---
title: "Classical: Vc and ic with 10 A source"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-capacitor-ic-vc-10a-20ohm"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-capacitor-ic-vc-10a-20ohm
site: BCT Engineering
---

# Classical: Vc and ic with 10 A source

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

- Asked in: 2068 Baisakh, 8 marks
- Video: [How to Solve Any RC Circuit Question](https://www.youtube.com/watch?v=iZJ8xqIQOoA) — ElectricalMath
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-capacitor-ic-vc-10a-20ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

At t = 0, switch is closed in the circuit of figure below. Find the Vc(t) and ic(t) using classical method.
[Figure: 10 A current source (arrow up) in parallel with a 20 Ω resistor; a switch (t = 0) connects this to a second 20 Ω resistor in parallel with a 10 μF capacitor; ic(t) is the current into the capacitor and Vc(t) its voltage]

## Short answer

After closing, the equivalent resistance seen by the capacitor is 10 Ohms, giving a time constant of 100 microseconds. The final capacitor voltage is 100 V. With initial voltage zero, Vc(t) = 100(1 - e^(-10000 t)) V and ic(t) = 10 e^(-10000 t) A.

## Answer

**Assumptions:** the 10 A current source is in parallel with a 20 Ω resistor. At t = 0 the switch connects this to a second 20 Ω resistor in parallel with C = 10 μF. The capacitor is initially uncharged.

#### Differential equation

After the switch closes, the 10 A source feeds the parallel combination of the two 20 Ω resistors and the capacitor. KCL at the top node:

```text
10 = v/20 + v/20 + C dv/dt
10⁻⁵ dv/dt + 0.1 v = 10
dv/dt + 10⁴ v = 10⁶
```

#### Solution

- Particular (final) value: v_p = 10⁶/10⁴ = 100 V (= 10 A × (20 ∥ 20) = 10 × 10).
- Complementary: s = −10⁴, so v_c = K e^(−10⁴t); time constant τ = 10 Ω × 10 μF = 100 μs.
- v_C(0⁺) = 0, so K = −100.

```text
V_c(t) = 100 (1 − e^(−10⁴ t)) V,   t ≥ 0
```

#### Capacitor current

```text
i_c(t) = C dv_C/dt = 10×10⁻⁶ × 100 × 10⁴ e^(−10⁴t) = 10 e^(−10⁴ t) A
```

Check: at t = 0⁺ the capacitor is a short, so the whole 10 A flows through it ✓. At t → ∞ i_c → 0 and the 10 A divides equally between the two 20 Ω resistors (5 A each, 100 V across them ✓).

**Answer: V_c(t) = 100(1 − e^(−10⁴t)) V and i_c(t) = 10e^(−10⁴t) A.**

## Check yourself

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

1. What is the differential equation after closing?
   A. dv/dt + 10^4 v = 10^6
   B. dv/dt + 5x10^3 v = 10^6
   C. dv/dt + 10^4 v = 10^5
   D. dv/dt + 10^2 v = 10^4
   Answer: A. 10 = v/20 + v/20 + C dv/dt, with C = 10 uF. (Key point: Equation)

2. What is v_C(infinity)?
   A. 200 V
   B. 100 V = 10 A x (20 || 20)
   C. 10 V
   D. 50 V
   Answer: B. At dc the capacitor is open, so 10 A flows through 10 ohm. (Key point: Final value)

3. What is the time constant?
   A. 20 ohm x 10 uF = 200 us
   B. 1 s
   C. 10 ohm x 10 uF = 100 us
   D. 1 ms
   Answer: C. The capacitor sees two 20 ohm in parallel. (Key point: Time constant)

4. What is i_c(0+) and why?
   A. 0 A
   B. 5 A
   C. 100 A
   D. 10 A, because the uncharged capacitor is a short
   Answer: D. i_c = C dv/dt = 10 e^(-10^4 t) A, starting at 10 A. (Key point: Capacitor current)

## Related in Electrical Circuit Theory

- [Classical: capacitor voltage, 10 A source, 1 F](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-voltage-capacitor-10a-1f-half) — Electrical Circuit Theory, chapter 3
- [Classical: capacitor voltage, 10 A source, 0.5 F](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-voltage-10a-half-ohm) — Electrical Circuit Theory, chapter 3
- [Classical: capacitor voltage after switch opens, 4 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-voltage-switch-opened-4v) — Electrical Circuit Theory, chapter 3
- [Classical: capacitor voltage, switch opened, 15 V and 60 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-switch-opened-15v-60v) — Electrical Circuit Theory, chapter 3, asked 2×
- [Classical: inductor current and capacitor voltage, 100 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-inductor-current-capacitor-voltage-100v) — Electrical Circuit Theory, chapter 3
- [Classical: capacitor voltage with 200 V source](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-capacitor-voltage-200v-100ohm) — Electrical Circuit Theory, chapter 3

## Related in other subjects

- [State equations of current-source RLC circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-current-source-1h-1f-2h) — Control Systems, chapter 8
- [Resistor and capacitor with half voltage](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aadjustable-resistor-25uf-0-8a) — Basic Electrical Engineering, chapter 6