---
title: "RMS, phase difference and power from v and i"
subject: "Basic Electrical Engineering"
chapter: 5
key: "semester_1/basic_electrical_engineering:v-500sin-400t-i-5sin-rms-phase-difference"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/v-500sin-400t-i-5sin-rms-phase-difference
site: BCT Engineering
---

# RMS, phase difference and power from v and i

Basic Electrical Engineering · Chapter 5: Alternating Quantities

- Asked in: 2068 Shrawan, 8 marks
- 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-500sin-400t-i-5sin-rms-phase-difference
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-5

## Question

The impressed voltage and the current flowing through a circuit are: v = 500 sin (400t + π/3) and i = 5 sin (400t – π/6). Determine the i) rms and average value of v and i, ii) Radian through which it's vector has gone when t = 0.001 sec, iii) Phase difference, and iv) Power consumed in the circuit.

## Short answer

RMS: Vrms=353.55 V, Irms=3.54 A; average half-cycle: Vav=318.3 V, Iav=3.18 A. At t=0.001 s, ωt=0.4 rad. Phase difference = 90° (voltage leads). Power consumed = 0 W (purely reactive).

## Answer

Given v = 500 sin(400t + π/3) V and i = 5 sin(400t − π/6) A. So Vm = 500 V, Im = 5 A, ω = 400 rad/s.

#### i) RMS and average values
- V_rms = Vm/√2 = 500/1.414 = **353.55 V**
- I_rms = 5/1.414 = **3.54 A**
- V_av (half cycle) = 2Vm/π = 2×500/π = **318.3 V**
- I_av (half cycle) = 2×5/π = **3.18 A**

(Over a full cycle the average of a sine wave is zero.)

#### ii) Radian passed by the vector at t = 0.001 s
Angle swept = ωt = 400 × 0.001 = **0.4 rad** (= 22.9°).

The actual phase angles at that instant are: voltage 0.4 + π/3 = 1.447 rad (82.9°), current 0.4 − π/6 = −0.124 rad (−7.1°).

#### iii) Phase difference
φ = π/3 − (−π/6) = π/2 rad = **90°**, with the voltage leading the current.

#### iv) Power consumed
P = V_rms I_rms cos φ = 353.55 × 3.536 × cos 90° = **0 W**

The circuit is purely reactive (an ideal inductor) so no real power is consumed. The reactive power is Q = 353.55 × 3.536 = 1250 VAR.

## Check yourself

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

1. What are the rms values of v = 500 sin(400t + π/3) and i = 5 sin(400t − π/6)?
   A. 250 V and 2.5 A
   B. 318.3 V and 3.18 A
   C. 353.55 V and 3.54 A
   D. 500 V and 5 A
   Answer: C. rms = peak/√2, giving 353.55 V and 3.54 A. (Key point: RMS of sine)

2. What does 2Vm/π = 318.3 V represent?
   A. Average over a half cycle
   B. Average over a full cycle
   C. Form factor
   D. RMS value
   Answer: A. Over a full cycle the average of a sine is zero, so the average is taken over a half cycle: 2Vm/π. (Key point: Average of half cycle)

3. What angle does the vector sweep at t = 0.001 s with ω = 400 rad/s?
   A. 1.047 rad
   B. 400 rad
   C. 0.4 rad
   D. 0.001 rad
   Answer: C. Angle swept = ωt = 400 × 0.001 = 0.4 rad (22.9°), independent of the initial phase. (Key point: Angle swept)

4. Voltage phase π/3 and current phase −π/6 give what phase difference and power?
   A. 90°, power = 0 W
   B. 90°, power = VI
   C. 30°, large power
   D. 60°, power = VI/2
   Answer: A. φ = π/3 − (−π/6) = π/2 = 90°, so P = VI cos 90° = 0 and the circuit is purely reactive. (Key point: Phase difference and power)

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