Alternating Current — CBSE Class 12 · 2019 Board Question with Solution

(a) In a series LCR circuit connected across an a.c. source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with the frequency of the a.c. source.
(b) What is the phase difference between the voltages across the inductor and the capacitor at resonance in the LCR circuit?
(c) When an inductor is connected to a 200 V d.c. voltage, a current of 1 A flows through it. When the same inductor is connected to a 200 V, 50 Hz a.c. source, only 0.5 A current flows. Explain why, and also calculate the self-inductance of the inductor.

Solution

(a) With

V_{R} = I R

(in phase),

V_{L} = I X_{L}

(leads by

\frac{π}{2}

) and

V_{C} = I X_{C}

(lags by

\frac{π}{2}

), the source voltage is

V = V_{R}^{2} + (V_{L} - V_{C})^{2} = I R^{2} + (X_{L} - X_{C})^{2}

, so

Z = R^{2} + (ω L - \frac{1}{ω C})^{2}

It is large at low and high

f

, dips to a minimum

Z = R

f_{r} = \frac{1}{2 π L C}

, giving a V-shaped curve.
(b) At resonance

V_{L}

leads and

V_{C}

lags the current each by

\frac{π}{2}

, so the phase difference between them is

π

(180°).
(c) With d.c. the inductor offers only its resistance (

X_{L} = 0

R = \frac{V}{I _{d c}} = \frac{200}{1} = 200 Ω

With a.c. the inductive reactance adds, raising the impedance and lowering the current:

Z = \frac{V}{I _{a c}} = \frac{200}{0.5} = 400 Ω

X_{L} = Z^{2} - R^{2} = 40 0^{2} - 20 0^{2} = 2003 Ω

. Since

X_{L} = ω L = 2 π f L

L = \frac{X _{L}}{2 π f} = \frac{200 3}{2 π \times 50} = \frac{2 3}{π} \approx 1.1 H

R = 200 Ω, L = \frac{2 3}{π} \approx 1.1 H

CBSE Class 12 Board · 2019 · Subjective

CBSE Class 12 · 2019 · Alternating CurrentLong Answer

Solution

More board questions on Alternating Current