A capacitor of capacitance is connected to an alternating source of emf given by The maximum value of current in the circuit is approximately equal to [2023]

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Solution

Q.
A capacitor of capacitance $150.0 μ F$ is connected to an alternating source of emf given by $E = 36 \sin (120 π t) V .$ The maximum value of current in the circuit is approximately equal to [2023]

1 $2 A$

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

3 $\sqrt{2} A$

4 $2 \sqrt{2} A$

Ans.
(1)

$Maximum current,$

$I_{0} = \frac{E_{0}}{X_{c}} = \frac{E_{0}}{\frac{1}{ω C}} = E_{0} ω C$

$\Rightarrow I_{0} = 36 \times 120 π \times 150 \times 10^{- 6} = 2 A$

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