JEE Main 2021 · 25 February · morning · Single correct · +4 / −1
The angular frequency of alternating current in a L-C-R circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.
- (a) 0.8 H and 150 μF
- (b) 0.8 H and 250 μF
- (c) 1.33 H and 150 μF
- (d) 1.33 H and 250 μF
Answer and solution
Answer (b) 0.8 H and 250 μF
Solution
Physics sir · 4 steps
The switch is open, so R' carries nothing: get the current from the 60 Ω resistor first.
- Ignore R′. The switch in the R' branch is open, so no current flows through R' = 40 Ω. That leaves C, R and L in series, and they all carry the same current.
- Find the current. The resistor has 15 V across it, so I = 15/60 = 0.25 A.
X_L = V_L/I = 20/0.25 = 80 Ω, X_C = V_C/I = 10/0.25 = 40 Ω
- Find L and C. With ω = 100 rad s⁻¹, L = X_L/ω = 80/100 = 0.8 H. Also C = 1/(ω X_C) = 1/(100 × 40) = 2.5 × 10⁻⁴ F = 250 μF.
- So. The coil has 0.8 H and the condenser has 250 μF, which is option (b).
The idea to keep. In a series AC circuit, one known resistor and its voltage give you the common current. Then each reactance is its own voltage divided by that current.
Answer from the NTA final answer key. Solution by PrepWiser.