Lcr Series Resonance
A series LCR circuit contains an inductor of 2 henry and a capacitor of 8 microfarads along with a resistor. At what angular frequency will this circuit exhibit electrical resonance and draw maximum current?
Select the correct option:
Solution
250rad/s
NCERT Class 12, Chapter 7 (Alternating Current) shows that a series LCR circuit resonates when the inductive and capacitive reactances become equal, so that ωL=ωC1, giving the resonant angular frequency ω0=LC1. At resonance the impedance is purely resistive and minimum, so the current is maximum. Substituting L=2 H and C=8×10−6 F: LC=2×8×10−6=1.6×10−5, so LC=4×10−3 s, and ω0=4×10−31=250 rad/s. The option 125 rad/s halves the result by a square-root error. The option 500 rad/s doubles it. The option 62.5 rad/s divides by an extra factor of four. As a plausibility check, the units of H⋅F1 reduce to rad/s, and resonance frequencies of this order are typical for such component values used in tuning circuits. An important insight is that the resonant frequency depends only on the inductance and capacitance, and is completely independent of the resistance in the circuit; the resistance instead controls how sharp the resonance peak is and how much current flows at resonance. At this special frequency the voltage across the inductor and the voltage across the capacitor are equal in magnitude but opposite in phase, so they cancel exactly, leaving the source to see only the resistance. This is why series resonance is sometimes called acceptor resonance, since the circuit most readily accepts current at precisely this frequency.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- lcr series resonance
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
250rad/s
NCERT Class 12, Chapter 7 (Alternating Current) shows that a series LCR circuit resonates when the inductive and capacitive reactances become equal, so that ωL=ωC1, giving the resonant angular frequency ω0=LC1. At resonance the impedance is purely resistive and minimum, so the current is maximum. Substituting L=2 H and C=8×10−6 F: LC=2×8×10−6=1.6×10−5, so LC=4×10−3 s, and ω0=4×10−31=250 rad/s. The option 125 rad/s halves the result by a square-root error. The option 500 rad/s doubles it. The option 62.5 rad/s divides by an extra factor of four. As a plausibility check, the units of H⋅F1 reduce to rad/s, and resonance frequencies of this order are typical for such component values used in tuning circuits. An important insight is that the resonant frequency depends only on the inductance and capacitance, and is completely independent of the resistance in the circuit; the resistance instead controls how sharp the resonance peak is and how much current flows at resonance. At this special frequency the voltage across the inductor and the voltage across the capacitor are equal in magnitude but opposite in phase, so they cancel exactly, leaving the source to see only the resistance. This is why series resonance is sometimes called acceptor resonance, since the circuit most readily accepts current at precisely this frequency.
This medium difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of lcr series resonance. It appeared in the 2025 exam.
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