---
title: "Derivatives at t=0+: 1 H, 1 F, 10 V"
subject: "Electrical Circuit Theory"
chapter: 2
key: "semester_3/electrical_circuit_theory:initial-values-1h-1f-10v"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/initial-values-1h-1f-10v
site: BCT Engineering
---

# Derivatives at t=0+: 1 H, 1 F, 10 V

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2074 Chaitra, 8 marks
- Video: [switch is closed at t=0. Determine 𝒊,  𝒅𝒊/𝒅𝒕,    (𝒅^𝟐 𝒊)/(𝒅𝒕^𝟐 )  ,  (𝒅^𝟑 𝒊)/(𝒅𝒕^𝟑 )    at  𝒕=𝟎^+.](https://www.youtube.com/watch?v=RBezQaCkMyg) — Engg-Course-Made-Easy
- 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-1h-1f-10v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2

## Question

Obtain the value of i1, i2, di1/dt, di2/dt, d²i1/dt² and d²i2/dt² at t = 0⁺, if the switch is closed at t = 0 in the circuit shown in the figure below.
[Figure: 10 V source; switch (closes at t = 0) and a 1 H inductor in the top wire of the left mesh (i1); a 10 Ω resistor in the middle vertical branch; a 10 Ω resistor in the top wire of the right mesh (i2) and a 1 F capacitor on the right]

## Short answer

At t = 0+, i1 = i2 = 0, di1/dt = 10 A/s, di2/dt = 5 A/s, d²i1/dt² = −50 A/s², d²i2/dt² = −25.25 A/s². Inductor current and capacitor voltage start at zero; derivatives from mesh equations.

## Answer

**Assumption:** the inductor and capacitor are initially relaxed. i₁ flows in the left mesh (10 V, switch, L = 1 H, common R = 10 Ω); i₂ flows in the right mesh (common 10 Ω, 10 Ω, C = 1 F).

#### Mesh equations (t > 0)

```text
Mesh 1:  10 = L di₁/dt + 10(i₁ − i₂)
Mesh 2:  10(i₂ − i₁) + 10 i₂ + (1/C)∫i₂ dt = 0
```

#### At t = 0⁺

- Inductor current cannot change: i₁(0⁺) = 0.
- Capacitor voltage cannot change: v_C(0⁺) = 0.

```text
Mesh 2:  20 i₂ − 10 i₁ + 0 = 0  →  i₂(0⁺) = 0
```

#### First derivatives

```text
Mesh 1:  10 = 1·i₁′ + 10(0 − 0)  →  di₁/dt(0⁺) = 10 A/s
Mesh 2 differentiated:  20 i₂′ − 10 i₁′ + i₂/C = 0
        20 i₂′ = 10(10) − 0  →  di₂/dt(0⁺) = 5 A/s
```

#### Second derivatives

```text
Mesh 1 differentiated:  0 = i₁″ + 10(i₁′ − i₂′)
        i₁″ = −10(10 − 5) = −50 A/s²
Mesh 2 differentiated twice:  20 i₂″ − 10 i₁″ + i₂′/C = 0
        20 i₂″ = 10(−50) − 5 = −505  →  i₂″ = −25.25 A/s²
```

| Quantity | Value at 0⁺ |
|---|---|
| i₁ | 0 |
| i₂ | 0 |
| di₁/dt | 10 A/s |
| di₂/dt | 5 A/s |
| d²i₁/dt² | −50 A/s² |
| d²i₂/dt² | −25.25 A/s² |

**Answer: as tabulated above.**

## Check yourself

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

1. What is i1(0+)?
   A. 1 A
   B. 0, since inductor current cannot change instantly
   C. 10 A
   D. 0.5 A
   Answer: B. The inductor is initially relaxed. (Key point: Inductor current)

2. What is di1/dt at 0+?
   A. 0
   B. 100 A/s
   C. 10 A/s = V/L
   D. 1 A/s
   Answer: C. With i1 = i2 = 0 mesh 1 gives 10 = L·di1/dt. (Key point: di1/dt)

3. What is di2/dt at 0+?
   A. 10 A/s
   B. 20 A/s
   C. 0
   D. 5 A/s
   Answer: D. From 20 i2′ = 10 i1′ = 100 you get i2′ = 5. (Key point: di2/dt)

## Close variants set in other exams

- [Derivatives at t=0+: 2 μF, 10 V, 1 H](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Ainitial-values-2uf-10v-10ohm) — Electrical Circuit Theory, chapter 2, asked 2×

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- [Initial values with switch open to closed (10 Ω, 15 Ω, 2 F)](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Ainitial-conditions-switch-open-10-15) — Electrical Circuit Theory, chapter 2

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