Question

A sinusoidal voltage of peak value 250 V 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 condition is 'x' kW. The value of x to the nearest integer is _________.

Answer 1

Hint: Before we proceed into the problem, it is important to know about series LCR circuits and resonant conditions. A LCR circuit consists of three components: an inductor (L), a capacitor (C), and a resistor (R). It is also called resonant circuit. A series LCR circuit is made up of these devices that are connected in series. All three component have same current flowing through them.
Resonance is a condition in LCR circuit where the magnitude of inductive reactance is equal to capacitive reactance but 180 degrees apart in phase.

Complete answer:
If the output of a circuit reaches its maximum at a frequency, it is said to be in resonance. The resonance phenomenon is connected with systems that have a tendency to oscillate at a specific frequency known as the natural frequency of the system.

The resistor is the only component in the circuit that consumes power; the inductor and capacitor do not. Therefore, power term only contains energy due to resistor.
Given:
Voltage peak value = $V = 250V$
Resistance = $R = 8\Omega $

Now at resonance condition, power is given by
$P = \dfrac{{{{({V_{rms}})}^2}}}{R}$
${V_{rms}} = \dfrac{V}{{\sqrt 2 }}$
$ \Rightarrow {V_{rms}} = \dfrac{{250}}{{\sqrt 2 }}$
$ \Rightarrow P = \dfrac{{{{(\dfrac{{250}}{{\sqrt 2 }})}^2}}}{8}$
$ \Rightarrow P = 3906.25$
$ \Rightarrow P \approx 4kW$

So, the value of power dissipated at resonant condition is 4kW.

Note: While attempting this question one should note that the value of capacitance and inductance given has no use as at the resonant condition only resistive part of circuit is present. Care must be taken while taking the value of ${V_{rms}}$. Do not take Peak voltage we take the root mean square of peak voltage while calculating power in the circuit.