Question
Download Solution PDFAn alternator is supplying a load of 300 kW at 0.6 p.f. lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFP = S cos ɸ
P = 300 kW, cos ɸ = 0.6
S = 300/0.6 = 500 kVA
Now, cos ɸ = 1, S = 500 kVA
P = 500 × 1 = 500 kW
More kilo watts can alternator supply for the same kVA loading = 500 – 300 = 200 kWLast 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.