Alternating Current

JEE Main level, single correct, medium

17 of 190

Question 17 of 190

A capacitor of capacitance 150.0μF150.0\mu F is connected to an alternating source of emf given by E=36sin⁡(120πt)VE=36\sin \left(120\pi t\right)V. The maximum value of current in the circuit is approximately equal to:

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Answer: (A) 2A2A

The formula for maximum current is given by
Imax=ωCVm...(i){I}_{max}=\omega C{V}_{m}...(i)
The given data is
ω=120π\omega =120\pi
Vm=36VC=150μF{V}_{m}=36V C=150\mu F
Substituting the values in equation (i)
Imax⁡=(120π)×(150×10−6)×36{I}_{\max }=\left(120\pi \right)\times \left(150\times {10}^{-6}\right)\times 36
=2.036A=2.036A
≈2A\approx 2A

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