Alternating Current

JEE Main level, single correct, medium

91 of 190

Question 91 of 190

A series LCR circuit has L=0.01H,R=10ΩL=0.01H,R=10\Omega and C=1μFC=1\mu F and it is connected to ac voltage of amplitude (Vm)50V\left({V}_{m}\right)50V. At frequency 6060% lower than resonant frequency, the amplitude of current will be approximately

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Answer: (C) 238mA238mA

For an LCR circuit. the resonant angular frequency is given by, ω0=1LC=104rads−1{\omega }_{0}=\dfrac{1}{\sqrt{LC}}={10}^{4}rad{s}^{-1}
The given frequency is 6060% lower than resonant frequency. Therefore,
ω′=0.4×104=4000rads−1{\omega }^{'}=0.4\times {10}^{4}=4000rad{s}^{-1}
Reactance of the capacitor at given frequency,
XC=(ω′C)−1=250Ω{X}_{C}={\left({\omega }^{'}C\right)}^{-1}=250\Omega.
Reactance of the inductor at given frequency,
XL=(ω′L)=40Ω{X}_{L}=\left({\omega }^{'}L\right)=40\Omega
Now the amplitude of the current in the given circuit will be,
i0=V0R2+(XC′−XL′)2=50102+(250−40)2=238mA{i}_{0}=\dfrac{{V}_{0}}{\sqrt{{R}^{2}+{\left({X}_{C}^{'}-{X}_{L}^{'}\right)}^{2}}}=\dfrac{50}{\sqrt{{10}^{2}+{\left(250-40\right)}^{2}}}=238mA

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