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PhysicsEMI/ACAC current and voltageMedium2 minPYQ_2023

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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:

Options:

Answer:

Solution:

The formula for maximum current is given by
$I_{m a x} = ω C V_{m} . . . (i)$
The given data is
$ω = 120 π$
$V_{m} = 36 V C = 150 μF$
Substituting the values in equation (i)
$I_{\max} = (120 π) \times (150 \times 10^{- 6}) \times 36$
$= 2.036 A$
$\approx 2 A$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:AC current and voltage

⏱ 2mℹ️ Source: PYQ_2023

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