A 230 V, 3-phase voltage is applied to a balance delta connected 3-phase load of phase impedance (15 + j 20) Ω. What is the power consumed per phase?

This question was previously asked in
SSC JE Electrical 11 Oct 2023 Shift 3 Official Paper-I
View all SSC JE EE Papers >
  1. 1269.6 W
  2. 2198.3 W
  3. 1161.6 W
  4. 3807.6 W

Answer (Detailed Solution Below)

Option 1 : 1269.6 W
Free
Electrical Machine for All AE/JE EE Exams Mock Test
7.7 K Users
20 Questions 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

Concept

The active power in a 3ϕ star-connected load is given by:

\(P=3V_pI_pcosϕ=\sqrt{3}V_LI_Lcosϕ\)

Per phase power, \(P=V_pI_pcosϕ\)

where, \(cosϕ={R\over \sqrt{R^2+X^2}}\)

Calculation

Given, Z = 15 + j20 Ω

\(cosϕ={15\over \sqrt{15^2+20^2}}=0.6\)

\(I_p={V_p\over Z}={230\over 25}=9.2\)

\(P=230\times 9.2\times 0.6\)

P = 1269.6 W

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.

Get Free Access Now
Hot Links: teen patti go teen patti gold real cash teen patti master gold download teen patti real teen patti royal - 3 patti