A capacitor of capacitance 150.0,μ F is connected to an alternating source of emf given by E =36 si

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7.Alternating Current

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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 $......\,A$

A

$2$

B

$\frac{1}{\sqrt{2}}$

C

$\sqrt{2}$

D

$2 \sqrt{2}$

(JEE MAIN-2023)

Solution

$I_0=\frac{E_0}{x_c}=\frac{E_0}{\frac{1}{\omega_c}}=E_0 \omega_c$

$\Rightarrow I_0=36 \times 120 \pi \times 150 \times 10^{-6}$

$\Rightarrow I_0=2.03$

$\simeq 2\,A$

Standard 12

Physics

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