1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A 50 Hz, 3−Φ synchronous generator has an inductance per phase of 1.75 mH and its neutral is grounded. It feeds a line through a circuit breaker. The total stray capacitance to ground of the generator and circuit breaker is 0.0025 μF. A fault occurs just beyond the circuit breaker which opens when the symmetrical SC current is 7500 A (rms). The value of maximum rate of rise of TRV is

A
3876 V/μS
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2786 V/μS
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1786 V/μS
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4786 V/μS
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2786 V/μSTransient recovery voltage, TRV=e=Em(1−cosωt) RRRV=dedt=Emωsinωt RRRVmax=Emω=Em√LC E=IωL =7500×314×1.75×10−3 =4.121kW RRRVmax=4121√2√1.75×10−3×25×10−10=2786V/μS

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Transformer
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program