---
title: "Classical: inductor current, 2 V charged 1 F capacitor"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-inductor-current-charged-capacitor-half"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-inductor-current-charged-capacitor-half
site: BCT Engineering
---

# Classical: inductor current, 2 V charged 1 F capacitor

Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution

- Asked in: 2070 Asar (old course), 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/classical-inductor-current-charged-capacitor-half
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

Using classical method find the expression for current through the inductor for t > 0 in the circuit shown in figure.
[Figure: single loop with a switch (t = 0) and a ½ H inductor in the top-left wire, then a 10 Ω resistor in parallel with a 1 F capacitor that has an initial voltage of 2 V (+ on the left plate)]

## Short answer

Using the homogeneous second-order ODE of the source-free parallel RLC, with ω_d=1.4133 rad/s and initial conditions i_L(0)=0, v(0)=2 V, the current is i_L(t)=2.83e^(−0.05t) sin(1.4133t) A.

## Answer

**Circuit reading.** After the switch closes, L, R = 10 Ω and C = 1 F (v_C(0) = 2 V) are connected across the same two nodes, so it is a source-free parallel RLC circuit. Initially the switch is open, so i_L(0) = 0. Take v as the voltage across the capacitor (+ on the left plate) and i_L as the inductor current driven by it.

#### Equations
```text
C dv/dt + v/R + i_L = 0
L di_L/dt = v
→ d²i_L/dt² + (1/RC) di_L/dt + (1/LC) i_L = 0
→ i_L'' + 0.1 i_L' + 2 i_L = 0
```

#### Roots
m² + 0.1m + 2 = 0 → m = −0.05 ± j√(2 − 0.0025) = −0.05 ± j1.4133

α = 1/(2RC) = 0.05 and ω₀ = 1/√(LC) = √2 = 1.414: underdamped (α < ω₀), ω_d = 1.4133 rad/s.

#### Constants
- i_L(0) = 0 → i_L = e^(−0.05t) (A cos ω_d t + B sin ω_d t) with A = 0
- i_L′(0) = v(0)/L = 2/0.5 = 4 A/s. With A = 0: B ω_d = 4 → B = 4/1.4133 = 2.830

**Answer: i_L(t) = 2.83 e^(−0.05t) sin(1.4133t) A, t > 0** (opposite capacitor polarity only reverses the sign).

The current is a slowly decaying oscillation (period 2π/ω_d = 4.45 s). Check: at t = 4 s the formula gives −1.365 A, the same as a numerical integration of the circuit equations; i_L(0) = 0 and i_L′(0) = 2.83 × 1.4133 = 4 A/s = v(0)/L ✓.

## Check yourself

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

1. Which values are compared to decide the type of response, and what is the result?
   A. alpha = 1/(2RC) = 0.05 and w0 = 1/sqrt(LC) = 1.414; alpha < w0 so underdamped
   B. alpha = R/2L = 10 and w0 = 1.414; overdamped
   C. alpha = 0.05 and w0 = 0.05; critically damped
   D. alpha = 1/(2RC) = 0.05 and w0 = 1.414; alpha < w0 so overdamped
   Answer: A. For a parallel RLC alpha = 1/(2RC); alpha < w0 gives complex roots, so underdamped. (Key point: Damping check)

2. Which two conditions fix the constants of iL(t)?
   A. iL(0) = 2 A and iL'(0) = 0
   B. iL(0) = 0 and iL'(0) = v(0)/L = 4 A/s
   C. iL(0) = 0 and iL'(0) = 0
   D. iL(0) = 2 A and iL'(0) = 4 A/s
   Answer: B. The switch was open so iL(0) = 0, and L diL/dt = v with v(0) = 2 V gives 4 A/s. (Key point: Initial conditions)

3. Which is the result for iL(t)?
   A. 2.83 e^(-0.05t) cos(1.4133t) A
   B. 2 e^(-0.05t) sin(1.414t) A
   C. 2.83 e^(-0.05t) sin(1.4133t) A
   D. 2.83 e^(-1.4133t) sin(0.05t) A
   Answer: C. A = 0 from iL(0) = 0, and B w_d = 4 gives B = 4/1.4133 = 2.83. (Key point: Final form)

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