CBSE Class 12 · 2016 · All India · Alternating CurrentLong Answer

In a series LCR circuit connected to an AC source of voltage

v = v_{m} sin ω t

, use a phasor diagram to derive an expression for the current in the circuit. Hence obtain the expression for the power dissipated in the circuit, and show that the power dissipated at resonance is maximum.

Solution

From the phasor diagram the source voltage is the resultant of

V_{R}

V_{L}

and

V_{C}

v_{m}^{2} = (i_{m} R)^{2} + (i_{m} X_{C} - i_{m} X_{L})^{2} \Rightarrow i_{m} = \frac{v _{m}}{Z}, Z = R^{2} + (X_{C} - X_{L})^{2}

so the current is

i = \frac{v _{m}}{Z} sin (ω t + ϕ)

with

tan ϕ = \frac{X _{C} - X _{L}}{R}

.
The instantaneous power is

p = v i

, and its average over a cycle (the oscillatory term averages to zero) is

P = \frac{v _{m} i _{m}}{2} cos ϕ = V_{rms} I_{rms} cos ϕ

At resonance

X_{L} = X_{C}

, so

Z = R

ϕ = 0

and

cos ϕ = 1

. Then the current is largest and

P_{m a x} = \frac{v _{rms}^{2}}{R}

showing the power dissipated is maximum at resonance.

CBSE Class 12 Board · 2016 · All India · Subjective

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CBSE Class 12 · 2016 · All India · Alternating CurrentLong Answer

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