CBSE Class 12 · 2014 · Delhi · Alternating CurrentLong Answer

A voltage

V = V_{0} sin ω t

is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is
(i) no power dissipated even though a current flows through the circuit,
(ii) maximum power dissipated in the circuit?

Solution

Let

V = V_{0} sin ω t

and

I = I_{0} sin (ω t - ϕ)

. The instantaneous power is

P = V_{0} I_{0} sin ω t sin (ω t - ϕ)

. With

2 sin A sin B = cos (A - B) - cos (A + B)

P = \frac{V _{0} I _{0}}{2} [cos ϕ - cos (2 ω t - ϕ)]

Averaging over a cycle (the oscillating term averages to zero):

P_{a v} = \frac{V _{0} I _{0}}{2} cos ϕ = V_{r m s} I_{r m s} cos ϕ

(i) No power is dissipated when

cos ϕ = 0

, i.e.

ϕ = \frac{π}{2}

— a purely reactive circuit with

R = 0

(wattless current).
(ii) Maximum power is dissipated when

cos ϕ = 1

, i.e.

ϕ = 0

, which occurs at resonance

X_{L} = X_{C}

where

Z = R

P_{a v} = V_{r m s} I_{r m s} cos ϕ

CBSE Class 12 Board · 2014 · Delhi · Subjective

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CBSE Class 12 · 2014 · Delhi · Alternating CurrentLong Answer

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