CBSE Class 12 · 2017 · Alternating CurrentLong Answer

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(a) Prove that the current flowing through an ideal inductor connected across an a.c. source lags the voltage in phase by

\frac{π}{2}

.
(b) An inductor of self-inductance 100 mH and a bulb are connected in series with an a.c. source of rms voltage 10 V, 50 Hz. It is found that the effective voltage of the circuit leads the current in phase by

\frac{π}{4}

. Calculate the average power dissipated in the circuit.

Answer: (a) $i = i_{m} sin (ω t - \frac{π}{2})$ , so the current lags by $\frac{π}{2}$ ;
(b) $P_{a v} \approx 1.59$ W.

Solution

(a) For an ideal inductor across

V = V_{m} sin ω t

, Kirchhoff's law gives

V_{m} sin ω t = L \frac{d I}{d t}

, so

d I = \frac{V _{m}}{L} sin ω t d t \Rightarrow I = - \frac{V _{m}}{ω L} cos ω t = \frac{V _{m}}{ω L} sin (ω t - \frac{π}{2})

Thus

I = i_{m} sin (ω t - \frac{π}{2})

— the current lags the voltage by

\frac{π}{2}

.
(b) The voltage leads the current by

ϕ = \frac{π}{4}

, so

tan ϕ = \frac{X _{L}}{R} = 1

, i.e.

R = X_{L}

. The inductive reactance is

X_{L} = ω L = 2 π \times 50 \times 0.1 = 10 π \approx 31.4 Ω = R

Impedance

Z = R^{2} + X_{L}^{2} = X_{L} 2 \approx 44.4 Ω

, so

I_{r m s} = \frac{V _{r m s}}{Z} = \frac{10}{44.4} \approx 0.225

P_{a v} = V_{r m s} I_{r m s} cos ϕ = 10 \times 0.225 \times \frac{1}{2} \approx 1.59 W

P_{a v} \approx 1.59 W

CBSE Class 12 Board · 2017 · Subjective

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CBSE Class 12 · 2017 · Alternating CurrentLong Answer

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