CBSE Class 12 · 2005 · Alternating CurrentShort Answer

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(i) Draw the graphs showing the variation of inductive reactance and capacitive reactance with the frequency of the applied a.c. source.
(ii) Can the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than the applied voltage of the a.c. source? Justify your answer.

Answer: (i) $X_{L} = 2 π f L$ rises linearly with $f$ ; $X_{C} = \frac{1}{2 π f C}$ falls as $1/ f$ .
(ii) Yes, it can.

Solution

(i) Inductive reactance

X_{L} = 2 π f L

is directly proportional to frequency, so its graph is a straight line through the origin. Capacitive reactance

X_{C} = \frac{1}{2 π f C}

is inversely proportional to frequency, so its graph is a rectangular hyperbola falling with

f

.
(ii) Yes. The voltages across

L

and

C

are not in phase with

V_{R}

—

V_{L}

leads and

V_{C}

lags the current by

\frac{π}{2}

. They add as phasors, not as ordinary numbers, so an individual reactive voltage can exceed the source voltage while the phasor sum

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

still equals it.

CBSE Class 12 Board · 2005 · Subjective

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CBSE Class 12 · 2005 · Alternating CurrentShort Answer

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