Alternating Current

JEE Main level, numerical answer, medium

98 of 190

Question 98 of 190

In the given circuit, the magnitude of VL{V}_{L} and VC{V}_{C} are twice that of VR{V}_{R}. Given that f=50Hzf=50Hz, the inductance of the coil is 1KπmH\dfrac{1}{K\pi }mH. The value of KK is _____ .

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Answer: 0

Since the voltage across the inductor and capacitor is same, the circuit is in resonance. Therefore, the current in the circuit will be,
i=VR=2205=44Ai=\dfrac{V}{R}=\dfrac{220}{5}=44A.
The voltage across the inductor is double that across the resistor. Therefore, VL=2VR=440V{V}_{L}=2{V}_{R}=440V.
We know that VL=iXL=i×2πfL⇒L=44044×100π=110πH=100πmH{V}_{L}=i{X}_{L}=i\times 2\pi fL\Rightarrow L=\dfrac{440}{44\times 100\pi }=\dfrac{1}{10\pi }H=\dfrac{100}{\pi }mH
Therefore, K=1100=0.01≈0K=\dfrac{1}{100}=0.01\approx 0.

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