A series LCR circuit connected to an AC source with voltage of the source \( v = v_m\sin\omega t \).
If ‘q’ is the charge on the capacitor and ‘i’ is the current, from Kirchhoff’s loop rule :
\( L\displaystyle\frac{di}{dt} + iR + \frac{q}{c} = V \)
The current in the circuit is given by \( i = I_m\sin(\omega t + \phi) \) where \( \phi \) is the phase difference between the voltage across the source and current in the circuit.
We know \( V_{Rm} = I_m R \); \( V_{Lm} = I_m X_L \); \( V_{Cm} = I_m X_C \); and
\( X_L = \omega L \); \( X_C = \displaystyle\frac{1}{\omega C} \)
Total impedance in the circuit regulates current. At resonance frequency of the LCR circuit current in the circuit is maximum.
Given below are two statements
Statement I : In an LCR series circuit, power dissipated is minimum at resonance.
Statement II : In an LCR circuit, power is dissipated only in the inductor and capacitor.
In the light of the above statements, choose the most appropriate answer from the options given below
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Both Statement I and Statement II are false
The average power in an ac circuit is
\( \displaystyle P = V_{\mathrm{rms}}\,I_{\mathrm{rms}}\cos\phi = I_{\mathrm{rms}}^{2}\,R \)
Statement I: In an LCR series circuit, power dissipated is minimum at resonance — False. At resonance the impedance is minimum (\( Z = R \)), so the current is maximum, and \( \phi = 0 \) gives \( \cos\phi = 1 \). The power dissipated is therefore maximum at resonance, not minimum.
Statement II: In an LCR circuit, power is dissipated only in the inductor and capacitor — False. In an ideal inductor or capacitor the current and voltage differ in phase by \( 90^\circ \), so their average power is zero; the energy they store in one quarter cycle is returned in the next. Power is dissipated only in the resistor.
So both statements are false.
NCERT (2026-27):
Power in AC Circuit: The Power Factor – “… Even in such cases, power is dissipated only in the resistor. … At resonance Xc – XL = 0, and φ = 0. Therefore, cosφ = 1 and P = I2Z = I2R. That is, maximum power is dissipated in a circuit (through R) at resonance.”
Points to Ponder – “… 7. There are no power losses associated with pure capacitances and pure inductances in an ac circuit. The only element that dissipates energy in an ac circuit is the resistive element.”