Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.3k views
in Physics by (63.6k points)

A 23-V, 50 Hz, 1-ph supply is feeding the following loads which are connected across it. 

(i) A motor load of 4 kW, 0.8 lagging p.f. 

(ii) A rectifier of 3 kW at 0.6 leading p.f. 

(ii) A lighter-load of 10 kVA at unity p.f. 

(iv) A pure capacitive load of 8 kVA 

Determine : Total kW, Total kVAR, Total kVA

1 Answer

+1 vote
by (64.9k points)
selected by
 
Best answer
S. No. Item  kW P.f kVA  kVAR I Is Ir
1 Motor 4 0.8 lag 5  3 − ve, Lag 21.74 17.4  13.04 Lag (−)
2 Rectifier 3  0.6 Lead 5 4 + ve, Lead 21.74  13.04 17.4 Lead (+)
3 Light-Load 10  1.0 10 zero 43.48  43.48 zero
4 Capacitive Load  Zero  0.0 Lead 8 8 + ve Lead 34.8 zero 34.8 Lead (+)
Total 17 - Phasor 
Addition 
required 
+ 9 + ve
Lead
Phasor 
Addition 
required
73.92 39.16 Lead (+)

Performing the calculations as per the tabular entries above, following answers are obtained 

Total kW = 17 Total kVAR = + 9, leading

Total kVA = √(172 + 92) = 19.2354

Overall circuit p.f = KW/KVA = 17/19.2354 = 0.884 leading

Overall Current = 19235.4/230 = 83.63 amp

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...