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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
1 $\Omega$
B
0.5 $\Omega$
C
1/3 $\Omega$
D
1/4 $\Omega$
Detailed Solution
For a cell of emf $\varepsilon$ and internal resistance r, $I = \frac{\varepsilon}{R + r}$
First case: $2 = \frac{\varepsilon}{2 + r}$, so $\varepsilon = 2(2 + r)$
Second case: $0.5 = \frac{\varepsilon}{9 + r}$, so $\varepsilon = 0.5(9 + r)$
Equating: $2(2 + r) = 0.5(9 + r)$
$4 + 2r = 4.5 + 0.5r$
$1.5r = 0.5$
$r = \frac{1}{3}\ \Omega$
First case: $2 = \frac{\varepsilon}{2 + r}$, so $\varepsilon = 2(2 + r)$
Second case: $0.5 = \frac{\varepsilon}{9 + r}$, so $\varepsilon = 0.5(9 + r)$
Equating: $2(2 + r) = 0.5(9 + r)$
$4 + 2r = 4.5 + 0.5r$
$1.5r = 0.5$
$r = \frac{1}{3}\ \Omega$