Alternating Current

JEE Main level, single correct, medium

121 of 190

Question 121 of 190

The angular frequency of alternating current in a L-C-R circuit is 100rads−1100rad{s}^{-1}. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

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Answer: (D) 0.8H0.8H and 250μF250\mu F

Current through 60Ω60\Omega resistance =1560=14A=\dfrac{15}{60}=\dfrac{1}{4}A
thus capacitor current =14A=\dfrac{1}{4}A
∵VC=IXC∵{V}_{C}=I{X}_{C}
10=14×1ωC10=\dfrac{1}{4}\times \dfrac{1}{\omega C}
∴C=140ω=14000=250μF∴C=\dfrac{1}{40\omega }=\dfrac{1}{4000}=250\mu F
Now,
current through 40Ω40\Omega resistance =2040=12A=\dfrac{20}{40}=\dfrac{1}{2}A
thus current through inductor =12−14=14A=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}A
VL=IXL=14×ωL{V}_{L}=I{X}_{L}=\dfrac{1}{4}\times \omega L
20=14×100×L20=\dfrac{1}{4}\times 100\times L
⇒L=0.8H\Rightarrow L=0.8H

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