Question

A series LCR circuit with $R = 20\Omega $, $L = 1 \cdot 5{\text{H}}$ and $C = 35\mu {\text{F}}$ is connected to a variable frequency $200{\text{V}}$ ac supply. Find the average power transferred to the circuit in one complete cycle when the frequency of the supply equals the natural frequency of the circuit.
A) $200{\text{W}}$
B) $2000{\text{W}}$
C) $100{\text{W}}$
D) $4000{\text{W}}$

Answer 1

Hint:When the supply frequency equals the natural frequency of the circuit, resonance is attained. At resonance, the inductor impedance will be equal to the capacitor impedance. The average power at resonance is to be determined and it is given as the product of the rms values of the current and the voltage.

Formulas used:
The average power in an LCR circuit at resonance over a cycle is given by, ${P_{avg}} = {V_{rms}}{I_{rms}}$ where ${V_{rms}}$ and ${I_{rms}}$ are the rms values of the current and voltage respectively.
The impedance in an LCR circuit is given by, $Z = \sqrt {{R^2} + \left( {{X_L}^2 - {X_C}^2} \right)} $ where $R$ is the resistance of the resistor, ${X_L}$ is the inductor impedance and ${X_C}$ is the capacitor impedance.
The rms value of current in an LCR circuit is given by, ${I_{rms}} = \dfrac{{{V_{rms}}}}{Z}$ where $Z$ is the impedance of the circuit and ${V_{rms}}$ is the rms value of voltage.

Complete step by step answer.
Step 1: List the parameters known from the question.
The resistance of the resistor is $R = 20\Omega $ .
The inductance of the inductor is $L = 1 \cdot 5{\text{H}}$ .
The capacitance of the capacitor is $C = 35\mu {\text{F}}$ .
The rms value of the voltage is given to be ${V_{rms}} = 200{\text{V}}$ .

Step 2: Express the relation for the impedance of the circuit.
The impedance in an LCR circuit is given by, $Z = \sqrt {{R^2} + \left( {{X_L}^2 - {X_C}^2} \right)} $ where $R$ is the resistance of the resistor, ${X_L}$ is the inductor impedance and ${X_C}$ is the capacitor impedance.
At resonance, we have ${X_L} = {X_C}$. So the impedance of the circuit will be $Z = \sqrt {{R^2}} = R$ .
Step 3: Express the relation for the rms value of the current in the circuit.
The rms value of current in an LCR circuit is given by, ${I_{rms}} = \dfrac{{{V_{rms}}}}{Z}$ where $Z$ is the impedance of the circuit and ${V_{rms}}$ is the rms value of voltage.
Since $Z = R$ at resonance, we have ${I_{rms}} = \dfrac{{{V_{rms}}}}{R}$ -------- (1)
Substituting for $R = 20\Omega $ and ${V_{rms}} = 200{\text{V}}$in equation (1) we get, ${I_{rms}} = \dfrac{{200}}{{20}} = 10{\text{A}}$
Thus the rms value of current in the LCR circuit is ${I_{rms}} = 10{\text{A}}$ .
Step 4: Express the relation for the average power at resonance of the circuit.
The average power in the given circuit over a cycle at resonance is given by,
${P_{avg}} = {V_{rms}}{I_{rms}}$ ----------- (2)
Substituting for ${V_{rms}} = 200{\text{V}}$ and ${I_{rms}} = 10{\text{A}}$ in equation (2) we get, ${P_{avg}} = 200 \times 10 = 2000{\text{W}}$ .
Thus the average power over one complete cycle at resonance is ${P_{avg}} = 2000{\text{W}}$ .

So the correct option is B.

Note:The average power over a cycle also depends on the phase angle $\phi $ between the current and the voltage and it can be expressed as ${P_{avg}} = {V_{rms}}{I_{rms}}\cos \phi $. But at resonance, $\phi = 0^\circ $ and so $\cos \phi = 1$. Thus we obtain the equation as ${P_{avg}} = {V_{rms}}{I_{rms}}$. As impedance equals the resistance of the resistor at resonance, the power dissipated will be maximum and will be through the resistance.