In the given circuit the AC source has $\omega=100 \mathrm{rad} \mathrm{s}^{-1}$. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Alternating Current

In the given circuit the AC source has

ω = 100 rad s^{- 1}

. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?

Answer: (B)

The circuit has an inductor and capacitor in parallel. For the inductor branch:

I_{L} = \frac{V}{R ^{2} + ( ω L ) ^{2}} = \frac{200 2}{5 0 ^{2} + ( 100 \times 0.50 ) ^{2}} = \frac{200 2}{2500 + 2500} = 4

A

For the capacitor branch:

I_{C} = \frac{V}{R ^{2} + ( 1/ ω C ) ^{2}} = \frac{200 2}{10 0 ^{2} + 10 0 ^{2}} = 2

A

The phase difference between the two currents is

90°

, so:

I = I_{L}^{2} + I_{C}^{2} = 16 + 4 = 20 \approx 4.47

A ≈ 4.24 A

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option