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:

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TestHub · Accepted Answer

The formula for maximum current is given by
$I_{ma x} = ω C V_{m} ... (i)$
The given data is
$ω = 120 π$
$V_{m} = 36 V C = 150 μF$
Substituting the values in equation (i)
$I_{ma x} = (120 π) \times (150 \times 1 0^{- 6}) \times 36$
$= 2.036 A$
$\approx 2 A$

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