Question 192 of 380

In the given circuit, an ideal voltmeter connected across the $10\Omega$ resistance reads $2V$ . The internal resistance $r$ , of each cell is:

A$0\Omega$B$1.5\Omega$C$0.5\Omega$D$1\Omega$

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Answer: (C) $0.5\Omega$

Current through battery $=\dfrac{2}{10}+\dfrac{2}{15}=\dfrac{6+4}{30}=\dfrac{1}{3}$ A
Also, net resistance ${R}_{eq}=\dfrac{15\times 10}{15+10}+2+2r$
$=8+2r$
$\Rightarrow \dfrac{1}{3}=\dfrac{1.5+1.5}{8+2r}$
$8+2r=9$
$r=\dfrac{1}{2}=0.5 \Omega$

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