Resistance Network Current and Potential Analysis

To analyze the resistance network, let us set the potential of the negative terminal (rightmost junction) as $V_Y = 0\text{ V}$ and the potential of the positive terminal (leftmost junction) as $V_X = 12\text{ V}$.

Let $V_P, V_Q, V_S,$ and $V_T$ denote the electric potentials at nodes $P, Q, S,$ and $T$, respectively.

1. Nodal Equations using Kirchhoff's Current Law (KCL)

At node $P$: $\frac{V_P - 12}{2} + \frac{V_P - V_Q}{1} + \frac{V_P - V_S}{2} = 0$ $V_P - 6 + 2(V_P - V_Q) + V_P - V_S = 0 \implies 4V_P - 2V_Q - V_S = 12 \quad \text{--- (1)}$

At node $Q$: $\frac{V_Q - 12}{4} + \frac{V_Q - V_P}{1} + \frac{V_Q - V_T}{4} = 0$ $V_Q - 12 + 4(V_Q - V_P) + V_Q - V_T = 0 \implies -4V_P + 6V_Q - V_T = 12 \quad \text{--- (2)}$

At node $S$: $\frac{V_S - V_P}{2} + \frac{V_S - V_T}{1} + \frac{V_S - 0}{2} = 0$ $V_S - V_P + 2(V_S - V_T) + V_S = 0 \implies -V_P + 4V_S - 2V_T = 0 \quad \text{--- (3)}$

At node $T$: $\frac{V_T - V_Q}{4} + \frac{V_T - V_S}{1} + \frac{V_T - 0}{4} = 0$ $V_T - V_Q + 4(V_T - V_S) + V_T = 0 \implies -V_Q - 4V_S + 6V_T = 0 \quad \text{--- (4)}$

2. Solving for Node Potentials

By bridge symmetry:

• Assume $V_P = V_Q$ and $V_S = V_T$.

From equation (1): $4V_P - 2V_P - V_S = 12 \implies 2V_P - V_S = 12 \quad \text{--- (5)}$

From equation (3): $-V_P + 4V_S - 2V_S = 0 \implies -V_P + 2V_S = 0 \implies V_P = 2V_S \quad \text{--- (6)}$

Substituting equation (6) into equation (5): $2(2V_S) - V_S = 12 \implies 3V_S = 12 \implies V_S = 4\text{ V}$

Using $V_S = 4\text{ V}$ in equation (6): $V_P = 8\text{ V}$

Thus, the node potentials are: $V_P = 8\text{ V}, \quad V_Q = 8\text{ V}, \quad V_S = 4\text{ V}, \quad V_T = 4\text{ V}$

Substituting these values back into equations (1)–(4) confirms that they uniquely satisfy all Kirchhoff's equations.

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3. Verification of Options

• Option (A): The electric current flowing through segment $PQ$ is: $I_{PQ} = \frac{V_P - V_Q}{1\,\Omega} = \frac{8 - 8}{1} = 0\text{ A}$ Therefore, Option (A) is correct.

• Option (B): The total current $I_1$ supplied by the $12\text{ V}$ battery is the sum of currents entering the top and bottom branches: $I_{\text{top}} = \frac{V_X - V_P}{2\,\Omega} = \frac{12 - 8}{2} = 2\text{ A}$ $I_{\text{bottom}} = \frac{V_X - V_Q}{4\,\Omega} = \frac{12 - 8}{4} = 1\text{ A}$ $I_1 = I_{\text{top}} + I_{\text{bottom}} = 2\text{ A} + 1\text{ A} = 3\text{ A}$ Therefore, Option (B) is correct.

• Option (C): Potential at node $S$ is $V_S = 4\text{ V}$, and potential at node $Q$ is $V_Q = 8\text{ V}$. Since $4\text{ V} < 8\text{ V}$, the potential at $S$ is lower than that at $Q$. Therefore, Option (C) is correct.

• Option (D): The current $I_2$ flowing through the segment $PS$ is: $I_2 = \frac{V_P - V_S}{2\,\Omega} = \frac{8 - 4}{2} = 2\text{ A}$ Therefore, Option (D) is correct.

Conclusion

Statements (A), (B), (C), and (D) are all valid.