A capacitor of capacitance 150.0,mu F is conn | vedclass

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(A) The given $emf$ is $E = E_0 \sin(\omega t)$,where $E_0 = 36 \, V$ and $\omega = 120 \pi \, rad/s$.The capacitive reactance is $X_C = \frac{1}{\ome

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$2$

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(A) The given $emf$ is $E = E_0 \sin(\omega t)$,where $E_0 = 36 \, V$ and $\omega = 120 \pi \, rad/s$.The capacitive reactance is $X_C = \frac{1}{\omega C}$.The maximum current $I_0$ is given by $I_0 = \frac{E_0}{X_C} = E_0 \omega C$.Substituting the values: $I_0 = 36 	imes 120 \pi 	imes 150 	imes 10^{-6} \, A$.$I_0 = 36 	imes 120 	imes 3.1416 	imes 150 	imes 10^{-6} \, A$.$I_0 = 648000 	imes 3.1416 	imes 10^{-6} \, A \approx 2.035 \, A$.Therefore,the maximum current is approximately $2 \, A$.

$A$ capacitor of capacitance $150.0\,\mu F$ is connected to an alternating source of $emf$ given by $E = 36 \sin(120 \pi t) \, V$. The maximum value of current in the circuit is approximately equal to $...... \, A$.

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