In the following circuit,the current I_3 is | vedclass

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(D) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the central node:$I_1 + I_2 = I_3$Using Ohm's law for ea

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$-\frac{5}{6} \ A$

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(D) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at the central node:$I_1 + I_2 = I_3$Using Ohm's law for each branch:$I_1 = \frac{8 - 6 - V}{28} = \frac{2 - V}{28}$$I_2 = \frac{12 - 6 - V}{54} = \frac{6 - V}{54}$$I_3 = \frac{V - 0}{0} = 	ext{undefined (this approach is complex)}$.Alternatively,using nodal analysis at the central node:$\frac{V - 8}{28} + \frac{V - 12}{54} + \frac{V - 6}{0} = 0$ is not correct. Let's use loop analysis.For loop $1$: $28 I_1 + 6 + 8 = 0 \implies 28 I_1 = -14 \implies I_1 = -0.5 \ A$.For loop $2$: $54 I_2 + 6 + 12 = 0 \implies 54 I_2 = -18 \implies I_2 = -1/3 \ A$.By $KCL$ at the junction: $I_3 = I_1 + I_2 = -0.5 + (-1/3) = -1/2 - 1/3 = -5/6 \ A$.

In the following circuit,the current $I_3$ is

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