---
title: "Initial values with charged capacitor, two meshes"
subject: "Electrical Circuit Theory"
chapter: 2
key: "semester_3/electrical_circuit_theory:initial-values-c1-charged-v0"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/initial-values-c1-charged-v0
site: BCT Engineering
---

# Initial values with charged capacitor, two meshes

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2075 Asoj, 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-c1-charged-v0
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2

## Question

In the given network, the capacitor C1 is charged to voltage V0 and switch K is closed at t = 0. When R1 = 2 MΩ, V0 = 1000 V, R2 = 1 MΩ, C1 = 10 μF and C2 = 20 μF, solve for i1, i2, di2/dt, d²i2/dt² at t = 0⁺.
[Figure: capacitor C1 (charged to V0) at the left; switch K in the top wire of the left mesh (closes at t = 0); resistor R2 in the middle vertical branch; resistor R1 in the top wire of the right mesh; capacitor C2 on the right; loop current i1 in the left mesh and i2 in the right mesh]

## Short answer

At t = 0+, i1 = 1.5 mA, i2 = 0.5 mA, di2/dt = −8.75×10⁻⁵ A/s, d²i2/dt² = 1.406×10⁻⁵ A/s². Initial capacitor voltages are V0 on C1 and zero on C2; derivatives obtained by differentiating mesh equations.

## Answer

**Assumptions:** C₂ is initially uncharged; C₁ is charged to V₀ (voltage polarity such that it drives i₁ in the direction shown); i₁ and i₂ are the mesh currents of the left and right loops, R₂ is the common branch.

#### Mesh equations for t > 0

```text
Mesh 1:  V₀ − (1/C₁)∫i₁ dt = R₂(i₁ − i₂)
Mesh 2:  R₂(i₂ − i₁) + R₁ i₂ + (1/C₂)∫i₂ dt = 0
```

#### At t = 0⁺ (v_C2 = 0, so ∫i₂ = 0; C₁ still has V₀)

```text
V₀ = R₂(i₁ − i₂)          (R₁ + R₂) i₂ = R₂ i₁
V₀ = R₂ i₁ · R₁/(R₁ + R₂)
i₁(0⁺) = V₀(R₁ + R₂)/(R₁R₂) = 1000 × 3×10⁶ /(2×10¹²) = 1.5 mA
i₂(0⁺) = R₂ i₁/(R₁ + R₂) = 1.5/3 = 0.5 mA
```

#### First derivatives

Differentiate Mesh 1: −i₁/C₁ = R₂(i₁′ − i₂′)

```text
i₁′ − i₂′ = −i₁/(R₂C₁) = −1.5×10⁻³/(10⁶ × 10⁻⁵) = −1.5×10⁻⁴ A/s
```

Differentiate Mesh 2: R₂(i₂′ − i₁′) + R₁ i₂′ + i₂/C₂ = 0

```text
R₁ i₂′ = R₂(i₁′ − i₂′) − i₂/C₂ = 10⁶(−1.5×10⁻⁴) − 0.5×10⁻³/(2×10⁻⁵)
       = −150 − 25 = −175
di₂/dt (0⁺) = −175/(2×10⁶) = −8.75×10⁻⁵ A/s
di₁/dt (0⁺) = −8.75×10⁻⁵ − 1.5×10⁻⁴ = −2.375×10⁻⁴ A/s
```

#### Second derivatives

Differentiate again:

```text
i₁″ − i₂″ = −i₁′/(R₂C₁) = 2.375×10⁻⁵ A/s²
R₁ i₂″ = R₂(i₁″ − i₂″) − i₂′/C₂
       = 10⁶(2.375×10⁻⁵) + 8.75×10⁻⁵/(2×10⁻⁵) = 23.75 + 4.375 = 28.125
d²i₂/dt² (0⁺) = 28.125/(2×10⁶) = 1.406×10⁻⁵ A/s²
d²i₁/dt² (0⁺) = 3.781×10⁻⁵ A/s²
```

**Answer: i₁(0⁺) = 1.5 mA, i₂(0⁺) = 0.5 mA, di₂/dt = −8.75×10⁻⁵ A/s, d²i₂/dt² = 1.406×10⁻⁵ A/s².**

## Check yourself

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

1. What are the capacitor voltages at t = 0+?
   A. Both zero
   B. Both V0
   C. v_C1 = V0 and v_C2 = 0
   D. v_C1 = 0 and v_C2 = V0
   Answer: C. Capacitor voltages cannot change instantly. (Key point: Initial state)

2. How are di/dt values at 0+ obtained?
   A. Differentiate the mesh equations and substitute the 0+ values
   B. Take the mesh equations at t = ∞
   C. Integrate the mesh equations
   D. Assume they are zero
   Answer: A. Each differentiation turns integrals into currents already known. (Key point: Derivatives)

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