Question

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 :

• A. $2.26 \times 10^3 \; J$
• B. $3.39 \times 10^3 \; J$
• C. $5.65 \times 10^2 \; J$
• D. $5.17 \times 10^2 \; J$

Answer 1

Answer: (D)

$5.17 \times 10^2 \; J$

Answer 2

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