Alternating Current

JEE Main level, single correct, medium

105 of 190

Question 105 of 190

A series LCR circuit driven by 300V300V at a frequency of 50Hz50Hz contains a resistance R=3kΩR=3k\Omega, an inductor of inductive reactance XL=250πΩ{X}_{L}=250\pi \Omega and an unknown capacitor. The value of capacitance to maximise the average power should be:
(takeπ2=10)\left(\text{take}{\pi }^{2}=10\right)

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Answer: (B) 4μF4\mu F

Angular frequency, ω=2πf=2π×50=100πrads−1\omega =2\pi f=2\pi \times 50=100\pi rad{s}^{-1}
At resonance,
XC=XL⇒1ωC=ωL⇒1ωC=250π⇒C=1100π×250πF=4μF{X}_{C}={X}_{L} \Rightarrow \dfrac{1}{\omega C}=\omega L \Rightarrow \dfrac{1}{\omega C}=250\pi \Rightarrow C=\dfrac{1}{100\pi \times 250\pi }F=4\mu F

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