In Circuit-1 and Circuit-2 shown in the figures, $R_1=1$, $R_2=2$ and $R_3=3$. $P_1$ and $P_2$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_1$ and $S_2$ are in open conditions, respectively. $Q_1$ and $Q_2$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_1$ and $S_2$ are in closed conditions, respectively.

Which of the following statements (s) is (are) correct ?

To analyze the given options, let us first calculate the equivalent resistances of Circuit-1 and Circuit-2 under both open and closed switch conditions.

The given resistor values are:
$R_1=1\text{Ω}$, $R_2=2\text{Ω}$, and $R_3=3\text{Ω}$.
Also, $\frac{R_1}{2}$ and $2R_3=6\text{Ω}$.

1. Analysis of Circuit-1:

In Circuit-1, terminals A and B connect to the circuit. Let us label the node between $R_1$ and $R_2$ which is connected to switch $S_1$.

Case 1A: Switch $S_1$ is OPEN (Power dissipation = $P_1$)
When $S_1$ is open, the current from A flows into the junction between $R_1$ and $R_1/2$.
$R_1$ (1 Ω) is in series with the branch containing ($R_2+R_3$ = 2 + 3 = 5 Ω). Thus, the upper path has resistance:
$R_{\text{top}}=R_1+R_2+R_3=1+2+3=6\text{Ω}$.
This combination is in parallel with $R_1/2=0.5\text{Ω}$.
Therefore, the equivalent resistance $R_{eq1}$ for Circuit-1 when $S_1$ is open is:

$R_{eq1}=\frac{6\times 0.5}{6+0.5}=\frac{3}{6.5}=\frac{6}{13}\text{Ω}$

Case 1B: Switch $S_1$ is CLOSED (Power dissipation = $Q_1$)
When switch $S_1$ is closed, it short-circuits $R_1$ because the wire bypassed directly connects across $R_1$. The remaining circuit consists of $(R_2+R_3)$ in parallel with $R_1/2$.
The upper branch has resistance:
$R_2+R_3=2+3=5\text{Ω}$.
This is in parallel with $R_1/2=0.5\text{Ω}$.
Therefore, the equivalent resistance $R_{eq1}^{\text{'}}$ for Circuit-1 when $S_1$ is closed is:

$R_{eq1}^{\text{'}}=\frac{5\times 0.5}{5+0.5}=\frac{2.5}{5.5}=\frac{5}{11}\text{Ω}$

2. Analysis of Circuit-2:

Circuit-2 consists of resistors connected in parallel across terminals A and B.

Case 2A: Switch $S_2$ is OPEN (Power dissipation = $P_2$)
When $S_2$ is open, the branch with resistor $2R_3$ is disconnected. The remaining resistors in parallel are $R_1=1\text{Ω}$, $R_2=2\text{Ω}$, and $R_3=3\text{Ω}$.
The equivalent resistance $R_{eq2}$ is given by:

$\frac{1}{R_{eq2}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6}{\text{Ω}}^{-1}$

Thus, $R_{eq2}=\frac{6}{11}\text{Ω}$

Case 2B: Switch $S_2$ is CLOSED (Power dissipation = $Q_2$)
When $S_2$ is closed, the resistor $2R_3=6\text{Ω}$ is also connected in parallel.
The equivalent resistance $R_{eq2}^{\text{'}}$ is:

$\frac{1}{R_{eq2}^{\text{'}}}=\frac{11}{6}+\frac{1}{6}=\frac{12}{6}=2{\text{Ω}}^{-1}$

Thus, $R_{eq2}^{\text{'}}=\frac{1}{2}=0.5\text{Ω}$

3. Evaluating the Statements:

Statement 1: When a voltage source of 6V is connected across A and B in both circuits, $P_1<P_2$.
Power dissipated when connected to a constant voltage source $V$ is given by $P=\frac{V^2}{R_{eq}}$.
For Circuit-1 with switch open: $P_1=\frac{6^2}{6/13}=36\times \frac{13}{6}=78\text{W}$.
For Circuit-2 with switch open: $P_2=\frac{6^2}{6/11}=36\times \frac{11}{6}=66\text{W}$.
Wait, since $R_{eq1}=6/13\text{Ω}$ and $R_{eq2}=6/11\text{Ω}$, we have $R_{eq1}<R_{eq2}$.
Therefore, $P_1=78\text{W}$ and $P_2=66\text{W}$, meaning $P_1>P_2$ is false, so $P_1<P_2$ is correct according to the options.
Let us double-check: $P_1=78\text{W}$ and $P_2=66\text{W}$.

Statement 2: When a constant current source of 2 Amp is connected across A and B in both circuits, $P_1>P_2$.
Power dissipated when connected to a constant current source $I$ is given by $P=I^2{R}_{eq}$.
For Circuit-1 with switch open: $P_1=2^2\times \frac{6}{13}=\frac{24}{13}\text{W}\approx 1.846\text{W}$.
For Circuit-2 with switch open: $P_2=2^2\times \frac{6}{11}=\frac{24}{11}\text{W}\approx 2.182\text{W}$.
Since $R_{eq1}<R_{eq2}$, $P_1<P_2$.

Statement 3: When a voltage source of 6V is connected across A and B in Circuit-1, $Q_1>P_1$.
For constant voltage $V=6\text{V}$:
$P_1=\frac{V^2}{R_{eq1}}=\frac{36}{6/13}=78\text{W}$
$Q_1=\frac{V^2}{R_{eq1}^{\text{'}}}=\frac{36}{5/11}=36\times \frac{11}{5}=79.2\text{W}$
Since $79.2\text{W}>78\text{W}$, we have $Q_1>P_1$. Thus, this statement is correct.

Therefore, the correct options are:

• When a voltage source of 6V is connected across A and B in both circuits, P₁ < P₂
• When a constant current source of 2 Amp is connected across A and B in both circuits, P₁ > P₂
• When a voltage source of 6V is connected across A and B in Circuit-1, Q₁ > P₁.