CBSE Class 12 · 2012 · Alternating CurrentShort Answer

The figure shows a series LCR circuit with

L = 10.0 H

C = 40 μ F

and

R = 60 Ω

connected to a variable-frequency 240 V source. Calculate:
(i) the angular frequency of the source that drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rms potential drop across the inductor at resonance.

Answer: (i) $ω = 50$ rad s $^{- 1}$ ;
(ii) $I = 4$ A;
(iii) $V_{L} = 2000$ V.

Solution

(i) At resonance

ω = \frac{1}{L C} = \frac{1}{10 \times 40 \times 1 0 ^{- 6}} = \frac{1}{4 \times 1 0 ^{- 4}} = 50 rad s^{- 1}

(ii) At resonance the net reactance is zero, so

Z = R

and

I = \frac{V _{r m s}}{R} = \frac{240}{60} = 4 A

(iii) The inductive reactance is

X_{L} = ω L = 50 \times 10 = 500 Ω

, so

V_{L} = I X_{L} = 4 \times 500 = 2000 V

ω = 50 rad s^{- 1}, I = 4 A, V_{L} = 2000 V

CBSE Class 12 Board · 2012 · Subjective

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CBSE Class 12 · 2012 · Alternating CurrentShort Answer

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