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(D) Let $V_0$ be the potential at point $O$. Applying the nodal analysis method at node $O$:$\frac{V_0 - V_1}{R_1} + \frac{V_0 - V_2}{R_2} + \frac{V_0

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(D) Let $V_0$ be the potential at point $O$. Applying the nodal analysis method at node $O$:$\frac{V_0 - V_1}{R_1} + \frac{V_0 - V_2}{R_2} + \frac{V_0 - V_3}{R_3} = 0$Substituting the given values:$\frac{V_0 - 20}{20} + \frac{V_0 - 30}{30} + \frac{V_0 - 60}{60} = 0$Multiplying the entire equation by $60$ to clear the denominators:$3(V_0 - 20) + 2(V_0 - 30) + 1(V_0 - 60) = 0$$3V_0 - 60 + 2V_0 - 60 + V_0 - 60 = 0$$6V_0 - 180 = 0$$6V_0 = 180 \implies V_0 = 30 \ V$The current $i_2$ flowing through resistance $R_2$ is given by:$i_2 = \frac{V_0 - V_2}{R_2} = \frac{30 - 30}{30} = 0 \ A$

Find the current flowing through the resistance $R_2$ of the circuit shown in the figure,if the resistances are $R_1 = 20 \ \Omega, R_2 = 30 \ \Omega$ and $R_3 = 60 \ \Omega$,and the potentials of points $1, 2$ and $3$ are $V_1 = 20 \ V, V_2 = 30 \ V$ and $V_3 = 60 \ V$ respectively. (in $A$)

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