The reading of wattmeter and ammeter is 1 kW and 10 A respectively in the three phase circuit given below. What is the value of power factor of the circuit, if the circuit is balanced?

This question was previously asked in
SSC JE EE Previous Year Paper 2 (Held On: 22 Jan 2018 Evening)
View all SSC JE EE Papers >
  1. 0.5
  2. 0.65
  3. 0.74
  4. 0.86

Answer (Detailed Solution Below)

Option 4 : 0.86
Free
Electrical Machine for All AE/JE EE Exams Mock Test
20 Qs. 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

Concept:

3-phase power measurement: for a balanced load 3-phase power measurement one wattmeter required

3-phase Load

Wattmeter method

Balanced load

1 wattmeter

Balanced + unbalanced load

2 wattmeter

Unbalanced load

3 wattmeter

3-phase power measurement of a balanced load by 1 wattmeter method

P = 3 Vph Iph cosϕ 

Vph = phase voltage

Iph = phase current

Calculation:

As per the given circuit diagram, a balanced load wattmeter reading will be

Iph = 10 A, Vph = V/√ 3 = 200/ √3 = 115.47

Power factor 

Important Point:

Power is measured according to Blondel's theorem

Phase/conductor

Wattmeter required

n

n-1

n with neutral wire

n

Latest SSC JE EE Updates

Last updated on Jul 1, 2025

-> SSC JE Electrical 2025 Notification is released on June 30 for the post of Junior Engineer Electrical, Civil & Mechanical.

-> There are a total 1340 No of vacancies have been announced. Categtory wise vacancy distribution will be announced later.

-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.

-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31. 

-> Candidates with a degree/diploma in engineering are eligible for this post.

-> The selection process includes Paper I and Paper II online exams, followed by document verification.

-> Prepare for the exam using SSC JE EE Previous Year Papers.

More Measurement of Power Questions

Hot Links: teen patti 3a teen patti master apk teen patti pro