CBSE Class 12 · 2006 · Alternating CurrentLong Answer

The given circuit diagram shows a series LCR circuit with

L = 5.0 H

C = 80 μ F

and

R = 40 Ω

connected to a variable-frequency 230 V source. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of the current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. (d) How do you explain the observation that the algebraic sum of the voltages across the three elements obtained in (c) is greater than the supplied voltage?

Solution

(a) At resonance

ω_{r} = \frac{1}{L C} = \frac{1}{5 \times 80 \times 1 0 ^{- 6}} = 50 rad s^{- 1} \Rightarrow f_{r} = \frac{ω _{r}}{2 π} \approx 7.96 Hz

(b) At resonance

X_{L} = X_{C}

, so the impedance is minimum:

Z = R = 40 Ω

I_{0} = \frac{V _{0}}{R} = \frac{2 V _{r m s}}{R} = \frac{2 \times 230}{40} \approx 8.13 A

(c)

I_{r m s} = \frac{V _{r m s}}{R} = \frac{230}{40} = 5.75

V_{R} = I_{r m s} R = 5.75 \times 40 = 230 V

V_{L} = I_{r m s} ω_{r} L = 5.75 \times 50 \times 5 = 1437.5 V

V_{C} = \frac{I _{r m s}}{ω _{r} C} = \frac{5.75}{50 \times 80 \times 1 0 ^{- 6}} = 1437.5 V

(d) The three voltages are not in phase.

V_{L}

leads and

V_{C}

lags the current by

\frac{π}{2}

, so they are in antiphase and, being equal at resonance, cancel exactly. The source voltage equals the phasor sum

V_{R}^{2} + (V_{L} - V_{C})^{2} = V_{R} = 230

V, not the algebraic sum.

CBSE Class 12 Board · 2006 · Subjective

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CBSE Class 12 · 2006 · Alternating CurrentLong Answer

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