---
title: "Rise and fall of current in RC circuit"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:rc-circuit-charge-discharge-current"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/rc-circuit-charge-discharge-current
site: BCT Engineering
---

# Rise and fall of current in RC circuit

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2073 Shrawan, 5 marks
- Video: [Capacitor charge time calculation - time constants](https://www.youtube.com/watch?v=FDmHVRC6pQk) — The Engineering Mindset
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/rc-circuit-charge-discharge-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

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.

## 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.

## Answer

Circuit: battery ε, resistor R, capacitor C in series. The time constant is τ = RC.

```text
        +---- R ----+
        |           |
      ε =|         ===  C
        |           |
        +---- S ----+
```

#### Charging (switch to battery)
1. Loop rule at time t: ε = iR + q/C, with i = dq/dt.
2. dq/(Cε − q) = dt/(RC). Integrating from q = 0 at t = 0: ln[(Cε − q)/Cε] = −t/RC
3. q = Cε (1 − e^(−t/RC))
4. 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)
1. Loop rule: the capacitor drives the current through R, so q/C − iR = 0, with i = −dq/dt (the charge decreases).
2. dq/q = −dt/RC. With q = q₀ at t = 0: q = q₀ e^(−t/RC)
3. 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
```text
  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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The RC time constant is
   A. R/C
   B. RC
   C. C/R
   D. 1/RC
   Answer: B. τ = RC. (Key point: Time constant)

2. In charging, the current is
   A. i0(1 − e^(−t/τ))
   B. i0 e^(−t/τ)
   C. i0 t/τ
   D. zero
   Answer: B. i = (ε/R)e^(−t/RC), starting at maximum. (Key point: Charging current)

3. At t = 0 the uncharged capacitor acts like
   A. An open circuit
   B. A short circuit
   C. A resistor R
   D. An inductor
   Answer: B. Current is ε/R. (Key point: Initial)

4. The discharge current compared with charging
   A. Same shape, reverse direction
   B. Increasing
   C. Same direction
   D. Constant
   Answer: A. i = i0 e^(−t/τ) flows opposite. (Key point: Discharge)

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