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(B) The circuit consists of two cells of $1.5\,V$ each in series,so the total $EMF$ is $E_{eq} = 1.5 + 1.5 = 3\,V$. The total internal resistance is $

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$0.5$

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(B) The circuit consists of two cells of $1.5\,V$ each in series,so the total $EMF$ is $E_{eq} = 1.5 + 1.5 = 3\,V$. The total internal resistance is $r_{eq} = r + r = 2r$.The external circuit has a $15\,\Omega$ and $10\,\Omega$ resistor in parallel,which are in series with a $2\,\Omega$ resistor.The equivalent resistance of the parallel part is $R_p = \frac{15 	imes 10}{15 + 10} = \frac{150}{25} = 6\,\Omega$.The total external resistance is $R_{ext} = 6 + 2 = 8\,\Omega$.The total resistance of the circuit is $R_{total} = 8 + 2r$.The total current in the circuit is $i = \frac{E_{eq}}{R_{total}} = \frac{3}{8 + 2r}$.The voltmeter is connected across the $10\,\Omega$ resistor,which is part of the parallel combination. The voltage across the parallel combination is $V_p = i 	imes R_p = i 	imes 6$.Given $V_p = 2\,V$,we have $2 = \frac{3}{8 + 2r} 	imes 6$.$2 = \frac{18}{8 + 2r} \Rightarrow 16 + 4r = 18 \Rightarrow 4r = 2 \Rightarrow r = 0.5\,\Omega$.

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

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