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(C) Let the potential at node $A$ be $10\, V$ and at node $C$ be $0\, V$. Let the potentials at nodes $B$ and $D$ be $x$ and $y$ respectively.Applying

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$4.87\, mA$

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(C) Let the potential at node $A$ be $10\, V$ and at node $C$ be $0\, V$. Let the potentials at nodes $B$ and $D$ be $x$ and $y$ respectively.Applying Kirchhoff's Current Law $(KCL)$ at node $B$:$\frac{x-10}{100} + \frac{x-y}{15} + \frac{x-0}{10} = 0$Multiplying by $300$:$3(x-10) + 20(x-y) + 30x = 0$$3x - 30 + 20x - 20y + 30x = 0$$53x - 20y = 30 \quad \dots(1)$Applying $KCL$ at node $D$:$\frac{y-10}{60} + \frac{y-x}{15} + \frac{y-0}{5} = 0$Multiplying by $60$:$(y-10) + 4(y-x) + 12y = 0$$y - 10 + 4y - 4x + 12y = 0$$-4x + 17y = 10 \quad \dots(2)$Solving equations $(1)$ and $(2)$:From $(2)$,$x = \frac{17y-10}{4}$. Substituting into $(1)$:$53(\frac{17y-10}{4}) - 20y = 30$$901y - 530 - 80y = 120$$821y = 650 \implies y \approx 0.7917\, V$$x = \frac{17(0.7917)-10}{4} \approx 0.8647\, V$The potential difference across the galvanometer is $V_B - V_D = x - y = 0.8647 - 0.7917 = 0.073\, V$.The current through the galvanometer is $I_g = \frac{V_B - V_D}{R_g} = \frac{0.073}{15} \approx 0.00487\, A = 4.87\, mA$.

The four arms of a Wheatstone bridge have resistances as shown in the figure. $A$ galvanometer of $15\, \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10\, V$ is maintained across $AC$.

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