A series LCR circuit has L=0.01H,R=10Ω and C=1μ F and it is connected to ac voltage of amplitude (V_m)50V. At frequency 60% lower than resonant…
JEE Main 2022 — Physics Electromagnetism
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A series LCR circuit has L=0.01H,R=10Ω and C=1μF and it is connected to ac voltage of amplitude (Vm)50V. At frequency 60 lower than resonant frequency, the amplitude of current will be approximately
Official previous-year question
Held on 27 Jul 2022 · Verified 6 Jul 2026.
Options
A
466mA
B
312mA
C
238mA
D
196mA
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Solution
For an LCR circuit. the resonant angular frequency is given by, ω0=LC1=104rads−1
The given frequency is 60 lower than resonant frequency. Therefore,
ω′=0.4×104=4000rads−1
Reactance of the capacitor at given frequency,
XC=(ω′C)−1=250Ω.
Reactance of the inductor at given frequency,
XL=(ω′L)=40Ω
Now the amplitude of the current in the given circuit will be,
i0=R2+(XC′−XL′)2V0=102+(250−40)250=238mA
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