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(C) From the circuit diagram,the total resistance $R_{total}$ is the sum of the resistors in series: $R_{total} = 20\,\Omega + 5\,\Omega + R = 25\,\Om

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$2.5$

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(C) From the circuit diagram,the total resistance $R_{total}$ is the sum of the resistors in series: $R_{total} = 20\,\Omega + 5\,\Omega + R = 25\,\Omega + R$.The net electromotive force $(EMF)$ in the circuit is the difference between the two cells: $E_{net} = 6\,V - 2\,V = 4\,V$.Using Ohm's law,$I = \frac{E_{net}}{R_{total}}$,we have $0.1 = \frac{4}{25 + R}$.$25 + R = \frac{4}{0.1} = 40$,which gives $R = 15\,\Omega$.Now,to find the potential difference between $X$ and $Y$,we traverse the path from $X$ to $Y$ through the left branch.The potential difference $V_{XY} = V_X - V_Y = I 	imes (20\,\Omega + 5\,\Omega)$.$V_{XY} = 0.1\,A 	imes 25\,\Omega = 2.5\,V$.

The current flowing in the given circuit is $0.1\,A$. The potential difference between the points $X$ and $Y$ is ................ $V$.

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