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The impedance of a circuit, when a resistance $R$ and an inductor of inductance $L$ are connected in series in an AC circuit of frequency $f$, is
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Verified Answer
The correct answer is:
$\sqrt{R^2+4 \pi^2 f^2 L^2}$
In $L-R$ circuit,
Impedance $Z=\sqrt{R^2+X_L^2}$
Here, $\quad X_L=\omega L=2 \pi f L$
$\therefore \quad Z=\sqrt{R^2+4 \pi^2 f^2 L^2}$
Impedance $Z=\sqrt{R^2+X_L^2}$
Here, $\quad X_L=\omega L=2 \pi f L$
$\therefore \quad Z=\sqrt{R^2+4 \pi^2 f^2 L^2}$
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