---
title: "Find Va(0-) and Va(0+) with 2 H inductor"
subject: "Electrical Circuit Theory"
chapter: 2
key: "semester_3/electrical_circuit_theory:va-initial-values-5v-10ohm-2h"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/va-initial-values-5v-10ohm-2h
site: BCT Engineering
---

# Find Va(0-) and Va(0+) with 2 H inductor

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2070 Asar (old course), 8 marks
- Video: [Numerical on initial conditions](https://www.youtube.com/watch?v=VWtFZnLcTzk) — 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/va-initial-values-5v-10ohm-2h
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2

## Question

In the network shown in the figure below, a steady state is reached with the switch open. At t = 0, the switch is closed. Determine the value of Va(0⁻) and Va(0⁺).
[Figure: 5 V source. A 10 Ω resistor runs along the top wire from the source positive terminal to the right top node (bridging). A second row has a 10 Ω resistor from the source positive terminal to node Va, then a 20 Ω resistor from Va to the right node. A 2 H inductor connects the right node to the bottom wire. A switch (closes at t = 0) in series with a 10 Ω resistor connects Va to the bottom wire.]

## Short answer

Va(0⁻) = 3.33 V, Va(0⁺) = 40/21 ≈ 1.905 V.

## Answer

**Assumptions from the figure:** 5 V source. A 10 Ω resistor connects the source + terminal directly to the right node R (the node at the top of the inductor). A second 10 Ω goes from the source + to node Va, then a 20 Ω from Va to R. The 2 H inductor connects R to ground. A switch in series with 10 Ω connects Va to ground and closes at t = 0.

#### At t = 0⁻ (switch open, dc steady state)

The inductor is a short, so V_R = 0. The path source → 10 Ω → Va → 20 Ω → ground carries

```text
I = 5/(10 + 20) = 1/6 A
Va(0⁻) = 20 × (1/6) = 3.333 V
```

Inductor current: through the direct 10 Ω: 5/10 = 0.5 A, plus 1/6 A, so

```text
i_L(0⁻) = 0.5 + 0.1667 = 0.6667 A
```

#### At t = 0⁺ (switch closed)

The inductor still carries 2/3 A (it acts as a 0.667 A current source into ground). Node equations, with Va and V_R unknown:

```text
Node Va:  (5 − Va)/10 = Va/10 + (Va − V_R)/20
          2(5 − Va) = 2Va + (Va − V_R)  →  V_R = 5Va − 10
Node R:   (5 − V_R)/10 + (Va − V_R)/20 = 2/3
          2(5 − V_R) + (Va − V_R) = 13.333
          Va − 3V_R = 3.333
```

Substitute V_R = 5Va − 10: Va − 15Va + 30 = 3.333, so

```text
−14 Va = −26.667
Va(0⁺) = 1.905 V (= 40/21),   V_R(0⁺) = −0.476 V
```

**Answer: Va(0⁻) = 3.333 V and Va(0⁺) = 1.905 V.**

## Check yourself

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

1. Why is the voltage at the inductor's top node zero at t = 0- (switch open)?
   A. In dc steady state the inductor is a short circuit
   B. The capacitor is open
   C. The switch is open so no current flows
   D. The 20 ohm resistor drops all 5 V
   Answer: A. The inductor acts as a short to ground, so the right node is at 0 V. (Key point: Inductor at 0-)

2. How is Va(0-) found?
   A. 5 V times 10/30 = 1.667 V
   B. Current 5/(10+20) = 1/6 A flows through 20 ohm to ground, so Va = 20/6 = 3.333 V
   C. Va = 5 V because the switch is open
   D. Va = 5/2 = 2.5 V
   Answer: B. Path source - 10 - Va - 20 - ground carries 1/6 A; the drop across 20 ohm is 3.333 V. (Key point: Va(0-))

3. What is i_L(0-) and what happens to it at 0+?
   A. 0.5 A; it drops to zero
   B. 0.167 A; it doubles
   C. 0.667 A; it stays 0.667 A and acts as a current source
   D. 0.667 A; it becomes zero because the switch closes
   Answer: C. i_L = 0.5 A (direct 10 ohm) + 0.1667 A = 0.667 A, and it cannot change instantly. (Key point: Inductor current)

4. How is Va(0+) found after the switch closes?
   A. Using the inductor as a short and repeating the t = 0- calculation
   B. Assuming Va stays at 3.333 V
   C. Using the Laplace transform with zero initial current
   D. Node equations at Va and the right node, with the inductor as a 2/3 A current source
   Answer: D. Solving the two node equations gives Va(0+) = 40/21 = 1.905 V; Va jumps because the switch closes. (Key point: Method at 0+)

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