In the following circuit, 5Ω resistor develops 45 J/s due to current flowing through it. The power developed across 12Ω resistor is

Power through $12\Omega =?$

$\text{p}={\text{i}}_{\text{1}}^{\text{2}}\text{R}$

$45={\text{i}}_{\text{1}}^{\text{2}}\left(5\right)\Rightarrow {\text{i}}_{\text{1}}^{\text{2}}=9\Rightarrow {\text{i}}_{\text{1}}=3\text{A}$

${\text{i}}_{\text{2}}=\text{i}-{\text{i}}_{\text{1}}$

$9\Omega ,6\Omega$ Are in series so ${\text{R}}_{\text{eff}}=15\Omega$

We get $15\Omega \&5\Omega$ are in parallel

So, ${\text{i}}_{\text{1}}{\text{R}}_{\text{1}}={\text{i}}_{\text{2}}{\text{R}}_{\text{2}}\left(\because {\text{i}}_{\text{1}}=\text{3A}\right)$

$3\left(5\right)={\text{i}}_{\text{2}}\left(15\right)$

${\text{i}}_{\text{2}}=1\text{then}\text{i}={\text{i}}_{\text{1}}+{\text{i}}_{\text{2}}=\text{4A}$

$\therefore \text{i}=4\text{A,}\text{R}=12\Omega$

$\therefore \text{Power}\text{in}12\Omega \text{is}$

$\text{P}\text{=}{\text{i}}^{\text{2}}\text{R}=4\times 4\times 12=19\text{2W}$