---
title: "Initial values with switch open to closed (10 Ω, 15 Ω, 2 F)"
subject: "Electrical Circuit Theory"
chapter: 2
key: "semester_3/electrical_circuit_theory:initial-conditions-switch-open-10-15"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/initial-conditions-switch-open-10-15
site: BCT Engineering
---

# Initial values with switch open to closed (10 Ω, 15 Ω, 2 F)

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2073 Shrawan
- Video: [For the network shown below switch is closed at t=0.  Determine 𝒊,    𝒅𝒊/𝒅𝒕](https://www.youtube.com/watch?v=py-HwjhEwq8) — 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-conditions-switch-open-10-15
- 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: 10 V battery (+ at top). The battery positive terminal connects through a 10 Ω resistor to node Va, and through a long 10 Ω resistor along the top wire to the right node Vb. Va connects to Vb through a 15 Ω resistor. A 2 F capacitor connects Vb to the bottom wire. From Va, a switch (closes at t = 0) in series with a 5 Ω resistor connects to the bottom wire.]

## Short answer

va(0⁻) = 10 V. At t = 0⁺, capacitor voltage remains 10 V; nodal analysis gives va(0⁺) = 50/11 ≈ 4.545 V.

## Answer

**Assumptions from the figure:** 10 V battery; 10 Ω from the + terminal to node Va; another 10 Ω along the top from the + terminal to Vb; 15 Ω between Va and Vb; C = 2 F from Vb to ground; the switch in series with 5 Ω connects Va to ground and closes at t = 0.

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

The capacitor is an open circuit in dc steady state, so no current flows in any resistor. Hence there is no voltage drop and

```text
Va(0⁻) = Vb(0⁻) = 10 V      v_C(0⁻) = 10 V
```

#### At t = 0⁺

The capacitor voltage cannot change suddenly: Vb(0⁺) = 10 V (the capacitor acts as a 10 V source). The switch is now closed, so the 5 Ω resistor connects Va to ground. KCL at node Va:

```text
(10 − Va)/10 + (Vb − Va)/15 = Va/5
(10 − Va)/10 + (10 − Va)/15 = Va/5
(10 − Va)(1/10 + 1/15) = Va/5
(10 − Va)/6 = Va/5
50 − 5Va = 6Va
Va(0⁺) = 50/11 = 4.545 V
```

**Answer: Va(0⁻) = 10 V and Va(0⁺) = 4.545 V.** Note that Va changes suddenly because it is not the capacitor voltage.

## Check yourself

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

1. With the switch open at steady state, what is Va(0−)?
   A. 0 V
   B. 5 V
   C. 4.545 V
   D. 10 V, because the capacitor blocks dc so no current flows
   Answer: D. No current means no voltage drops, so Va = Vb = 10 V. (Key point: Steady state)

2. Which quantity is continuous at t = 0?
   A. Capacitor voltage: Vb(0+) = 10 V
   B. Va
   C. Resistor currents
   D. Va and Vb
   Answer: A. The capacitor voltage cannot change instantly. (Key point: Continuity)

3. What is Va(0+)?
   A. 10 V
   B. 50/11 = 4.545 V
   C. 5 V
   D. 0 V
   Answer: B. KCL: (10 − Va)/6 = Va/5. (Key point: Va after closing)

4. Why can Va change suddenly?
   A. Capacitor voltage jumps
   B. Inductor present
   C. It is not a capacitor voltage
   D. Source changes
   Answer: C. Only capacitor voltage and inductor current are continuous. (Key point: Sudden change)

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