Power dissipated in LCR series circuit at resonance

Question

A series LCR circuit has $R = 50\,\Omega$ , $L = 0.2\,\text{H}$ , and $C = 50\,\mu\text{F}$ . It is connected to an AC source of 220 V (rms) at the resonant frequency. Find the power dissipated in the circuit at resonance.

(JEE Main 2022, similar pattern)

Solution — Step by Step

At resonance, the inductive reactance equals the capacitive reactance: $X_L = X_C$ . They cancel each other out. The impedance of the circuit becomes purely resistive:

Z = R

This is the critical insight — at resonance, the circuit behaves as if only the resistance exists.

Since $Z = R$ at resonance:

I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{R} = \frac{220}{50} = 4.4\,\text{A}

This is the maximum possible current in the circuit — resonance gives you peak current.

Power is dissipated only in the resistance (L and C store and return energy, they don’t dissipate):

P = I_{rms}^2 \cdot R = (4.4)^2 \times 50 = 19.36 \times 50 = \mathbf{968\,\text{W}}

Alternatively, at resonance the power factor $\cos\phi = 1$ , so:

P = \frac{V_{rms}^2}{R} = \frac{(220)^2}{50} = \frac{48400}{50} = \mathbf{968\,\text{W}}

Why This Works

In an LCR circuit, the inductor and capacitor create opposing reactances. At resonance frequency $\omega_0 = 1/\sqrt{LC}$ , these exactly cancel. The voltage across L and voltage across C are equal in magnitude but opposite in phase — they neutralise each other.

The result: the full source voltage appears across the resistance alone. Current is maximum, and since the phase angle $\phi = 0$ , the power factor is unity. The circuit draws maximum power from the source.

JEE Main frequently tests three things about LCR resonance: (1) the resonant frequency formula $\omega_0 = 1/\sqrt{LC}$ , (2) the fact that impedance = R at resonance, and (3) the quality factor $Q = \omega_0 L / R$ , which tells you how “sharp” the resonance peak is.

Alternative Method

Use the power formula with power factor directly:

P = V_{rms} \cdot I_{rms} \cdot \cos\phi

At resonance, $\cos\phi = 1$ (voltage and current are in phase). So:

P = 220 \times 4.4 \times 1 = 968\,\text{W}

Quick check for resonance problems: if the question says “at resonance” or “at resonant frequency,” immediately set $Z = R$ and $\cos\phi = 1$ . You do not need to calculate $X_L$ , $X_C$ , or the resonant frequency itself unless explicitly asked.

Common Mistake

Students sometimes calculate $X_L$ and $X_C$ separately and then compute $Z = \sqrt{R^2 + (X_L - X_C)^2}$ . While correct in general, at resonance $X_L = X_C$ , so $Z$ simplifies to $R$ . The mistake is wasting time computing these reactances when the question already states the circuit is at resonance.

A more serious error: assuming power is dissipated in L and C. Ideal inductors and capacitors dissipate zero average power — they alternately store and release energy. All power dissipation in an ideal LCR circuit occurs in the resistance.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next