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Question

# In the given figure, a series L−C−R circuits is connected to a variable frequency source of 230 V. The impedance and amplitude of the current at the resonating frequency will be

A
20Ω and 4.2 A
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B
30Ω and 6.9 A
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C
25Ω and 5.8 A
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D
40Ω and 5.75 A
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Solution

## The correct option is D 40Ω and 5.75 AInductance L=5.0H Capacitance C=80×10−6F Resistance R=40Ω Effective voltage Ev=230V Now, the resonance angular frequency ωr=1√LC ωr=1√5×80×10−6 ωr=1√400×10−6 ωr=120×10−3 ωr=50rad/s Now, impedance of the circuit Z=√R2+(ωL−1ωC)2 At resonance ωL=1ωC So, Z=√R2 Z=R=40Ω Now, amplitude of current at resonating frequency Peak value of current I0=E0Z I0=√2×EvZ I0=√2×23040 I0=8.13A Now, Rms value of current Iv=I0√2 Iv=8.13√2 Iv=5.75A Hence, the impedance and amplitude of the current at the resonating frequency will be 40 Ω and 5.75 A

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