In Circuit-1 and Circuit-2 shown in the figures, 𝑅�=1 Ω, 𝑅�=2Ω and 𝑅�=3 Ω. � �� and 𝑃� are the power dissipations in Circuit-1 and Circuit-2 when the switches S1 and S2 are in open conditions, respectively. � ��and 𝑄� are the power dissipations in Circuit-1 and Circuit-2 when the switches S1 and S2 are in closed conditions, respectively. Which of the following statement(s) is(are) correct? (A) When a voltage source of 6 𝑉 is connected across A and B in both circuits, 𝑃�<𝑃�. (B) When a constant current source of 2 𝐴𝑚𝑝 is connected across A and B in both circuits, 𝑃�>𝑃�. (C) When a voltage source of 6 𝑉 is connected across A and B in Circuit-1, 𝑄�>𝑃�. (D) When a constant current source of 2 𝐴𝑚𝑝 is connected across A and B in both circuits, � ��<𝑄�.

 Alright, let's tackle this problem step by step. We have two circuits, Circuit-1 and Circuit-2, with resistors 

$R_1=1\mathrm{\Omega }$, $R_2=2\mathrm{\Omega }$, and $R_3=3\mathrm{\Omega }$. There are switches S1 and S2 in each circuit that can be open or closed, affecting the power dissipation in the circuits. We need to analyze the power dissipations $P$ and $Q$ under different conditions and determine which of the given statements (A) through (D) are correct.

Understanding the Circuits

First, let's visualize the circuits based on the description. Since the actual figures aren't provided, I'll make reasonable assumptions based on typical circuit configurations involving switches and resistors.

Assumed Configurations:

• Circuit-1:

  • When S1 is open: Maybe $R_1$ and $R_2$ are in series.

  • When S1 is closed: Perhaps $R_3$ is added in parallel or series, changing the total resistance.

• Circuit-2:

  • When S2 is open: Possibly $R_2$ and $R_3$ are in series.

  • When S2 is closed: Maybe $R_1$ is added in some configuration, altering the total resistance.

However, without the exact figures, it's challenging. Let's assume a common setup where:

• In Circuit-1:

  • Open S1: $R_1$ and $R_2$ in series.

  • Closed S1: $R_3$ is connected in parallel to $R_2$, making $R_2$ and $R_3$ parallel, then in series with $R_1$.

• In Circuit-2:

  • Open S2: $R_2$ and $R_3$ in series.

  • Closed S2: $R_1$ is connected in parallel to $R_2$, making $R_1$ and $R_2$ parallel, then in series with $R_3$.

This is a guess; the actual configurations might differ. For the sake of proceeding, let's proceed with these assumptions. If the actual configurations are different, the answers would change accordingly.

Calculating Resistances in Different States

Circuit-1:

1. S1 Open:

   • $R_1$ and $R_2$ in series.

   • Total resistance, $R_{P1}=R_1+R_2=1+2=3\mathrm{\Omega }$.

2. S1 Closed:

   • $R_2$ and $R_3$ in parallel, then in series with $R_1$.

   • Parallel combination: $\frac{1}{R_{parallel}}=\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$.

   • $R_{parallel}=\frac{6}{5}\mathrm{\Omega }$.

   • Total resistance, $R_{Q1}=R_1+R_{parallel}=1+\frac{6}{5}=\frac{11}{5}\mathrm{\Omega }$.

Circuit-2:

1. S2 Open:

   • $R_2$ and $R_3$ in series.

   • Total resistance, $R_{P2}=R_2+R_3=2+3=5\mathrm{\Omega }$.

2. S2 Closed:

   • $R_1$ and $R_2$ in parallel, then in series with $R_3$.

   • Parallel combination: $\frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2}$.

   • $R_{parallel}=\frac{2}{3}\mathrm{\Omega }$.

   • Total resistance, $R_{Q2}=R_{parallel}+R_3=\frac{2}{3}+3=\frac{11}{3}\mathrm{\Omega }$.

Power Dissipation Calculations

Power dissipation in a resistor (or circuit) can be calculated using:

• For a voltage source: $P=\frac{V^2}{R}$.

• For a current source: $P=I^{2}R$.

Let's evaluate each statement.

Statement (A): 6V voltage source, $P_1<P_2$.

• $P_1$ (Circuit-1, S1 open): $\frac{6^2}{3}=\frac{36}{3}=12W$.

• $P_2$ (Circuit-2, S2 open): $\frac{6^2}{5}=\frac{36}{5}=7.2W$.

Is $12<7.2$? No, $12>7.2$. So, (A) is incorrect.

Statement (B): 2A current source, $P_1>P_2$.

• $P_1$: $I^2{R}_{P1}=2^2\times 3=4\times 3=12W$.

• $P_2$: $I^2{R}_{P2}=4\times 5=20W$.

Is $12>20$? No, $12<20$. So, (B) is incorrect.

Statement (C): 6V in Circuit-1, $Q_1>P_1$.

• $P_1$: $\frac{36}{3}=12W$ (as above).

• $Q_1$ (Circuit-1, S1 closed): $\frac{36}{11\mathrm{/}5}=\frac{36\times 5}{11}=\frac{180}{11}\approx 16.36W$.

Is $16.36>12$? Yes. So, (C) is correct.

Statement (D): 2A current source in both, $P_2<Q_2$.

• $P_2$: $4\times 5=20W$ (as above).

• $Q_2$ (Circuit-2, S2 closed): $I^2{R}_{Q2}=4\times \frac{11}{3}=\frac{44}{3}\approx 14.67W$.

Is $20<14.67$? No, $20>14.67$. So, (D) is incorrect.

Re-evaluating Circuit Assumptions

Wait a minute, I think I might have made incorrect assumptions about the circuit configurations. Maybe the switches bypass resistors when closed, changing the circuit topology differently. Let me reconsider the possible configurations.

Alternative Assumption:

• Circuit-1:

  • Open S1: All resistors in some configuration, say $R_1$ and $R_2$ in series.

  • Closed S1: Maybe S1 shorts $R_2$, leaving only $R_1$.

• Circuit-2:

  • Open S2: $R_2$ and $R_3$ in series.

  • Closed S2: S2 shorts $R_3$, leaving only $R_2$.

Let's try this:

Circuit-1:

1. S1 Open: $R_1$ and $R_2$ in series: $R_{P1}=3\mathrm{\Omega }$.

2. S1 Closed: $R_2$ is shorted, only $R_1$: $R_{Q1}=1\mathrm{\Omega }$.

Circuit-2:

1. S2 Open: $R_2$ and $R_3$ in series: $R_{P2}=5\mathrm{\Omega }$.

2. S2 Closed: $R_3$ is shorted, only $R_2$: $R_{Q2}=2\mathrm{\Omega }$.

Now recalculate powers.

Statement (A): 6V, $P_1<P_2$.

• $P_1=\frac{36}{3}=12W$.

• $P_2=\frac{36}{5}=7.2W$.

Is $12<7.2$? No. So (A) incorrect.

Statement (B): 2A, $P_1>P_2$.

• $P_1=4\times 3=12W$.

• $P_2=4\times 5=20W$.

Is $12>20$? No. So (B) incorrect.

Statement (C): 6V in Circuit-1, $Q_1>P_1$.

• $P_1=12W$.

• $Q_1=\frac{36}{1}=36W$.

Is $36>12$? Yes. So (C) correct.

Statement (D): 2A in both, $P_2<Q_2$.

• $P_2=20W$.

• $Q_2=4\times 2=8W$.

Is $20<8$? No. So (D) incorrect.

This also leads to only (C) being correct.

Given that in both assumptions, only (C) is correct, it's likely the right answer. However, since the exact circuit configurations aren't provided, there's some ambiguity. But based on standard interpretations, (C) seems consistently correct.

Final Answer

After carefully analyzing both possible circuit configurations, the only statement that is consistently correct is:

(C) When a voltage source of 6 𝑉 is connected across A and B in Circuit-1, 𝑄�>𝑃�.

Therefore, the correct choice is:

\boxed{C}