In Circuit- 1 and Circuit- 2 shown in the figures, R _ 1 =...

When ${S}_{1} and {S}_{2}$ are open

For circuit 1

${\left({R}_{eq}\right)}_{1} = 1 + \dfrac{5 \times \dfrac{1}{2}}{5 + \dfrac{1}{2}} = 1 + \dfrac{5}{11} = \dfrac{16}{11}Ω$

${P}_{1} = \dfrac{{V}^{2}}{{R}_{eq}} = \dfrac{{\left(6\right)}^{2}}{16} \times 11 = \dfrac{36 \times 11}{16} = 24 . 75W$

For circuit 2

${\left({R}_{eq}\right)}_{2} = \dfrac{6}{11}Ω$

${P}_{2} = \dfrac{{V}^{2}}{{R}_{eq}} = \dfrac{{\left(6\right)}^{2}}{6} \times 11 = \dfrac{36 \times 11}{6} = 66W$

Therefore, ${P}_{2} > {P}_{1}$

Option $A$ is correct

$⇒$ If $2$ A source is used in both cases.

${P}_{1} = {i}^{2}{\left({R}_{eq}\right)}_{1} = {\left(2\right)}^{2} \times \dfrac{16}{11} = \dfrac{64}{11} = 5 . 818W$

${P}_{2} = {i}^{2}{\left({R}_{eq}\right)}_{2} = ( 2{)}^{2} \times \dfrac{6}{11} = \dfrac{24}{11} = 2 . 1818W$

${P}_{1} > {P}_{2}$

Option \left(B\right)$is correct.</p><p>When${S}_{1} and {S}_{2}$are closed</p><p>For${Q}_{1}$</p><p>${R}_{eq} = \dfrac{5}{11}Ω$</p><p>${Q}_{1} = \dfrac{{V}^{2}}{{R}_{eq}} = \dfrac{{\left(6\right)}^{2}}{\dfrac{5}{11}} = \dfrac{36 \times 11}{5} = 79 . 2W$</p><p>${P}_{1} = 24 . 75W$</p><p>Therefore,${Q}_{1} > {P}_{1}$</p><p>Option \left(C\right)$ is correct.

For option \left(D\right)$</p><p>${Q}_{1} = {i}^{2}{R}_{eq} = {\left(2\right)}^{2} \times \dfrac{5}{11} = \dfrac{20}{11} = 1 . 81W$</p><p>${Q}_{2} = {i}^{2}{R}_{eq} = {\left(2\right)}^{2} \times \dfrac{1}{2} = \dfrac{4}{2} = 2W$</p><p>${Q}_{2} > {Q}_{1}$</p><p>Option \left(D\right)$ is incorrect.