In Circuit-1 and Circuit-2 shown in the figures, $R_1=1 \Omega$, $R_2=2 \Omega$ and $R_3=3 \Omega$. $P_1$ 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 ?

The correct statements are:
(1) When a voltage source of 6V is connected across A and B in both circuits, P₁ < P₂
(2) When a constant current source of 2 Amp is connected across A and B in both circuits, P₁ > P₂
(3) When a voltage source of 6V is connected across A and B in Circuit-1, Q₁ > P₁

First, let us carefully read the circuit diagrams from the image.

Circuit-1 (Switch S₁ open): The top branch has R₁, R₂, R₃ all in series. S₁ is connected in parallel with R₁ (it bridges the two nodes at either end of R₁). A resistor R₁/2 is in the bottom branch connected between the middle node (between R₁ and R₂) and terminal A/B side. When S₁ is open, no current flows through S₁ or R₁/2. The circuit between A and B is simply R₁ + R₂ + R₃ in series.

R_{\mathrm{eq},1}\left(\mathrm{open}\right)=R_1+R_2+R_3=1+2+3=6\text{Ω}

Circuit-2 (Switch S₂ open): R₁, R₂, R₃ are three separate parallel branches between A and B. S₂ is in series with 2R₃ forming a fourth branch — but since S₂ is open, that branch carries no current. So only R₁, R₂, R₃ in parallel remain.

\frac{1}{R_{\mathrm{eq},2}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6}{6}+\frac{3}{6}+\frac{2}{6}=\frac{11}{6}

R_{\mathrm{eq},2}\left(\mathrm{open}\right)=\frac{6}{11}\text{Ω}

Now we also need the equivalent resistances when the switches are closed.

Circuit-1 (Switch S₁ closed): When S₁ is closed, it short-circuits R₁, so no current flows through R₁. Also, the bottom branch with R₁/2 is now connected (since S₁ closes the path). The current now flows through S₁ (zero resistance) in parallel with R₁, effectively making the parallel combination = 0 Ω. Then in series with R₂ and R₃, we also have R₁/2 in the bottom branch. Let us be precise: S₁ shorts R₁ at the top-left, and R₁/2 connects the node between R₁ and R₂ down to the bottom rail. With S₁ closed, the node to the left of R₂ is connected directly to the bottom rail through S₁. This means R₁/2 becomes connected between the same node (junction of R₁ and R₂) and the bottom rail — i.e., R₁/2 is now in parallel with R₂ + R₃.

With R₁ = 1 Ω, so R₁/2 = 0.5 Ω. R₂ + R₃ = 2 + 3 = 5 Ω.

R_{\mathrm{eq},1}\left(\mathrm{closed}\right)=\frac{\frac{R_1}{2}\times (R_2+R_3)}{\frac{R_1}{2}+(R_2+R_3)}=\frac{0.5\times 5}{0.5+5}=\frac{2.5}{5.5}=\frac{5}{11}\text{Ω}

Circuit-2 (Switch S₂ closed): Now all four branches are active in parallel: R₁, R₂, R₃, and 2R₃. With R₃ = 3 Ω, so 2R₃ = 6 Ω.

\frac{1}{R_{\mathrm{eq},2}}\left(\mathrm{closed}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{6}{6}+\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{12}{6}=2

R_{\mathrm{eq},2}\left(\mathrm{closed}\right)=\frac{1}{2}\text{Ω}

Let us now compile the four equivalent resistances:

R_{\mathrm{eq},1}\left(\mathrm{open}\right)=6\text{Ω}\text{,}{R}_{\mathrm{eq},2}\left(\mathrm{open}\right)=\frac{6}{11}\text{Ω}

R_{\mathrm{eq},1}\left(\mathrm{closed}\right)=\frac{5}{11}\text{Ω}\text{,}{R}_{\mathrm{eq},2}\left(\mathrm{closed}\right)=\frac{1}{2}\text{Ω}

Statement (1): Voltage source of 6V connected across A and B. Compare P₁ and P₂ (switches open).

Power dissipated by a voltage source V across resistance R is:

P=\frac{V^2}{R}

Since the voltage is fixed at 6V, a larger equivalent resistance means less power dissipated.

P_1=\frac{6^2}{6}=\frac{36}{6}=6\text{W}

P_2=\frac{6^2}{\frac{6}{11}}=\frac{36}{1}\times \frac{11}{6}=66\text{W}

Since 6 W < 66 W, we get P₁ < P₂. ✓ Statement (1) is CORRECT.

Statement (2): Constant current source of 2 A connected across A and B. Compare P₁ and P₂ (switches open).

Power dissipated by a constant current source I through resistance R is:

P=I^{2}R

Since the current is fixed at 2 A, a larger resistance means more power dissipated.

P_1=2^2\times 6=4\times 6=24\text{W}

P_2=2^2\times \frac{6}{11}=4\times \frac{6}{11}=\frac{24}{11}\approx 2.18\text{W}

Since 24 W > 2.18 W, we get P₁ > P₂. ✓ Statement (2) is CORRECT.

Statement (3): Voltage source of 6V across A and B in Circuit-1. Compare Q₁ (switch closed) and P₁ (switch open).

When S₁ is closed in Circuit-1:

Q_1=\frac{6^2}{\frac{5}{11}}=36\times \frac{11}{5}=\frac{396}{5}=79.2\text{W}

We already computed P₁ = 6 W. Since 79.2 W > 6 W, Q₁ > P₁. ✓ Statement (3) is CORRECT.

This makes intuitive sense: closing S₁ drastically reduces the total resistance (from 6 Ω to 5/11 Ω ≈ 0.45 Ω), so with the same 6V supply, much more current flows and power dissipation increases enormously.

Statement (4): Constant current source of 2 A across A and B in both circuits. Compare Q₂ (S₂ closed) and Q₁ (S₁ closed).

With 2 A constant current:

Q_1=2^2\times \frac{5}{11}=4\times \frac{5}{11}=\frac{20}{11}\approx 1.818\text{W}

Q_2=2^2\times \frac{1}{2}=4\times \frac{1}{2}=2\text{W}

Since 2 W > 1.818 W, we get Q₂ > Q₁, meaning Q₂ < Q₁ is FALSE. ✗ Statement (4) is INCORRECT.

Summary Table:

Therefore, statements (1), (2), and (3) are correct.