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…

When $S_1$ and $S_2$ are open

For circuit 1

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

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

For circuit 2

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

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

Therefore, $P_2>P_1$

Option $A$ is correct

$\Rightarrow$  If $2$ A source is used in both cases.

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

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

$P_1>P_2$

Option $\left(B\right)$ is correct.

When $S_1$ and $S_2$ are closed

For $Q_1$

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

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

$P_1=24.75 \mathrm{W}$

Therefore, $Q_1>P_1$

Option $\left(C\right)$ is correct.

For option $\left(D\right)$

$Q_1=i^2{R}_{\mathrm{eq}}={\left(2\right)}^2\times \frac{5}{11}=\frac{20}{11}=1.81 \mathrm{W}$

$Q_2=i^2{R}_{\mathrm{eq}}={\left(2\right)}^2\times \frac{1}{2}=\frac{4}{2}=2 \mathrm{W}$

$Q_2>Q_1$

Option $\left(D\right)$ is incorrect.