---
title: "Loop currents at t=0+ with two 1 F capacitors"
subject: "Electrical Circuit Theory"
chapter: 2
key: "semester_3/electrical_circuit_theory:loop-currents-at-0plus-two-1f"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/loop-currents-at-0plus-two-1f
site: BCT Engineering
---

# Loop currents at t=0+ with two 1 F capacitors

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2071 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/loop-currents-at-0plus-two-1f
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2

## Question

In the circuit shown in the following figure, find the loop currents i1, i2, i3 at t = 0⁺.
[Figure: 100 V source in series with a 5 Ω resistor forming the outer loop (current i1). Inside, on the left, two 1 F capacitors stacked vertically between the top and bottom wires. A switch (t = 0) connects the junction of the two capacitors to the left end of a 10 Ω resistor; the right end of this 10 Ω resistor is a node joined to the top wire through a 1 H inductor and to the bottom wire through a 10 Ω resistor. Loop i2 is the upper loop (top 1 F capacitor, switch, 10 Ω, 1 H) and loop i3 is the lower loop (bottom 1 F capacitor, switch, 10 Ω, 10 Ω).]

## Short answer

i1 = i2 = i3 = 20 A. Uncharged capacitors act as shorts, unfluxed inductor as open; all points have same potential so full 100 V across 5 Ω.

## Answer

**Assumptions (figure description):** the capacitors are uncharged and the inductor carries no current before t = 0. Take all mesh currents clockwise: i₁ in the source loop (100 V, 5 Ω), i₂ in the upper loop (top 1 F, switch, 10 Ω, 1 H) and i₃ in the lower loop (bottom 1 F, switch, 10 Ω, 10 Ω). Call the top wire T, the bottom wire B, the capacitor junction M and the node at the right end of the switched 10 Ω resistor X.

#### Equivalent circuit at t = 0⁺

- Uncharged capacitors act as short circuits: T, M and B are at the same potential.
- The inductor with zero initial current acts as an open circuit: i_L = 0.

#### Using the conditions

Current through the inductor (T to X): i_L = i₂ − i₁ = 0, so **i₂ = i₁**.

The two 10 Ω resistors are in series between M and B (through X). With M and B at the same potential, no voltage appears across the series pair, so

```text
10(i₃ − i₂) + 10(i₃ − i₁) = 0  →  i₃ = (i₁ + i₂)/2 = i₁
```

So all three mesh currents are equal, and no current flows in the two 10 Ω resistors or the inductor.

The 100 V source now sees only the 5 Ω resistor, since the capacitors short the rest of the circuit:

```text
i₁(0⁺) = 100/5 = 20 A
```

**Answer: i₁ = i₂ = i₃ = 20 A at t = 0⁺** (the current circulates through the 5 Ω resistor and the two shorted capacitors).

## Check yourself

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

1. How does an uncharged capacitor behave at 0+?
   A. Open circuit
   B. Source
   C. As a short circuit
   D. Inductor
   Answer: C. v_C = 0 so it has zero voltage. (Key point: Uncharged capacitors)

2. How does an inductor with zero initial current behave at 0+?
   A. Short circuit
   B. Source
   C. Resistor
   D. As an open circuit
   Answer: D. i_L = 0 persists. (Key point: Inductor)

3. What are the loop currents?
   A. i1 = i2 = i3 = 20 A
   B. i1 = 20 A, others 0
   C. 10 A each
   D. 6.67 A each
   Answer: A. Source sees only the 5 Ω: 100/5 = 20 A. (Key point: Result)

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