---
title: "Initial currents i1, i2, i3 at t=0+ with C1, C2, L"
subject: "Electrical Circuit Theory"
chapter: 2
key: "semester_3/electrical_circuit_theory:initial-values-three-loop-currents-v"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/initial-values-three-loop-currents-v
site: BCT Engineering
---

# Initial currents i1, i2, i3 at t=0+ with C1, C2, L

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2068 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/initial-values-three-loop-currents-v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2

## Question

For the circuit shown in the following figure, find the current i1, i2, i3 at t = 0⁺.
[Figure: source V in series with a resistor (value not labelled) forming the outer loop (current i1). Inside, the left branch has R1 (upper) in series with L (lower); the right branch has C1 (upper) in series with C2 (lower). A switch (t = 0) connects the junction of R1 and L to the junction of C1 and C2. Loop i2 is the upper mesh (R1, C1) and loop i3 is the lower mesh (L, C2).]

## Short answer

i1 = i2 = i3 = V/R. Initially relaxed inductor is open, capacitors are shorts; all nodes equal potential, so V appears only across the series resistor R.

## Answer

**Assumptions (figure values not given, so the answer is in symbols):** the series resistor in the source branch is R (unlabelled); all energy-storing elements are initially relaxed (i_L = 0, v_C1 = v_C2 = 0) and the switch closes at t = 0 with V applied. Mesh currents are clockwise: i₁ in the source loop (V, R, R₁, L), i₂ in the upper loop (R₁, C₁) and i₃ in the lower loop (L, C₂). The current in R₁ is i₁ − i₂ and in L is i₁ − i₃.

#### Equivalent circuit at t = 0⁺

- L with zero current: open circuit, so i_L = i₁ − i₃ = 0.
- C₁ and C₂ uncharged: short circuits, so the top wire, the capacitor junction and the bottom wire are at the same potential.

#### Solution

1. The switch connects the R₁-L junction to the C₁-C₂ junction. With both capacitors shorted, the top wire, the bottom wire and both junctions are at the same potential: V_top = V_bottom.
2. So the whole source voltage V appears across the series resistor R in the source branch:

```text
V = R·i₁   →   i₁(0⁺) = V/R
```

3. The voltage across R₁ is zero (its two ends are joined through the shorted C₁ and the closed switch), so its current is zero: i₁ − i₂ = 0, hence i₂ = i₁.
4. The inductor is open: i₁ − i₃ = 0, hence i₃ = i₁.

**Answer: i₁(0⁺) = i₂(0⁺) = i₃(0⁺) = V/R.** There is no current in R₁ or L; the whole current V/R flows through the series resistor and the short-circuited capacitors.

## Check yourself

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

1. How do relaxed capacitors behave at 0+?
   A. Opens
   B. As shorts
   C. Sources
   D. Resistors
   Answer: B. v_C = 0. (Key point: Capacitors)

2. How does a relaxed inductor behave at 0+?
   A. Short
   B. Source
   C. As an open circuit, so i1 = i3
   D. Capacitor
   Answer: C. i_L = i1 − i3 = 0. (Key point: Inductor)

3. What is i1(0+)?
   A. V/(R + R1)
   B. 0
   C. V/R1
   D. V/R
   Answer: D. Everything else is at the same potential. (Key point: i1)

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