An L – C – R circuit of $R = 100\ \Omega$ is connected to an AC source 100 V, 50 Hz. The magnitude of phase difference between current and voltage is $30{^\circ}$. The power dissipated in the L – C – R circuit is

Question

Dr. Abubacker · Accepted Answer

(b) Average power dissipated in an AC circuit
$P_{av} = V_{rms}I_{rms}\cos\phi$ …(i)
Where the term $\cos\phi$ is known as power factor.
Given, $V_{rms} = 100\ V,\ R = 100\ \Omega,\ \phi = 30{^\circ}$
$	herefore\ \ \ \ \ \ \ \ \ \ \ I_{rms} = \frac{V_{rms}}{R} = \frac{100}{100} = 1\ $A
Putting the values in Eq. (i), we get
$P_{av} = 100 	imes 1 	imes \cos{30{^\circ}}$
$= 100\frac{\sqrt{3}}{2}$
$= 50\sqrt{3} = 86.6\ $W

Q. An L – C – R circuit of $R = 100\ \Omega$ is connected to an AC source 100 V, 50 Hz. The magnitude of phase difference between current and voltage is $30{^\circ}$. The power dissipated in the L – C – R circuit is

Explanation

Reviewed by Dr. Abubacker

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