JEE Main 2021 · 27 July · morning · Single correct · +4 / −1
A capacitor of capacitance C = 1 μF is suddenly connected to a battery of 100 volt through a resistance R = 100 Ω. The time taken for the capacitor to be charged to get 50 V is :
[Take ln 2 = 0.69]
- (a) 1.44 × 10⁻⁴ s
- (b) 3.33 × 10⁻⁴ s
- (c) 0.69 × 10⁻⁴ s
- (d) 0.30 × 10⁻⁴ s
Answer and solution
Answer (c) 0.69 × 10⁻⁴ s
Solution
Physics sir · 3 steps
Half the battery voltage means the capacitor is half charged, and that takes exactly RC ln 2.
- Time constant. For a capacitor charging through a resistor, τ = RC = 100 × 1 × 10⁻⁶ = 1 × 10⁻⁴ s.
V = V₀(1 - e^(-t/RC))
- Put in numbers. With V = 50 V and V₀ = 100 V, we get 1 - e^(-t/RC) = 1/2, so e^(-t/RC) = 1/2. Taking the natural log of both sides gives t = RC ln 2.
- So. t = 1 × 10⁻⁴ × 0.69 = 0.69 × 10⁻⁴ s, which is option (c).
The idea to keep. A charging capacitor reaches half its final voltage after t = RC ln 2. This half-time is the same whatever the battery voltage, so learn it as 0.69 RC.
Answer from the NTA final answer key. Solution by PrepWiser.