In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is (a) 2.5 × 10⁻² V (b) 40 V (c) 250 V (d) 4 × 10⁻³ V

Answer: (c). I = VR/R = (100)/(1000) = 0.1 A At resonance XL = XC = (1)/(ω C) = 1200×2×10⁻⁶ = 2500 Ω VL = IXL = 0.1×2500 = 250 V So the answer is (c) 250 V.

In a series resonant LCR circuit, the voltage across R is 100 volts… | JEE Main 2006 (AIEEE)

Q109JEE Main 2006 · AIEEEDifficult

Resonance, Bandwidth & Q Factor

In a series resonant $LCR$ circuit, the voltage across $R$ is $100$ volts and $R = 1\text{ k}\Omega$ with $C = 2\ \mu\text{F}.$ The resonant frequency $\omega$ is $200\text{ rad/s}$ . At resonance the voltage across $L$ is

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Answer: (c)

$I = \frac{V_R}{R} = \frac{100}{1000} = 0.1\text{ A}$
At resonance $X_L = X_C = \frac{1}{\omega C} = \frac{1}{200\times2\times10^{-6}} = 2500\ \Omega$
$V_L = IX_L = 0.1\times2500 = 250\text{ V}$
So the answer is (c) 250 V.

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