- 2073 Shrawan · 5 marks
Rise and fall of current in RC circuit
Short answer
Charging: current i = (ε/R)e^(−t/RC) decays from initial maximum ε/R. Discharging: i = (V₀/R)e^(−t/RC) flows opposite. Both curves are exponential with time constant τ = RC, falling to 1/e after τ and nearly zero after 5τ.
Derive the relation for rise and fall of current in charging and discharging of a capacitor through a resistor. Plot graphs between current and time and explain the figures.
Circuit: battery ε, resistor R, capacitor C in series. The time constant is τ = RC.
+---- R ----+
| |
ε =| === C
| |
+---- S ----+
Charging (switch to battery)
- Loop rule at time t: ε = iR + q/C, with i = dq/dt.
- dq/(Cε − q) = dt/(RC). Integrating from q = 0 at t = 0: ln[(Cε − q)/Cε] = −t/RC
- q = Cε (1 − e^(−t/RC))
- Current: i = dq/dt = (ε/R) e^(−t/RC) = i₀ e^(−t/τ), with i₀ = ε/R
The current starts at its maximum i₀ and decays to zero as the capacitor charges.
Discharging (battery removed, plates joined through R)
- Loop rule: the capacitor drives the current through R, so q/C − iR = 0, with i = −dq/dt (the charge decreases).
- dq/q = −dt/RC. With q = q₀ at t = 0: q = q₀ e^(−t/RC)
- Current: i = (q₀/RC) e^(−t/RC) = i₀ e^(−t/τ), with i₀ = q₀/RC = V₀/R, in the direction opposite to the charging current.
Graphs
i i
i0|\ i0|\
| \ charging | \ discharging
| `. | `.
| ` . . | ` . .
|_________ t |_________ t
0 tau 2tau 0 tau 2tau
Charging current i = i₀e^(−t/τ) (flows in the charging direction); discharging current has the same exponential shape but flows in the reverse direction.
Explanation
- At t = 0 the uncharged capacitor acts like a short circuit and current is ε/R.
- After t = τ = RC, the current falls to 36.8% (1/e) of i₀.
- After about 5τ, the current is practically zero.
- A larger R or C increases τ and slows the process.
✓ Answer checked by Claude Opus 5.5 (high effort)
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