In the circuit shown,what is the potential di | vedclass

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(A) Let the potential at the junction between the $4 \ V$ battery and the bottom wire be $0 \ V$. Let the potential at node $R$ be $V_R = 4 \ V$.Apply

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$+ 3 \ V$

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(A) Let the potential at the junction between the $4 \ V$ battery and the bottom wire be $0 \ V$. Let the potential at node $R$ be $V_R = 4 \ V$.Applying Kirchhoff's Current Law at node $R$:$\frac{V_P - 4 - 2}{2 + 1} + \frac{V_Q - 4}{3 + 2} = 0$$\frac{V_P - 6}{3} + \frac{V_Q - 4}{5} = 0$$5(V_P - 6) + 3(V_Q - 4) = 0 \implies 5V_P + 3V_Q = 42 \quad (1)$Also,the potential difference $V_{PQ} = V_P - V_Q$. From the left loop,$V_P - V_R = i_1(2+1) + 2 = 3i_1 + 2$. Since $V_R = 4 \ V$,$V_P = 3i_1 + 6$.From the right loop,$V_Q - V_R = i_2(3+2) - 1 = 5i_2 - 1$. Since $V_R = 4 \ V$,$V_Q = 5i_2 + 3$.Using nodal analysis,let $V_R = 4 \ V$. Current leaving $R$ towards $P$ is $i_1 = \frac{4 - (V_P - 2)}{3} = \frac{6 - V_P}{3}$. Current leaving $R$ towards $Q$ is $i_2 = \frac{4 - (V_Q + 1)}{5} = \frac{3 - V_Q}{5}$.At node $R$,$i_1 + i_2 = 0 \implies \frac{6 - V_P}{3} + \frac{3 - V_Q}{5} = 0 \implies 30 - 5V_P + 9 - 3V_Q = 0 \implies 5V_P + 3V_Q = 39$.Solving for $V_P - V_Q$: The potential difference $V_{PQ} = V_P - V_Q = 3 \ V$.

In the circuit shown,what is the potential difference $V_{PQ}$?

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