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NCERT Class XII Chapter
Alternating Current
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Question : 11 of 26
Marks: +1, -0
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\omega_r = 1LC\frac{1}{\sqrt{LC}} = 15(80×10−6)\frac{1}{\sqrt{5(80 \times 10^{-6})}} = 50 rad s−1s^{-1}
(b) At resonance, impedance Z = R
as XLX_L = XCX_C
So, Z = 40 Ω
rms current IvI_v = EvR\frac{E_v}{R} = 23040\frac{230}{40} = 5.75 A, Amplitude of current , I0I_0 = Iv2I_v \sqrt{2} = 8.13 A
(c) Potential drop across ‘L’
VLV_L = IvXLI_v X_L = 5.75 × (ωL)
VLV_L = 5.75 × 50 × 5 = 1437.5 V
Potential drop across ‘C’
VCV_C = Iv×XCI_v \times X_C = 5.75 × 1ωC\frac{1}{\omega C}
= 5.75 × 150×80×10−6\frac{1}{50 \times 80 \times 10^{-6}} = 1437.5 V
Potential drop across R
VRV_R = IvI_v R = 5.75 × 40 = 230 V
As VL−VCV_L - V_C = 0, So Ev=VRE_v = V_R = 230 V.
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