alternating-currents Question 18

Question: Q. 2. (i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase : current or voltage?

(ii) Without making any other change, find the value of the additional capacitor C1, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.

U] [Delhi I, II, III 2017]

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Solution:

Ans. (i) Calculation of phase difference between current and voltage

Name of quantity which leads

(ii) Calculation of value of ’ C ‘, is to be connected in parallel

(i) XL=ωL=(1000×100×103)Ω

=100Ω

XC=1ωC=(11000×2×106)Ω

=500Ω

Phase angle

tanϕ=XLXCR tanϕ=100500400=1 ϕ=π4

As XC>XL (phase angle is negative), hence current leads voltage (ii) To make power factor unity

XC=XL 1ωC=100 C=10μF C=C+C1 10=2+C1 C1=8μF

[CBSE Marking Scheme 2017]

[AI Q. 3. A sinusoidal voltage of peak value 10 V is applied to a series LCR circuit in which resistance, capacitance and inductance have values of 10Ω, 1μF and 1H respectively. Find (i) the peak voltage across the inductor at resonance (ii) quality factor of the circuit.

U] [CBSE SQP 2018-19]

Ans.

I0=V0/R=10/10=1 A1/2

ωr=1LC=1(1×1×106)1/2

=103rad/s

V0=I0XL=I0ωrL

=1×103×1=103 V

Q=ωrL/R

=(103×1)/10=1001/2

[CBSE Marking Scheme 2018]



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