What is the value of i_3 in the given circuit | vedclass

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(D) Let the potential at the top junction be $V$ and the bottom junction be $0 \, V$. Applying Kirchhoff's Current Law $(KCL)$ at the top junction:$\f

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$-5/6 \, A$

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(D) Let the potential at the top junction be $V$ and the bottom junction be $0 \, V$. Applying Kirchhoff's Current Law $(KCL)$ at the top junction:$\frac{V - 8}{28} + \frac{V - 12}{54} + \frac{V - 6}{0} = 0$ is not applicable directly due to the branch with only a battery. Instead,use nodal analysis.Let the potential at the top node be $V_x$ and the bottom node be $0 \, V$.The current $i_3$ flows downwards through the $6 \, V$ battery.Using the node voltage method:$\frac{V_x - 8}{28} + \frac{V_x - 12}{54} + \frac{V_x - 6}{0}$ is incorrect. Let's use loop analysis.In loop $1$ (left): $28 i_1 + 6 + 8 = 0 \Rightarrow 28 i_1 = -14 \Rightarrow i_1 = -0.5 \, A$.In loop $2$ (right): $54 i_2 + 6 + 12 = 0 \Rightarrow 54 i_2 = -18 \Rightarrow i_2 = -1/3 \, A$.At the central node,by $KCL$: $i_3 = i_1 + i_2 = -0.5 - 1/3 = -3/6 - 2/6 = -5/6 \, A$.

What is the value of $i_3$ in the given circuit?

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