JEE Main 2020 · 8 January · evening · Single correct · +4 / −1
A shown in the figure, a battery of emf ε is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = t c (t c is the time constant of the circuit) is :
- (a) (ε L)/eR²
- (b) (ε L)/R²
- (c) (ε L)/R²( 1 - 1/e )
- (d) (ε R)/eL²
Answer and solution
Answer (a) (ε L)/eR²
Solution
Physics sir · 4 steps
Current is not constant while it grows, so add up charge as the integral of i dt up to one time constant.
- Growing current. After the switch closes, the current in a series LR circuit rises as i = ε/R(1 - e^(-t/τ)), where τ = t_c = L/R.
- Charge from current. Charge is dq = i dt. Integrate from t = 0 to t = τ.
Q = ε/R∫₀^τ(1 - e^(-t/τ))dt = ε/R[τ - τ(1 - 1/e)] = ε/R·τ/e
- Put in tau. With τ = L/R, this gives Q = ε/R·L/eR = (ε L)/eR².
- So. The charge that flows from the battery in one time constant is (ε L)/eR², option (a).
The idea to keep. Whenever current changes with time, charge is the area under the i–t graph, so integrate. In LR growth, the integral up to t = τ comes out as (ε/R)(τ/e).
Answer from the NTA final answer key. Solution by PrepWiser.