JEE Main 2021 · 26 August · evening · Single correct · +4 / −1
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?
- (a) 5.9 A
- (b) 3.16 A
- (c) 0.94 A
- (d) 6 A
Answer and solution
Answer (b) 3.16 A
Solution
Physics sir · 3 steps
The two branch currents are 90° apart, so add them like perpendicular vectors, not like plain numbers.
- Find the reactances. X_C = 1/(ω C) = 1/(100 × 100 × 10⁻⁶) = 100 Ω and X_L = ω L = 100 × 0.5 = 50 Ω. The branches are in parallel, so each has the full 200 V across it.
Branch Impedance Current Phase with voltage C + 100 Ω √(100² + 100²) = 100√2 Ω 200/100√2 = √2 A leads by 45° (X_C = R) L + 50 Ω √(50² + 50²) = 50√2 Ω 200/50√2 = 2√2 A lags by 45° (X_L = R) - Combine as phasors. One current is 45° ahead of the voltage and the other is 45° behind, so they are 90° apart. Their resultant is the hypotenuse of a right triangle.
I = √(I₁² + I₂²) = √((√2)² + (2√2)²) = √(2 + 8) = √10 A
- So. I = √10 ≈ 3.16 A, which is option (b).
The idea to keep. In parallel AC branches, the voltage is common and the currents add as phasors. Find each branch current and its phase angle, then add them as vectors.
Answer from the NTA final answer key. Solution by PrepWiser.