CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 13
Total: 27
SECTION -C

A capacitor (C) and resistor (R) are connected in series with an ac source of voltage of frequency 50‌Hz. The potential difference across C and R are respectively 120V,90V, and the current in the circuit is 3A. Calculate (i) the impedance of the circuit (ii) the value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity.
OR
The figure shows a series LCR circuit connected to a variable frequency 230V source.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Calculate the impedance of the circuit and amplitude of current at resonance.
(c) Show that potential drop across LC combination is zero at resonating frequency.
Solution:  
Ans. Calculation of impedance Calculation of inductance
(i) ‌Z=√R2+XC2
‌R=‌
VR
IR
=30Ω

‌XC=‌
VC
IC
=‌
120
30
=40Ω

‌Z=√(30)2+(40)2=50Ω
‌XC=XL
(ii) As power factor =1
‌100πL=40
‌L=‌
2
5Ï€
‌ henry
‌

Determining the source frequency
Calculating impedance
For showing potential drop across LC
(a) ‌ω=‌
1
√LC
=‌
1
√5×80×10−6
=‌
1
√400×10−6

‌ω=‌
1000
20
=50‌Hz

‌ (b) ‌Z‌=R=40Ω
ω‌=‌
1000
20
=50‌Hz

Immax‌=‌
230√2
R
=‌
230√2
40
=8.1A
Vc=ImmaxXc‌=‌
230√2
40
×
‌
1
ωC
=2033‌ volt ‌
VL=ImmaxXL‌=‌
230√2
40
×2πvL
=2033‌ volt ‌
(c) VC−VL=0
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