Q. A series AC circuit containing an inductor (20 mH), a capacitor $(120 \; \mu F) $and a resistor $(60 \Omega)$ is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is :

Solution:

$R = 60 \Omega \, f = 50 Hz , \omega = 2\pi f = 100 \pi $
$ x_{C} = \frac{1}{\omega C} = \frac{1}{100 \pi \times120 \times10^{-6}} $
$ x_{C} 26.52 \Omega $
$ x_{L} =\omega L = 100 \pi \times20 \times10^{-3} = 2 \pi\Omega$
$ x_{C} - x_{L} = 20.24 \approx 20 $
$ z = \sqrt{R^{2} + \left(x_{C} -x_{L}\right)^{2}} $
$ z = 20 \sqrt{10} \Omega$
$ \cos\phi = \frac{R}{z} = \frac{3}{\sqrt{10} } $
$ P_{avg } = VI \cos \phi , I = \frac{v}{z} $
$ = \frac{v^{2}}{z} \cos \phi $
$ = 8.64 $ watt
$Q = P.t = 8.64 \times 60 = 5.18 \times 10^2$

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