(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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(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: $I_0 = 36 \times 120 \pi \times 150 \times 10^{-6} \, A$.$I_0 = 36 \times 120 \times 3.1416 \times 150 \times 10^{-6} \, A$.$I_0 = 648000 \times 3.1416 \times 10^{-6} \, A \approx 2.035 \, A$.Therefore,the maximum current is approximately $2 \, A$.

Class 12 Physics Alternating Current Only Inductor, Only Capacitor and Only Resistor Circuit

Medium

$A$ 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 $...... \, A$.

JEE MAIN 2023 All JEE MAIN

A
$2$
B
$\frac{1}{\sqrt{2}}$
C
$\sqrt{2}$
D
$2 \sqrt{2}$

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Only Inductor, Only Capacitor and Only Resistor Circuit

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$A$ 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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