In a series resonant LCR circuit, the voltage across R is 100 V and R = 1 kΩ with C = 2 μF . The resonant frequency ω is 200 rad/s . At resonance, the - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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In a series resonant LCR circuit, the voltage across R is 100 V and R = 1 k Ω with C = 2 μ F . The resonant frequency ω is 200 rad/s . At resonance, the voltage across L is:

Solution & Step-by-step Explanation

At resonance, V_{L} = V_{C} .Also, current I = V_{R} / R = 100/1000 = 0.1 A .Voltage across capacitor V_{C} = I X_{C} = I \frac{1}{ω C} . V_{C} = 0.1 \times \frac{1}{200 \times 2 \times 1 0 ^{- 6}} V_{C} = 0.1 \times \frac{1}{400 \times 1 0 ^{- 6}} = 0.1 \times \frac{1 0 ^{6}}{400} V_{C} = \frac{1 0 ^{5}}{400} = \frac{1000}{4} = 250 V Since V_{L} = V_{C} at resonance, V_{L} = 250 V .

Try it yourself before checking the explanation above.

In a series resonant LCR circuit, the voltage across R is 100 V and R = 1 k Ω with C = 2 μ F . The resonant frequency ω is 200 rad/s . At resonance, the voltage across L is:

4 \times 1 0^{- 3} V

2.5 \times 1 0^{- 2} V

40 V

250 V

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