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  • A capacitor of capacitance 150.0 F is connected to an alternating source of emf given by E = 36 sin  <img

A capacitor of capacitance 150.0 \muF is connected to an alternating source of emf given by E = 36 sin  (120 \pi \mathrm{t})  V.

The maximum value of current in the circuit is approximatively equal to :

Option: 1

\sqrt{2A}


Option: 2

2\sqrt{2A}


Option: 3

\frac{1}{\sqrt{2}}A


Option: 4

2A


Answers (1)

best_answer

Given alternating AC source E = 36 sin (120 \pi \mathrm{t})  v & capacitor  C = 150 \muF

using Q = CV 
we can write Q = \left(\mathrm{CE}_0 \sin \omega \mathrm{t}\right)  

Current i    =\frac{\mathrm{dQ}}{\mathrm{dt}}=\left(\mathrm{CE}_0 \omega \cos \omega \mathrm{t}\right)                         

max. value of current {i }_{0}=\mathrm{CE}_0 \omega

i_{0} \begin{aligned} & \text { or } \mathrm{i}_0=150 \times 10^{-6} \times 36 \times 120 \pi \\ & =2.03 \mathrm{~A} \end{aligned}

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