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In the circuit given below, the current through inductor is $0.9 \mathrm{~A}$ and through the capacitor is $0.6 \mathrm{~A}$. The current drawn from the a.c. source is
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The correct answer is:
$0.3 \mathrm{~A}$
As the currents in an inductor and capacitor are out of phase by $180^{\circ}$, we can write Current through the capacitor $\mathrm{I}_{\mathrm{C}}=0.9 \mathrm{~A}$ Current through the inductor $\mathrm{I}_{\mathrm{L}}=-(0.6 \mathrm{~A})$
$\begin{aligned}
\therefore \quad \text { Total current drawn from the source } & =\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{L}} \\
& =0.9-0.6 \\
& =0.3 \mathrm{~A}
\end{aligned}$
$\begin{aligned}
\therefore \quad \text { Total current drawn from the source } & =\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{L}} \\
& =0.9-0.6 \\
& =0.3 \mathrm{~A}
\end{aligned}$
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