Question

A sinusoidal voltage of peak value $250V$ is applied to a series $LCR$ circuit, in which $R=8\Omega$, $L=24mH$ and$C=60\mu F$. The value of power dissipated at resonant conditions is $\text{'}x\text{'}kW$. The value of$x$ to the nearest integer is _________.

Answer 1

Step 1. Given Data for series LCR circuit :

$R=8\Omega$, $L=24mH$ and$C=60\mu F$

Peak Voltage, $V_{peak}=250V$

Step 2. Finding the power dissipation at resonant condition:

At Resonant condition the capacitive and inductive reactance are equal, that is $X_c=X_L$ and phase angle $\varphi =0^o$

The Power dissipated is $P=V_{rms}.I_{rms}\cos \varphi =V_{rms}.\frac{V_{rms}}{R}.\cos 0^o$

$V_{rms}=\frac{V_{peak}}{\sqrt{2}}=\frac{250}{\sqrt{2}}$

$P={\left(\frac{V_{rms}}{\sqrt{2}}\right)}^2\frac{1}{R}=\frac{250}{2}\times \frac{1}{8}=4kW$

Hence, the value of x is $4$.