In the given figure,the value of V_{0} will b | vedclass

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(B) To find the value of $V_{0}$,we apply Kirchhoff's Current Law $(KCL)$ at the node where $V_{0}$ is defined.Let the node voltage be $V_{0}$. The cu

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$4$

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(B) To find the value of $V_{0}$,we apply Kirchhoff's Current Law $(KCL)$ at the node where $V_{0}$ is defined.Let the node voltage be $V_{0}$. The currents leaving the node through the three parallel branches must sum to zero.Using nodal analysis:$\frac{V_{0}-2}{1 	ext{ k}\Omega} + \frac{V_{0}-4}{1 	ext{ k}\Omega} + \frac{V_{0}-6}{1 	ext{ k}\Omega} = 0$Since the resistances are equal,we can multiply the entire equation by $1 	ext{ k}\Omega$:$(V_{0}-2) + (V_{0}-4) + (V_{0}-6) = 0$$3V_{0} - 12 = 0$$3V_{0} = 12$$V_{0} = 4 	ext{ V}$

In the given figure,the value of $V_{0}$ will be $.....V$.

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