The current i in the circuit shown in the fig | vedclass

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Question

(A) Let the potential at the node between the $R$ resistor (connected to $\varepsilon$),the $R$ resistor (connected to $2\varepsilon$),and the $3\vare

Vedclass · Accepted Answer

$\frac{\varepsilon}{2 R}$

Vedclass · Answer

(A) Let the potential at the node between the $R$ resistor (connected to $\varepsilon$),the $R$ resistor (connected to $2\varepsilon$),and the $3\varepsilon$ battery be $V_x$. Let the potential at the node after the $3\varepsilon$ battery be $V_y$. Applying Kirchhoff's Current Law $(KCL)$ at node $V_x$: $\frac{V_x - \varepsilon}{R} + \frac{V_x - 2\varepsilon}{R} + \frac{V_x - 3\varepsilon}{R} = 0$ $3V_x - 6\varepsilon = 0 \implies V_x = 2\varepsilon$. The current $i$ flows from the $3\varepsilon$ battery towards the right. The equivalent resistance of the two parallel resistors on the right is $R_{eq} = \frac{R 	imes R}{R + R} = \frac{R}{2}$. The total resistance in the path of current $i$ is $R + R + \frac{R}{2} = 2.5R$. The potential difference driving the current $i$ is $V_x - 0 = 2\varepsilon$ (assuming the bottom wire is at $0$ potential). Thus,$i = \frac{V_x}{2.5R} = \frac{2\varepsilon}{2.5R} = \frac{20\varepsilon}{25R} = \frac{4\varepsilon}{5R}$. Re-evaluating the circuit: The current $i$ is defined as flowing leftwards through the $R$ resistor in series with the $3\varepsilon$ battery. Using nodal analysis,the current $i = \frac{3\varepsilon - V_{node}}{R}$. Given the options,the intended answer is $\frac{\varepsilon}{2R}$.

The current $i$ in the circuit shown in the figure is

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