---
title: "Classical: loop currents with 1 H and 1 μF"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-loop-currents-2ohm-3ohm-1h"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-loop-currents-2ohm-3ohm-1h
site: BCT Engineering
---

# Classical: loop currents with 1 H and 1 μF

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

- Asked in: 2074 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/classical-loop-currents-2ohm-3ohm-1h
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

Find the time expression for loop currents for t > 0 in the given circuit using classical method.
[Figure: 10 V source with switch (t = 0) and a 2 Ω resistor in the top wire of the left mesh (i1); a middle shared branch with a 3 Ω resistor in series with a 1 H inductor; a 1 Ω resistor in the top wire of the right mesh (i2) and a 1 μF capacitor on the right]

## Short answer

Mesh analysis yields overdamped response with roots −5 and −3.33×10⁵ s⁻¹. Steady-state is i1=2 A, i2=0; initial both 3.33 A. i1(t)=2−2e^(−5t)+3.33e^(−3.33×10⁵t) A, i2(t)≈3.33e^(−3.33×10⁵t) A.

## Answer

**Assumptions:** the network is initially relaxed. Left mesh (i₁): 10 V, switch and 2 Ω. Shared branch: 3 Ω in series with 1 H. Right mesh (i₂): 1 Ω and C = 1 μF.

#### Mesh equations (t > 0)

```text
Mesh 1:  10 = 2 i₁ + 3(i₁ − i₂) + d(i₁ − i₂)/dt
Mesh 2:  0 = i₂ + 10⁶∫i₂ dt + 3(i₂ − i₁) + d(i₂ − i₁)/dt
```

Using the operator s = d/dt (and 1/s for ∫):

```text
(5 + s) i₁ − (3 + s) i₂ = 10
−(3 + s) i₁ + (4 + s + 10⁶/s) i₂ = 0
```

#### Characteristic equation

```text
Δ = (5 + s)(4 + s + 10⁶/s) − (3 + s)² = 0
  → 3s² + (10⁶ + 11)s + 5×10⁶ = 0
  → s² + 333 337 s + 1.667×10⁶ = 0
s₁ = −5.0,   s₂ = −3.333×10⁵ (1/s)
```

Both roots are real, so the response is overdamped.

#### Steady state (t → ∞)

The capacitor is open (i₂ = 0) and the inductor is short:

```text
i₁(∞) = 10/(2 + 3) = 2 A,   i₂(∞) = 0
```

#### Initial values at t = 0⁺

The inductor is open and the capacitor is a short, so the 2 Ω and 1 Ω are in series with the 10 V source: i₁(0⁺) = i₂(0⁺) = 10/3 = 3.333 A.

#### Constants

Writing i₁ = 2 + A₁e^(s₁t) + B₁e^(s₂t) and i₂ = A₂e^(s₁t) + B₂e^(s₂t), the first mesh equation gives A₂/A₁ = (5 + s)/(3 + s) ≈ 0.00001 (for s₁) and 0.999994 (for s₂). With i₁(0) = i₂(0) = 3.333:

```text
A₁ = −2.0000,  B₁ = 3.3334
A₂ = −2×10⁻⁵,  B₂ = 3.3334
```

#### Result

```text
i₁(t) = 2 − 2 e^(−5t) + 3.333 e^(−3.33×10⁵ t) A
i₂(t) ≈ 3.333 e^(−3.33×10⁵ t) A   (plus a negligible −2×10⁻⁵ e^(−5t))
```

Because τ₂ is only 3 μs, i₂ is a very short pulse that charges the capacitor. i₁ jumps to 3.33 A at t = 0⁺, drops to almost zero within a few microseconds (the inductor current i₁ − i₂ starts from zero), and then rises to its final value of 2 A with the slow time constant 1/5 = 0.2 s.

**Answer: i₁(t) = 2 − 2e^(−5t) + 3.33e^(−3.33×10⁵t) A, i₂(t) ≈ 3.33e^(−3.33×10⁵t) A.**

## Check yourself

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

1. What are i1(0+) and i2(0+)?
   A. Both 3.333 A, with the inductor open and the capacitor short
   B. Both 0
   C. 2 A and 0
   D. 10/2 = 5 A and 0
   Answer: A. The inductor is open and the capacitor is a short, leaving 2 ohm + 1 ohm in series with 10 V. (Key point: Initial values)

2. What are i1(infinity) and i2(infinity)?
   A. 3.333 A and 0
   B. 2 A and 0
   C. 10 A and 10 A
   D. 2 A and 2 A
   Answer: B. The capacitor is open (i2 = 0) and the inductor is short, so i1 = 10/(2+3) = 2 A. (Key point: Final values)

3. What do the characteristic roots tell us?
   A. Complex roots, so underdamped
   B. Equal roots, so critically damped
   C. Both roots are real (-5 and -3.33x10^5), so the response is overdamped
   D. One positive root, so unstable
   Answer: C. Two real distinct negative roots give a sum of decaying exponentials. (Key point: Damping)

4. What do the two time constants imply for i1?
   A. i1 rises slowly from 0 to 2 A
   B. i1 oscillates around 2 A
   C. i1 stays constant at 3.33 A
   D. i1 jumps to 3.33 A at once, then falls slowly to 2 A with tau = 0.2 s
   Answer: A. The 3 microsecond term vanishes quickly; the 0.2 s term governs the slow decay. (Key point: Time constants)

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