_2-I_1 R_1}{I_1-I_2}$
BETA
