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(A) Let the current in the circuit be $I$. The circuit consists of two cells of $EMF$ $E$ connected in series,but with opposing polarities. The total

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(A) Let the current in the circuit be $I$. The circuit consists of two cells of $EMF$ $E$ connected in series,but with opposing polarities. The total $EMF$ of the circuit is $E_{net} = E - E = 0$. However,looking at the diagram,the cells are connected such that they support each other in driving current through the loop. The total $EMF$ is $2E$ and the total resistance is $R_{total} = r + 1 + 2 = r + 3$.Using Ohm's law,the current $I = \frac{2E}{r+3}$.The voltmeter is connected across the cell with internal resistance $r$. The terminal voltage $V$ across this cell is given by $V = E - Ir$.Given that $V = \frac{E}{2}$,we have $\frac{E}{2} = E - Ir$,which implies $Ir = \frac{E}{2}$,so $I = \frac{E}{2r}$.Equating the two expressions for current: $\frac{2E}{r+3} = \frac{E}{2r}$.$4r = r + 3 \implies 3r = 3 \implies r = 1 \ \Omega$.

In the given circuit,the reading of the ideal voltmeter is $\frac{E}{2}$. Find the internal resistance $r$ of the cell in $\Omega$.

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