Q132 / 150 · MHT CET 2026 (19 April Shift 1) · Physics
In an LCR circuit at resonance, the a.c. source current is
A) maximum in a series LCR circuit only.
B) maximum in a parallel LCR circuit only.
C) maximum in both series and parallel LCR circuits.
D) minimum in both series and parallel LCR circuits.
The correct answer is
A) maximum in a series LCR circuit only.
B) maximum in a parallel LCR circuit only.
C) maximum in both series and parallel LCR circuits.
D) minimum in both series and parallel LCR circuits.
The correct answer is
- AB
- BD
- CC
- Correct answerDA
Solution
In a series LCR circuit, the impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance, $X_L = X_C$, which makes the impedance minimum ($Z = R$).
Since $I = \dfrac{V}{Z}$, the current is maximum in a series LCR circuit at resonance.
In a parallel LCR circuit, the admittance is given by $Y = \sqrt{\left(\dfrac{1}{R}\right)^2 + \left(\dfrac{1}{X_C} - \dfrac{1}{X_L}\right)^2}$.
At resonance, $X_L = X_C$, which makes the admittance minimum ($Y = \dfrac{1}{R}$) and the impedance maximum.
Since $I = V Y$, the current is minimum in a parallel LCR circuit at resonance.
Thus, the a.c. source current is maximum in a series LCR circuit only.
Answer: A
At resonance, $X_L = X_C$, which makes the impedance minimum ($Z = R$).
Since $I = \dfrac{V}{Z}$, the current is maximum in a series LCR circuit at resonance.
In a parallel LCR circuit, the admittance is given by $Y = \sqrt{\left(\dfrac{1}{R}\right)^2 + \left(\dfrac{1}{X_C} - \dfrac{1}{X_L}\right)^2}$.
At resonance, $X_L = X_C$, which makes the admittance minimum ($Y = \dfrac{1}{R}$) and the impedance maximum.
Since $I = V Y$, the current is minimum in a parallel LCR circuit at resonance.
Thus, the a.c. source current is maximum in a series LCR circuit only.
Answer: A
Hence, the correct answer is option: 'D'.