Generally, distribution transformers rating is 5 kVA, 10 kVA, 15 kVA, 25 kVA, 37.5 kVA, 50 kVA, 100 kVA, 200 kVA, 250kVA, 500kVA, 1000kVA. The protection method used for distribution transformers depends on the transformer ratings.
Above 500 kVA distribution transformers HT side (11 kV) use Drop Out Fuse Cut-out (DOFC) and Lightning arrester and LT side (0.415 kV) use Molded Case Circuit Breaker (MCCB). Neutral of LT should be earthed properly with low resistance wire. It is mention that Buchholz relays is use above 500 kVA Transformer.
Below 500 KVA distribution transformers HT side (11kV) use Drop Out Fuse Cut-out (DOFC) and Lightning arrester and LT side (0.415 kV) use Molded Case Circuit Breaker (MCCB). Neutral of LT should be earthed properly with low resistance wire.
Related Question
View All
Maximum Power Transfer Theorem :
In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a power source with internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.
Let Impedance of,
Xc near the terminal X is Z1=-j10
10 ohm resistance, Z2=10
3 ohm resistance, Z3=3
XL, Z4 =j14
Xc near to source, Z5=-j10
Now,
Zth=[Z5 || (Z3 +Z4) ] + Z2 +Z1
= [ -j10 || (3+j4)] + 100- j10
= 112 - j36
For t<0, the switch is closed.
Capacitor acts as open to DC.
Voltage across the capacitor,
v(0-)= {(12||4)×V}÷{(12||4)+6} = 8VFor t=0, the switch is opened,Voltage across the Capacitor cannot change instantaneously,So, v(0-)= v(0) = 8VAt t>0, The capacitor is discharging.Rth = (12||4) = 3 ohmC= 1/6 FTime constant, Π = Rth×C = 0.5 So, v(t) = v(0)e-t/Π = 8e-t/0.5 =8e-2t V (Ans.)
The initial energy stored in the Capacitor,
W=(1/2)CVo2
= 0.5×0.167×82
= 5.333J (Ans)
১ ক্লিকে প্রশ্ন, শীট, সাজেশন ও
অনলাইন পরীক্ষা তৈরির সফটওয়্যার!
শুধু প্রশ্ন সিলেক্ট করুন — প্রশ্নপত্র অটোমেটিক তৈরি!
Related Question
Complete Exam
Preparation
Learn, practice, analyse and improve
