A capacitor of capacitance $150.0 \mu\mathrm{F}$ is connected to an alternating source of emf given by $\mathrm{E}=36 \sin (120 \pi \mathrm{t})$ V. The maximum value of current in the circuit is approximately equal to :

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Alternating Current

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 :

Answer: (A)

For a purely capacitive circuit the capacitive reactance is

X_{C} = \frac{1}{ω C}

, so the peak current is

I_{0} = \frac{E _{0}}{X _{C}} = E_{0} ω C

. With

E_{0} = 36

ω = 120 π

rad s

^{- 1}

and

C = 150 \times 1 0^{- 6}

I_{0} = 36 \times 120 π \times 150 \times 1 0^{- 6} \approx 2.03

A. The maximum value of the current is approximately

2

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option