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

# Derivatives at t=0+: 2 μF, 10 V, 1 H

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2072 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-2uf-10v-10ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2

## Question

In the circuit shown in figure below, switch K is closed at time t = 0. Find the values of i1, i2, di1/dt, di2/dt, d²i1/dt², d²i2/dt² at t = 0⁺.
[Figure: 10 V source; switch K and a 2 μF capacitor in the top wire of the left mesh (current i1); a 10 Ω resistor in the middle vertical branch; a 10 Ω resistor in the top wire of the right mesh and a 1 H inductor on the right (current i2)]

## Short answer

At 0⁺, capacitor short, inductor open: i1 = 10/10 = 1 A, i2 = 0. Derivatives: di1/dt = −49 990 A/s, di2/dt = 10 A/s; d²i1/dt² ≈ 2.499×10⁹ A/s², d²i2/dt² = −500 100 A/s².

## Answer

**Assumptions:** the capacitor (C = 2 μF) and inductor (L = 1 H) are initially relaxed. Left mesh (current i₁): 10 V, switch K, C, common 10 Ω. Right mesh (current i₂): common 10 Ω, 10 Ω and L.

#### Mesh equations

```text
Mesh 1:  10 = (1/C)∫i₁ dt + 10(i₁ − i₂)
Mesh 2:  10(i₂ − i₁) + 10 i₂ + L di₂/dt = 0
```

#### At t = 0⁺ (v_C = 0, i_L = 0)

```text
i₂(0⁺) = 0
Mesh 1:  10 = 0 + 10(i₁ − 0)  →  i₁(0⁺) = 1 A
```

#### First derivatives

```text
Mesh 2:  10(0 − 1) + 0 + 1·i₂′ = 0  →  di₂/dt(0⁺) = 10 A/s
Mesh 1 differentiated:  0 = i₁/C + 10(i₁′ − i₂′)
        10(i₁′ − 10) = −1/(2×10⁻⁶) = −5×10⁵
        di₁/dt(0⁺) = 10 − 5×10⁴ = −49 990 A/s
```

#### Second derivatives

```text
Mesh 2 differentiated:  10(i₂′ − i₁′) + 10 i₂′ + i₂″ = 0
        i₂″ = −[10(10 + 49 990) + 100] = −500 100 A/s²
Mesh 1 differentiated twice:  0 = i₁′/C + 10(i₁″ − i₂″)
        i₁″ = i₂″ − i₁′/(10C) = −500 100 + 49 990/(2×10⁻⁵)
            = 2.499×10⁹ A/s²   (2 498 999 900)
```

| Quantity | Value at 0⁺ |
|---|---|
| i₁ | 1 A |
| i₂ | 0 |
| di₁/dt | −49 990 A/s |
| di₂/dt | 10 A/s |
| d²i₁/dt² | 2.499×10⁹ A/s² |
| d²i₂/dt² | −5.001×10⁵ 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. 0
   B. 0.5 A
   C. 10 A
   D. 1 A, since the uncharged capacitor acts as a short and i2 = 0
   Answer: D. 10 = 10(i1 − 0). (Key point: i1 at 0+)

2. What is di2/dt at 0+?
   A. 10 A/s
   B. 0
   C. 1 A/s
   D. 100 A/s
   Answer: A. Mesh 2: 10(0 − 1) + i2′ = 0. (Key point: di2/dt)

3. Why is di1/dt large and negative (−49 990 A/s)?
   A. Inductor opposing
   B. i1/C = 5×10^5 must be balanced because the capacitor voltage rises
   C. Because source is negative
   D. Because R is small
   Answer: B. 0 = i1/C + 10(i1′ − i2′). (Key point: di1/dt)

## Close variants set in other exams

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

## Related in Electrical Circuit Theory

- [Laplace: i1, i2 in unenergised two-mesh network](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-two-mesh-unenergised-100v-10ohm) — Electrical Circuit Theory, chapter 4, asked 3×
- [Laplace: i1, i2 with 2 V charged capacitor](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-i1-i2-charged-c-2v) — Electrical Circuit Theory, chapter 4, asked 2×
- [Derivatives at t=0+: 2 H, 100 V, 1 μF](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Ainitial-values-2h-100v-1uf) — Electrical Circuit Theory, chapter 2
- [Initial values with charged capacitor, two meshes](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Ainitial-values-c1-charged-v0) — Electrical Circuit Theory, chapter 2
- [Classical: mesh currents and capacitor voltage, 50 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-mesh-currents-50v-2uf) — Electrical Circuit Theory, chapter 3
- [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

## Related in other subjects

- [Mesh current method with current source](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amesh-current-10-2a-3-50v) — Basic Electrical Engineering, chapter 2
- [Series-parallel circuit with I1, I2, I3 phasors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-ac-v100-z1-z2-z3) — Basic Electrical Engineering, chapter 6, asked 2×
- [Mesh analysis with 3 A source](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Amesh-analysis-3a-source) — Basic Electrical and Electronics Engineering, chapter 1
- [Mesh analysis, current in 5Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Amesh-current-5-ohm-8v-10v-12v) — Basic Electrical Engineering, chapter 2