---
title: "Time for battery current to reach 500 mA"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:battery-current-reaches-500ma-classical"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/battery-current-reaches-500ma-classical
site: BCT Engineering
---

# Time for battery current to reach 500 mA

Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution

- Asked in: 2074 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/battery-current-reaches-500ma-classical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

In the circuit shown in the figure below, if the switch is closed at t = 0, find the time when the current from the battery reaches 500 mA. Use classical method.
[Figure: 10 V battery with a switch (t = 0) and a 70 Ω resistor in series; a 50 Ω resistor and a 100 μF capacitor connected in parallel after the 70 Ω]

## Short answer

The current starts at 142.9 mA and decays to 83.3 mA; with a 10 V source it never reaches 500 mA. For the likely intended value 100 mA, t = τ ln[(I0−I∞)/(I−I∞)] = 3.71 ms.

## Answer

**Assumptions:** 10 V battery and switch in series with 70 Ω; the 50 Ω resistor and the 100 μF capacitor are in parallel after the 70 Ω; the capacitor is initially uncharged; the switch closes at t = 0.

#### Battery current as a function of time

Let v be the voltage across the parallel combination: i = (10 − v)/70 and

```text
C dv/dt + v/50 = (10 − v)/70
```

Equivalent: v rises from 0 to the final value with time constant τ = (70 ∥ 50) C.

```text
R_eq = 70 × 50 /(120) = 29.17 Ω
τ = R_eq C = 29.17 × 100×10⁻⁶ = 2.917 ms
v(∞) = 10 × 50/120 = 4.167 V
v(t) = 4.167 (1 − e^(−t/τ))
```

Battery current:

```text
i(t) = (10 − v)/70 = 83.3 mA + (142.9 − 83.3) mA · e^(−t/τ)
     = 0.0833 + 0.0595 e^(−343 t) A
```

#### Check of the 500 mA value

The battery current starts at 10/70 = 142.9 mA (capacitor acts as a short) and decays to 10/120 = 83.3 mA. It lies between these two limits at all times, so **with the data as given (10 V) the current can never reach 500 mA.** The value is probably misprinted in the question (for example, 100 mA with 10 V, or 500 mA with a 50 V battery), so the general formula is shown:

```text
t = τ ln[(I₀ − I∞)/(I − I∞)]
```

For I = 100 mA with the 10 V data: t = 2.917 ms × ln[(142.86 − 83.33)/(100 − 83.33)] = 2.917 ms × ln(3.571) = 3.71 ms.

(For 500 mA with a 50 V battery the ratio is identical, so t = 3.71 ms again.)

**Answer: i(t) = 83.3 mA + 59.5 mA e^(−343t); the current is always below 142.9 mA, so 500 mA is never reached with 10 V. The corresponding time for the likely intended 100 mA is t ≈ 3.71 ms.**

## Check yourself

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

1. What is the battery current at t = 0+?
   A. 10/70 = 142.9 mA, because the uncharged capacitor acts as a short
   B. 10/120 = 83.3 mA
   C. 500 mA
   D. 0, because the capacitor is uncharged
   Answer: A. The capacitor short bypasses the 50 ohm, leaving only the 70 ohm resistor. (Key point: Initial battery current)

2. What is the time constant of the circuit?
   A. (70 + 50) x C = 12 ms
   B. (70 || 50) x C = 2.917 ms
   C. 50 x C = 5 ms
   D. 70 x C = 7 ms
   Answer: B. The capacitor sees the two resistors in parallel: 29.17 ohm x 100 uF. (Key point: Time constant)

3. What is the final battery current?
   A. 10/70 = 142.9 mA
   B. 10/50 = 200 mA
   C. 10/120 = 83.3 mA
   D. 0 mA
   Answer: C. At dc the capacitor is open, so the current flows through 70 + 50 ohm. (Key point: Final current)

4. Why can 500 mA never be reached with 10 V?
   A. The time constant is too large
   B. The capacitor stores too little energy
   C. The current grows only slowly toward 500 mA
   D. The current always lies between 142.9 mA and 83.3 mA
   Answer: D. The current decays from 142.9 mA to 83.3 mA, so it never exceeds 142.9 mA. (Key point: Conclusion)

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