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NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 11
Total: 26
Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H. C = 80 µF, R = 40 Ω

(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Solution:  
(a) Condition for resonance is when applied frequency matches with natural frequency.
Resonant frequency ωr =
1
√LC
 =
1
√5(80×10−6)
 = 50 rad s−1
(b) At resonance, impedance Z = R
as XL = XC
So, Z = 40 Ω
rms current Iv =
Ev
R
 =
230
40
 = 5.75 A, Amplitude of current , I0 = Iv√2 = 8.13 A
(c) Potential drop across ‘L’
VL = IvXL = 5.75 × (ωL)
VL = 5.75 × 50 × 5 = 1437.5 V

Potential drop across ‘C’
VC = Iv×XC = 5.75 ×
1
ωC
= 5.75 ×
1
50×80×10−6
 = 1437.5 V
Potential drop across R
VR = Iv R = 5.75 × 40 = 230 V
As VL–VC = 0, So Ev=VR = 230 V.
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