CBSE Class 12 · 2016 · Delhi · Alternating CurrentLong Answer

(a) An AC source of voltage

V = V_{0} sin ω t

is connected to a series combination of

L

C

and

R

. Use a phasor diagram to obtain expressions for the impedance of the circuit and the phase angle between the voltage and the current. Find the condition for the current to be in phase with the voltage, and state what the circuit is then called.
(b) In a series LR circuit

X_{L} = R

and the power factor of the circuit is

P_{1}

. When a capacitor with capacitance

C

such that

X_{L} = X_{C}

is put in series, the power factor becomes

P_{2}

. Calculate

P_{1} / P_{2}

Solution

(a) In the phasor diagram

V_{R} = I R

is along the current,

V_{L} = I X_{L}

leads by

9 0^{\circ}

and

V_{C} = I X_{C}

lags by

9 0^{\circ}

. The resultant source voltage is

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

so the impedance is

Z = R^{2} + (X_{L} - X_{C})^{2}, tan ϕ = \frac{X _{L} - X _{C}}{R}

The current is in phase with the voltage when

X_{L} = X_{C}

, i.e. at resonance; the circuit is then called a series resonant (acceptor) circuit, with

Z = R

.
(b) For the LR circuit with

X_{L} = R

Z_{1} = R^{2} + R^{2} = R 2

, so

P_{1} = cos ϕ_{1} = \frac{R}{Z _{1}} = \frac{R}{R 2} = \frac{1}{2}

Adding

C

with

X_{C} = X_{L}

gives resonance,

Z_{2} = R

, so

P_{2} = \frac{R}{R} = 1

. Hence

\frac{P _{1}}{P _{2}} = \frac{1}{2}

CBSE Class 12 Board · 2016 · Delhi · Subjective

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CBSE Class 12 · 2016 · Delhi · Alternating CurrentLong Answer

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