Question Number 216153 by Tawa11 last updated on 28/Jan/25
Answered by AntonCWX last updated on 29/Jan/25
$${i}={i}_{{L}} =\frac{\mathrm{12}{V}}{\mathrm{1}\Omega+\mathrm{5}\Omega}=\mathrm{2}{A} \\ $$$${v}_{{C}} =\left(\mathrm{1}\Omega+\mathrm{4}\Omega\right)\left(\mathrm{2}{A}\right)=\mathrm{10}{V} \\ $$$$ \\ $$$${W}_{{C}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{Cv}_{{C}} ^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\left(\mathrm{10}^{\mathrm{2}} \right)\:=\:\mathrm{50}{J} \\ $$$${W}_{{L}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{Li}_{{L}} ^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\left(\mathrm{2}^{\mathrm{2}} \right)\:=\:\mathrm{4}{J} \\ $$
Commented by Tawa11 last updated on 29/Jan/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$