---
title: "i(0+) and di/dt(0+) with sinusoidal source"
subject: "Electrical Circuit Theory"
chapter: 2
key: "semester_3/electrical_circuit_theory:initial-values-ac-source-switch-k"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/initial-values-ac-source-switch-k
site: BCT Engineering
---

# i(0+) and di/dt(0+) with sinusoidal source

Electrical Circuit Theory · Chapter 2: Initial Conditions

- Asked in: 2067 Asar (old course), 4 marks
- Video: [Sept-2020-QP-Problem on initial conditions](https://www.youtube.com/watch?v=s_dSHq1xdPE) — 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-values-ac-source-switch-k
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-2

## Question

In the given circuit, switch K is closed at time t = 0. Find i1(0⁺), i2(0⁺), di1(0⁺)/dt and di2(0⁺)/dt.
[Figure: source Vo sin(ωt + φ) with switch K (closes at t = 0), feeding two parallel branches: a resistor R in series with a capacitor C (current i1), and a resistor R in series with an inductor L (current i2)]

## Short answer

i1(0⁺) = V0 sin φ / R, i2(0⁺) = 0, di1/dt(0⁺) = (V0/R)[ω cos φ − sin φ/(RC)], di2/dt(0⁺) = V0 sin φ / L.

## Answer

**Assumptions:** the source is v(t) = V₀ sin(ωt + φ), the capacitor is initially uncharged, and the inductor current is initially zero. Branch 1 is R + C (current i₁), branch 2 is R + L (current i₂). Both branches are connected directly across the source after K closes.

#### Values at t = 0⁺

- Capacitor voltage v_C(0⁺) = 0 (continuous).
- Inductor current i₂(0⁺) = 0 (continuous).
- Source voltage v(0⁺) = V₀ sin φ.

**Branch 1 (R–C):** v = R i₁ + v_C

```text
i₁(0⁺) = [v(0⁺) − v_C(0⁺)]/R = V₀ sin φ / R
```

**Branch 2 (R–L):** v = R i₂ + L di₂/dt

```text
i₂(0⁺) = 0
di₂/dt (0⁺) = [v(0⁺) − R i₂(0⁺)]/L = V₀ sin φ / L
```

#### Derivative of i₁

Differentiate v = R i₁ + v_C, noting dv_C/dt = i₁/C:

```text
dv/dt = R di₁/dt + i₁/C
di₁/dt = (1/R)[dv/dt − i₁/C]
```

with dv/dt (0⁺) = ωV₀ cos φ:

```text
di₁/dt (0⁺) = ωV₀ cos φ / R − V₀ sin φ /(R²C)
```

**Answer:**
- i₁(0⁺) = V₀ sin φ / R
- i₂(0⁺) = 0
- di₁/dt (0⁺) = (V₀/R)[ω cos φ − sin φ/(RC)]
- di₂/dt (0⁺) = V₀ sin φ / L

## Check yourself

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

1. Which values are used as known quantities at t = 0+ in this problem?
   A. v_C(0+) = 0 and i2(0+) = 0
   B. i1(0+) = 0 and v_C(0+) = V0
   C. i2(0+) = V0 sin(phi)/R and v_C(0+) = 0
   D. v_L(0+) = 0 and i1(0+) = 0
   Answer: A. Capacitor voltage and inductor current are continuous, so both start at zero. (Key point: Continuity at 0+)

2. How is i1(0+) obtained?
   A. v(0+)/(R + L)
   B. [v(0+) - v_C(0+)]/R = V0 sin(phi)/R
   C. V0/R regardless of phi
   D. 0, since the capacitor is uncharged
   Answer: B. With v_C(0+) = 0 the capacitor acts as a short, so the full source voltage V0 sin(phi) appears across R. (Key point: i1(0+))

3. Why is di2/dt(0+) = V0 sin(phi)/L?
   A. Because the voltage across R is V0 sin(phi) at 0+
   B. Because di2/dt = 0 when i2 = 0
   C. i2(0+) = 0, so the whole source voltage appears across L
   D. Because the inductor acts as a short at 0+
   Answer: C. From v = R i2 + L di2/dt with i2(0+) = 0, L di2/dt equals v(0+). (Key point: di2/dt(0+))

4. Which relation is differentiated to find di1/dt(0+), and what extra term appears?
   A. dv/dt = R di1/dt, ignoring the capacitor
   B. dv/dt = L di1/dt + i1/C
   C. dv/dt = R di1/dt + C i1
   D. dv/dt = R di1/dt + i1/C, using dv_C/dt = i1/C and dv/dt(0+) = wV0 cos(phi)
   Answer: D. Since dv_C/dt = i1/C, di1/dt = (1/R)[dv/dt - i1/C], giving (V0/R)[w cos(phi) - sin(phi)/(RC)]. (Key point: di1/dt(0+))

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