Power Dissipation Comparison in Circuit Configurations with Switches

To determine the correct statement(s), we first analyze the equivalent resistance of both circuits under open and closed switch conditions.

Given: $R_1 = 1\ \Omega, \quad R_2 = 2\ \Omega, \quad R_3 = 3\ \Omega$

---

1. Analysis of Circuit-1

• When switch $\text{S}_1$ is OPEN:

  • The upper branch consists of $R_1$, $R_2$, and $R_3$ connected in series: $R_{\text{upper, open}} = R_1 + R_2 + R_3 = 1 + 2 + 3 = 6\ \Omega$
  • The middle branch has resistance $R_1/2 = 0.5\ \Omega$, connected in parallel with the upper branch across terminals $\text{A}$ and $\text{B}$.
  • The net equivalent resistance $R_{1,\text{open}}$ is: $R_{1,\text{open}} = \frac{6 \times 0.5}{6 + 0.5} = \frac{3}{6.5} = \frac{6}{13}\ \Omega$

• When switch $\text{S}_1$ is CLOSED:

  • Switch $\text{S}_1$ short-circuits resistor $R_1$. The upper branch now consists of only $R_2$ and $R_3$ in series: $R_{\text{upper, closed}} = R_2 + R_3 = 2 + 3 = 5\ \Omega$
  • The net equivalent resistance $R_{1,\text{closed}}$ is: $R_{1,\text{closed}} = \frac{5 \times 0.5}{5 + 0.5} = \frac{2.5}{5.5} = \frac{5}{11}\ \Omega$

---

2. Analysis of Circuit-2

• When switch $\text{S}_2$ is OPEN:

  • The active branches connected in parallel across terminals $\text{A}$ and $\text{B}$ are $R_1$, $R_2$, and $R_3$: $\frac{1}{R_{2,\text{open}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}\ \Omega^{-1}$ $R_{2,\text{open}} = \frac{6}{11}\ \Omega$

• When switch $\text{S}_2$ is CLOSED:

  • An additional parallel branch containing resistor $2R_3 = 2(3) = 6\ \Omega$ is included: $\frac{1}{R_{2,\text{closed}}} = \frac{1}{R_{2,\text{open}}} + \frac{1}{2R_3} = \frac{11}{6} + \frac{1}{6} = \frac{12}{6} = 2\ \Omega^{-1}$ $R_{2,\text{closed}} = \frac{1}{2}\ \Omega$

---

3. Verification of the Options

• Option (A): Voltage source of $V = 6\text{ V}$ connected across A and B in both circuits when switches are open. $P_1 = \frac{V^2}{R_{1,\text{open}}} = \frac{6^2}{6/13} = 78\text{ W}$ $P_2 = \frac{V^2}{R_{2,\text{open}}} = \frac{6^2}{6/11} = 66\text{ W}$ Since $P_1 = 78\text{ W} > P_2 = 66\text{ W}$, statement (A) ($P_1 < P_2$) is incorrect.

• Option (B): Constant current source of $I = 2\text{ A}$ connected across A and B in both circuits when switches are open. $P_1 = I^2 R_{1,\text{open}} = 2^2 \times \frac{6}{13} = \frac{24}{13}\text{ W} \approx 1.85\text{ W}$ $P_2 = I^2 R_{2,\text{open}} = 2^2 \times \frac{6}{11} = \frac{24}{11}\text{ W} \approx 2.18\text{ W}$ Since $P_1 < P_2$, statement (B) ($P_1 > P_2$) is incorrect.

• Option (C): Voltage source of $V = 6\text{ V}$ connected across A and B in Circuit-1. $P_1 = 78\text{ W}$ $Q_1 = \frac{V^2}{R_{1,\text{closed}}} = \frac{6^2}{5/11} = \frac{396}{5} = 79.2\text{ W}$ Since $Q_1 = 79.2\text{ W} > P_1 = 78\text{ W}$, statement (C) ($Q_1 > P_1$) is correct.

• Option (D): Constant current source of $I = 2\text{ A}$ connected across A and B in both circuits when switches are closed. $Q_1 = I^2 R_{1,\text{closed}} = 2^2 \times \frac{5}{11} = \frac{20}{11}\text{ W} \approx 1.82\text{ W}$ $Q_2 = I^2 R_{2,\text{closed}} = 2^2 \times \frac{1}{2} = 2\text{ W}$ Since $Q_2 = 2\text{ W} > Q_1 = \frac{20}{11}\text{ W}$, statement (D) ($Q_2 < Q_1$) is incorrect.

---

Conclusion

The only correct statement is (C).