Question

A series R-L-C circuit consists of an $8\,\Omega $ resistor, a \[5.0{\text{ }}\mu F\] capacitor, and a 50.0 mH inductor. A variable frequency source applies an emf of 400 V (RMS) across the combination. The power delivered to the circuit when the frequency is equal to one-half the resonance frequency is:-
(A) 52 W
(B) 57 W
(C) 63 W
(D) 69 W

Answer 1

Hint: In this solution, we will first calculate the resonant frequency of the circuit. Then we will calculate the net reactance of the circuit and the power delivered to the circuit for a frequency half of the resonant frequency.
Formula used: In this solution, we will use the following formula:
-Resonant frequency of series LCR circuit: $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{1}{{LC}}} $ where $L$ is the inductance and $C$ is the capacitance of the circuit.
1 - Capacitive reactance: ${X_C} = \dfrac{1}{{2\pi fC}}$
2- Inductive reactance: ${X_L} = 2\pi fL$
3- Magnitude of Impedance of a series LCR circuit: \[\left| z \right| = \sqrt {{R^2} + {{\left( {X_L^2 - X_C^2} \right)}^2}} \] where $R$ is the resistance, ${X_L}$ is the inductive impedance, and ${X_C}$ is the capacitive inductance.

Complete step by step answer:
In a series LCR circuit, the applied frequency is one-half of the resonant frequency as given to us. Let us start by finding the resonant frequency of the circuit which is calculated as
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{1}{{LC}}} $
Substituting $L = 50 \times {10^{ - 3}}\,H$ and $C = 5 \times {10^{ - 6}}\,F$, we get
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{1}{{10 \times {{10}^{ - 10}}}}} $
Which gives us
$f = \dfrac{{{{10}^5}}}{{2\pi \sqrt {10} }}$
Now the net reactance of the circuit will be
\[\left| z \right| = \sqrt {{R^2} + {{\left( {X_L^2 - X_C^2} \right)}^2}} \]
Substituting the value of ${X_C} = \dfrac{1}{{2\pi fC}}$ and ${X_L} = 2\pi fL$ and $R = 8\,\Omega $, we get the net impedance as
$\left| z \right| = 150.21\,{\text{ohm}}$
Then the current in the circuit will be calculated as the ratio of the RMS voltage and the net impedance as determined from ohm’s law as:
$i = \dfrac{{{E_{RMS}}}}{{\left| z \right|}}$
$ \Rightarrow i = \dfrac{{400}}{{150.21}} = 2.66\,A$
Then the average power delivered to the circuit will be
$P = {i^2}R$
$ \Rightarrow P = {(2.66)^2} \times 8$
Which can be simplified to

$P = 56.7\,W \approx 57\,W$ which corresponds to option (B).

Note: The formulae that we have used is only valid for a series LCR circuit connected with a sinusoidally varying input voltage. We should be aware of the formulae of reactance and net impedance as well as other basic concepts of circuits to answer this question.