A capacitor is discharging through a resistor R. Consider in time t₁, the energy stored in the capacitor reduces to half of its initial value and in time t₂, the charge stored reduces to one eighth of its initial value. The ratio t₁/t₂ will be

Solution

Physics sir · 4 steps

Energy goes as q², so it decays twice as fast as charge. Write both as exponentials and compare.

1. Charge decay. While discharging, q = q₀ e^(-t/τ) with τ = RC. Energy is U = q²/2C, so U = U₀ e^(-2t/τ). The factor 2 in the exponent comes from squaring q.
2. Energy halves. Set e^(-2t₁/τ) = 1/2. Then 2t₁/τ = ln 2, so t₁ = (τ ln 2)/2.
3. Charge to one eighth. Set e^(-t₂/τ) = 1/8. Then t₂/τ = ln 8 = 3 ln 2, so t₂ = 3τ ln 2.

   t₁/t₂ = (τ ln 2 / 2)/(3τ ln 2) = 1/6

4. So. The ratio t₁/t₂ is 1/6, which is option (d).

The idea to keep. In RC discharge, charge, current and voltage all decay with time constant τ, but energy goes as q² and decays with τ/2. Check which quantity the question asks about.

Answer from the NTA final answer key. Solution by PrepWiser.