CBSE Class 12 · 2017 · Delhi · Alternating CurrentShort Answer

For the series LCR circuit shown, with

L = 100 mH

C = 2 μ F

and

R = 400 Ω

driven by

V = V_{0} sin (1000 t + ϕ)

:
(i) find the phase difference between the current and the voltage, and state which one leads in phase.
(ii) Without making any other change, find the value of the additional capacitor

C_{1}

to be connected in parallel with the capacitor

C

in order to make the power factor of the circuit unity.

Answer: (i) $∣ ϕ ∣ = \frac{π}{4}$ (45°), the current leads the voltage.
(ii) $C_{1} = 8 μ F$ .

Solution

Here

ω = 1000

rad s

^{- 1}

X_{L} = ω L = 1000 \times 0.1 = 100 Ω, X_{C} = \frac{1}{ω C} = \frac{1}{1000 \times 2 \times 1 0 ^{- 6}} = 500 Ω

(i)

tan ϕ = \frac{X _{L} - X _{C}}{R} = \frac{100 - 500}{400} = - 1 \Rightarrow ϕ = - 4 5^{\circ}

Since

X_{C} > X_{L}

the circuit is capacitive, so the current leads the voltage by

\frac{π}{4}

.
(ii) For unity power factor the circuit must be at resonance,

X_{C}^{'} = X_{L} = 100 Ω

C^{'} = \frac{1}{ω X _{C}^{'}} = \frac{1}{1000 \times 100} = 10 μ F

With

C

and

C_{1}

in parallel,

C^{'} = C + C_{1}

, so

C_{1} = C^{'} - C = 10 - 2 = 8 μ F

ϕ = \frac{π}{4} (current leads), C_{1} = 8 μ F

CBSE Class 12 Board · 2017 · Delhi · Subjective

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CBSE Class 12 · 2017 · Delhi · Alternating CurrentShort Answer

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