A capacitor of capacitance 150 . 0 is connected to an alternating...

The formula for maximum current is given by

{I}_\max = \omegaC{V}_{m} . . . ( i )

The given data is

$\omega = 120\pi$

${V}_{m} = 36VC = 150\mu F$

Substituting the values in equation (i)

{I}_\max = \left(120\pi\right) \times \left(150 \times {10}^{- 6}\right) \times 36

$= 2 . 036A$

$\approx 2A$