Alternating Current

JEE Main level, numerical answer, medium

79 of 190

Question 79 of 190

An oscillating LCLCcircuit consists of a 75mH75mH inductor and a 1.2µF1.2µF capacitor. If the maximum charge to the capacitor is 2.7µC2.7µC. The maximum current in the circuit will be _______mA.

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Answer: 9

When the capacitor is fully discharged, the current is maximum and all the energy resides in the inductor. Hence, the energy of the capacitor is equal to the energy of the inductor.
Li2m2=Q2m2C...(i)\dfrac{L{{i}^{2}}_{m}}{2}=\dfrac{{{Q}^{2}}_{m}}{2C}...(i)
where im{i}_{m} is the maximum current and Qm{Q}_{m} is the maximum charge.
The given values are
Qm=2.7×10−6CL=75×10−3HC=1.2×10−6F{Q}_{m}=2.7\times {10}^{-6}C L=75\times {10}^{-3}H C=1.2\times {10}^{-6}F
The maximum current is found from equation (i)
im=1LCQm⇒im=2.7×10−675×10−3×1.2×10−6=9mA{i}_{m}=\sqrt{\dfrac{1}{LC}}{Q}_{m} \Rightarrow {i}_{m}=\dfrac{2.7\times {10}^{-6}}{\sqrt{75\times {10}^{-3}\times 1.2\times {10}^{-6}}}=9mA

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