---
title: "Power factor and parameters from v and i"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:v-10sin500t-60-i-6sin500t-10"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/v-10sin500t-60-i-6sin500t-10
site: BCT Engineering
---

# Power factor and parameters from v and i

Basic Electrical Engineering · Chapter 6: Single-phase AC Circuits

- Asked in: 2069 Asar, 6 marks
- Video: [RLC Series Circuit: Impedance, Power Factor, Active Power, & Reactive Power Analysis with AC Supply](https://www.youtube.com/watch?v=QpRtsF6RlBQ) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/v-10sin500t-60-i-6sin500t-10
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

When a voltage v= 10sin(500t–60°)v is applied to a series ac circuit, the current is i=6sin(500t–10°). Find (i) Power factor (ii) Circuit parameters (iii) Apparent, active and reactive power (iv) Also indicate whether the circuit is capacitive or inductive and why?

## Short answer

Power factor 0.643 leading. Circuit parameters: R=1.07 Ω, C=1566 μF (series RC). Apparent power 30 VA, active power 19.3 W, reactive power 23.0 VAR leading. The circuit is capacitive because current leads voltage.

## Answer

Given v = 10 sin(500t − 60°) V and i = 6 sin(500t − 10°) A. So Vm = 10 V, Im = 6 A, ω = 500 rad/s.

#### (i) Power factor
Phase angle φ = (−60°) − (−10°) = −50°, so the current leads the voltage by 50°.

**Power factor = cos 50° = 0.643 leading**

#### (ii) Circuit parameters
- |Z| = Vm/Im = 10/6 = 1.667 Ω
- R = Z cos φ = 1.667 × 0.643 = **1.07 Ω**
- X = Z sin φ = 1.667 × 0.766 = 1.277 Ω (capacitive)
- X_C = 1/(ωC) ⟹ C = 1/(500 × 1.277) = 1.566 × 10⁻³ F = **1566 μF**

The circuit is a series R–C: R = 1.07 Ω, C ≈ 1566 μF.

#### (iii) Powers
V = 10/√2 = 7.07 V, I = 6/√2 = 4.24 A

- Apparent: S = VI = **30 VA**
- Active: P = VI cos φ = 30 × 0.643 = **19.3 W**
- Reactive: Q = VI sin φ = 30 × 0.766 = **23.0 VAR** (leading)

#### (iv) Nature of the circuit
The circuit is **capacitive**, because the current leads the voltage (the current wave reaches its peak 50° earlier than the voltage wave). Inductive circuits give lagging current.

## Check yourself

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

1. With v at −60° and i at −10°, the current relative to voltage:
   A. In phase
   B. Leads by 50°
   C. Lags by 50°
   D. Leads by 70°
   Answer: B. Current phase is greater (less negative), so it leads by 50°; pf = cos 50° = 0.643 leading. (Key point: Phase angle)

2. How are R and X obtained?
   A. X = Z cos φ
   B. R = Z sin φ
   C. Z = Vm/Im = 1.667 Ω; R = Z cos φ = 1.07 Ω, X = Z sin φ = 1.277 Ω
   D. R = Vm/Im
   Answer: C. Use the impedance triangle. (Key point: Impedance parameters)

3. Why is C = 1/(ωX) used?
   A. R is capacitive
   B. It is inductive
   C. ω is zero
   D. The reactance is capacitive, X_C = 1/(ωC), giving about 1566 μF
   Answer: D. X = 1.277 Ω is capacitive, so C = 1/(500 × 1.277). (Key point: Find C)

4. Why is the circuit capacitive?
   A. Current lags voltage
   B. Current leads voltage
   C. Resistance is small
   D. Voltage is larger
   Answer: B. Leading current indicates capacitive behaviour; lagging would be inductive. (Key point: Nature of circuit)

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