Question

In a series LCR circuit connected to an a.c. source of voltage $v=v_{m}sin\omega t$, use phasor diagram to derive an expression for the current in the circuit. Hence, obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum.

Answer 1

Phasor diagram is shown for a series LCR circuit.

Using definition of reactance,

Voltage across capacitor is $V_C=I{X}_C=\frac{I}{\omega C}$

Voltage across inductor is $V_L=I{X}_L=I\omega L$

Voltage across resistor is $V_R=IR$

From definition of impedance, $V_S=IZ$

From the phasor diagram,

$V_S^2=V_R^2+(V_L-V_C{)}^2$

${\left(\frac{V_m}{\sqrt{2}}\right)}^2=I^2(R^2+(\omega L-\frac{1}{\omega C}{)}^2)$

$I=\frac{V_m}{\sqrt{2(R^2+(\omega L-\frac{1}{\omega C}{)}^2)}}$

Power dissipated in the circuit is:

$P=I^{2}R$

$P=\frac{V_m^{2}R}{2(R^2+(\omega L-\frac{1}{\omega C}{)}^2)}$

At resonance,

$\omega L=\frac{1}{\omega C}$

Denominator of power is minimum and hence, power is maximum.

Maximum power is given by:

$P_{max}=\frac{V_m^2}{2R}$