CBSE Class 12 · 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 AC 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 AC source? Justify your answer.

Solution

(i) Inductive reactance

X_{L} = 2 π f L

is directly proportional to

f

, so its graph is a straight line through the origin.
Capacitive reactance

X_{C} = \frac{1}{2 π f C}

is inversely proportional to

f

, so its graph is a rectangular hyperbola falling with increasing

f

.
(ii) Yes. The voltages across

L

and

C

are

9 0^{\circ}

out of phase with the resistor voltage and

18 0^{\circ}

out of phase with each other, so they partly cancel. The source voltage is the phasor sum

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

, not the arithmetic sum; hence

V_{L}

V_{C}

individually (especially near resonance) can exceed the applied voltage.

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CBSE Class 12 · Alternating CurrentShort Answer

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