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Question Number 216153 by Tawa11 last updated on 28/Jan/25

Answered by AntonCWX last updated on 29/Jan/25

$$i=i_L=\frac{12V}{1\mathrm{\Omega }+5\mathrm{\Omega }}=2A$$$$v_C=(1\mathrm{\Omega }+4\mathrm{\Omega })\left(2A\right)=10V$$$$$$$$W_C=\frac{1}{2}{Cv}_C^2=\frac{1}{2}\left(1\right)\left({10}^2\right)=50J$$$$W_L=\frac{1}{2}{Li}_L^2=\frac{1}{2}\left(2\right)\left(2^2\right)=4J$$

Commented by Tawa11 last updated on 29/Jan/25

$$\mathrm{Thanks}\mathrm{sir}.$$

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