CBSE Class 12 · 2017 · Alternating CurrentLong Answer

In the circuit shown, a resistor

R = 10 Ω

and an inductor

L = 200 mH

are connected in series with a capacitor

C

and an a.c. source of rms voltage 50 V and angular frequency

100 s^{- 1}

. Calculate
(a) the capacitance of the capacitor if the power factor of the circuit is unity,
(b) the Q-factor of this circuit and state the significance of the Q-factor in an a.c. circuit, and
(c) the average power dissipated in the circuit.

Answer: (a) $C = 0.5 mF = 500 μ F$ ;
(b) $Q = 2$ (it measures the sharpness of resonance);
(c) $P = 250$ W.

Solution

(a) Unity power factor means the circuit is at resonance,

X_{L} = X_{C}

, so

ω = \frac{1}{L C}

C = \frac{1}{ω ^{2} L} = \frac{1}{( 100 ) ^{2} \times 0.2} = \frac{1}{2000} = 5 \times 1 0^{- 4} F = 0.5 mF

(b)

Q = \frac{1}{R} \frac{L}{C} = \frac{1}{10} \frac{0.2}{5 \times 1 0 ^{- 4}} = \frac{1}{10} 400 = \frac{1}{10} \times 20 = 2

The Q-factor measures the sharpness (selectivity) of resonance — a higher

Q

gives a narrower, sharper resonance peak.
(c) At resonance

Z = R

, so

I_{r m s} = \frac{V _{r m s}}{R} = \frac{50}{10} = 5

A and

cos ϕ = 1

P_{a v} = V_{r m s} I_{r m s} cos ϕ = 50 \times 5 \times 1 = 250 W

C = 0.5 mF, Q = 2, P_{a v} = 250 W

CBSE Class 12 Board · 2017 · Subjective

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CBSE Class 12 · 2017 · Alternating CurrentLong Answer

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