---
title: "Two-wattmeter method star load"
subject: "Basic Electrical and Electronics Engineering"
chapter: 1
key: "semester_1/basic_electrical_and_electronics_engineering:two-wattmeter-star-load"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/two-wattmeter-star-load
site: BCT Engineering
---

# Two-wattmeter method star load

Basic Electrical and Electronics Engineering · Chapter 1: Fundamentals of Electrical and Electronics Circuits

- Asked in: 2081 Baishakh (new course), 6 marks
- Video: [Measuring Three Phase Power - 2 wattmeter Method](https://www.youtube.com/watch?v=FA9D1FuV5yA) — StudyYaar.com
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/two-wattmeter-star-load
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1

## Question

The two-wattmeter method produces wattmeter readings $W_1 = 2100$ W and $W_2 = 1550$ W when connected to a star-connected load. If the line voltage is 400 V, calculate: (i) the per-phase average power, (ii) the per phase reactive power, (iii) the power factor, and (iv) the phase impedance.

## Short answer

Total power P = 3650 W, total Q = √3(2100-1550) = 952.63 VAR. Per-phase: P_ph = 1216.67 W, Q_ph = 317.54 VAR. Power factor cosφ = 0.9676 lagging. Phase voltage = 230.94 V, line current = 5.445 A, phase impedance = 42.41 Ω (41.04 + j10.71 Ω).

## Answer

**Given:** $W_1=2100$ W, $W_2=1550$ W, $V_L=400$ V, star connection.

**(i) Per-phase average power.** Total power is the sum of the readings:

$$
P = W_1+W_2 = 3650\ \text{W}, \qquad P_{ph}=\frac{3650}{3}=1216.67\ \text{W}
$$

**(ii) Per-phase reactive power.** Total reactive power is $\sqrt3\,(W_1-W_2)$:

$$
Q=\sqrt3(2100-1550)=952.63\ \text{VAR}, \qquad Q_{ph}=\frac{952.63}{3}=317.54\ \text{VAR}
$$

**(iii) Power factor.**

$$
\tan\phi=\frac{Q}{P}=\frac{952.63}{3650}=0.261\ \Rightarrow\ \phi=14.63^\circ
$$

$$
\cos\phi=0.9676\ \text{(lagging, for an inductive load)}
$$

**(iv) Phase impedance.** Line current from $P=\sqrt3 V_L I_L\cos\phi$:

$$
I_L=\frac{3650}{\sqrt3\times400\times0.9676}=5.445\ \text{A}=I_{ph}
$$

$$
V_{ph}=\frac{400}{\sqrt3}=230.94\ \text{V}, \qquad Z_{ph}=\frac{230.94}{5.445}=42.41\ \Omega
$$

Then $R_{ph}=Z\cos\phi=41.04\ \Omega$ and $X_{ph}=Z\sin\phi=10.71\ \Omega$, so $Z_{ph}=41.04+j10.71\ \Omega$.

**Answer: (i) 1216.67 W, (ii) 317.54 VAR, (iii) 0.9676 lagging, (iv) 42.41 Ω ($41.04+j10.71\ \Omega$).**

## Check yourself

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

1. How is the per-phase average power obtained from W1 = 2100 W and W2 = 1550 W?
   A. (W1 - W2)/3 = 183.3 W
   B. sqrt(3)(W1 + W2)/3 = 2107 W
   C. (W1 + W2)/3 = 1216.67 W
   D. W1 + W2 = 3650 W per phase
   Answer: C. Total power is the sum of the readings (3650 W); a balanced load shares it equally over three phases. (Key point: Total and phase power)

2. Which expression gives the total reactive power?
   A. Q = sqrt(3)(W1 - W2)
   B. Q = W1 - W2
   C. Q = sqrt(3)(W1 + W2)
   D. Q = (W1 + W2)/sqrt(3)
   Answer: A. Q = sqrt(3)(W1 - W2) = 952.63 VAR total, i.e. 317.54 VAR per phase. (Key point: Reactive power)

3. How is the power factor found from the readings?
   A. cos(phi) = W2/W1
   B. tan(phi) = P/Q, then cos(phi)
   C. tan(phi) = Q/P = 0.261, so phi = 14.63° and pf = 0.9676
   D. pf = (W1 - W2)/(W1 + W2)
   Answer: C. tan(phi) = Q/P = 952.63/3650 = 0.261, giving phi = 14.63° and cos(phi) = 0.9676 lagging. (Key point: Power factor)

4. For the star load, how is Zph found?
   A. Vph = 400 V and Iph = 5.445 A, so Z = 73.5 Ω
   B. Vph = 400/sqrt(3) = 230.94 V and Iph = IL = 5.445 A, so Z = 42.41 Ω
   C. Vph = 400/sqrt(3) and Iph = IL/sqrt(3), so Z = 73.5 Ω
   D. Vph = 400 V and Iph = IL x sqrt(3), so Z = 24.5 Ω
   Answer: B. In star, Vph = VL/sqrt(3) and Iph = IL, so Z = 230.94/5.445 = 42.41 Ω. (Key point: Phase impedance)

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