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 statement(s) is(are) correct?

Case 1 : When $$S_1$$ and $$S_2$$ are open

$$(R_{eq})_1 = 1 + \frac{5 \times \frac{1}{2}}{5 + \frac{1}{2}} = 1 + \frac{5}{11} = \frac{16}{11}$$

$$P_1 = \frac{V^2}{(R_{eq})_1} = \frac{(6)^2}{\frac{16}{11}} = \frac{36 \times 11}{16} = 24.75\text{ W}$$

$$(R_{eq})_2 = \frac{6}{11}\ \Omega$$

$$P_2 = \frac{V^2}{(R_{eq})_2} = \frac{(6)^2}{\frac{6}{11}} = \frac{36 \times 11}{6} = 66\text{ W}$$

$$P_2 > P_1$$

Option (A) is correct.

Case 2 : If $$\text{I} = 2\text{ A}$$ source is used in both circuits, then

$$P_1 = i^2(R_{eq})_1 = (2)^2 \times \frac{16}{11} = \frac{64}{11} = 5.818\text{ W}$$

$$P_2 = i^2(R_{eq})_2 = (2)^2 \times \frac{6}{11} = \frac{24}{11} = 2.1818\text{ W}$$

$$P_1 > P_2$$

Option (B) is correct.

Case 3 : For $$Q_1$$

$$R_{eq} = \frac{5}{11}\ \Omega$$

$$Q_1 = \frac{V^2}{R_{eq}} = \frac{(6)^2}{\frac{5}{11}} = \frac{36 \times 11}{5} = 79.2\text{ W}$$

$$P_1 = 24.75\text{ W}$$

$$Q_1 > P_1$$

Option (C) is correct.

Case 4 : For option (D)

$$Q_1 = i^2R_{eq} = (2)^2 \times \frac{5}{11} = \frac{20}{11} = 1.81\text{ W}$$

$$Q_2 = i^2R_{eq} = (2)^2 \times \frac{1}{2} = \frac{4}{2} = 2\text{ W}$$

$$Q_2 > Q_1$$

Option (D) is incorrect.