CBSE Class 12 · 2010 · Alternating CurrentLong Answer

A series LCR circuit is connected to a source having voltage

v = v_{m} sin ω t

. Derive the expression for the instantaneous current

I

and its phase relationship to the applied voltage. Obtain the condition for resonance to occur. Define 'power factor' and state the conditions under which it is
(i) maximum and
(ii) minimum.

Solution

Using the phasor method,

V_{R} = I R

is in phase with

I

V_{L} = I X_{L}

leads by

\frac{π}{2}

, and

V_{C} = I X_{C}

lags by

\frac{π}{2}

. The resultant source voltage has magnitude

v_{m} = I_{m} R^{2} + (X_{L} - X_{C})^{2} = I_{m} Z

so the instantaneous current is

I = \frac{v _{m}}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}} sin (ω t - ϕ), tan ϕ = \frac{X _{L} - X _{C}}{R}

The current lags the voltage by

ϕ

when

X_{L} > X_{C}

and leads when

X_{C} > X_{L}

.
Resonance: when

X_{L} = X_{C}

, i.e.

ω L = \frac{1}{ω C} \Rightarrow ω = \frac{1}{L C}

then

Z = R

(minimum) and

I

is maximum.
Power factor is

cos ϕ = \frac{R}{Z}

. It is (i) maximum (

= 1

) when the circuit is purely resistive or at resonance, and (ii) minimum (

= 0

) for a purely inductive or purely capacitive circuit.

ω_{r} = \frac{1}{L C}, cos ϕ = \frac{R}{Z}

CBSE Class 12 Board · 2010 · Subjective

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CBSE Class 12 · 2010 · Alternating CurrentLong Answer

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