Electronics - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 84)
84.
What is the maximum source current for a transformer rated at 10 kVA, 1000/500 60 Hz?
Discussion:
5 comments Page 1 of 1.
KIRAN V said:
9 years ago
1000/500 means that the voltage applied to the primary is 1000 Volt and so the secondary voltage is 500 Volt. If this is true, then the maximum current that you can deliver from the secondary is:
Vp = 1000.
Vs = 500.
(Apparent Power = V x I).
I = P/Vs.
I = 10 * 10^3 / 500,
I = 20 Amp.
Vp = 1000.
Vs = 500.
(Apparent Power = V x I).
I = P/Vs.
I = 10 * 10^3 / 500,
I = 20 Amp.
Nisarkhan said:
10 years ago
As P apparent = Irms*Vrms.
As given value of we is Vmax. So must be divided by under root 2.
Do same for I because we need Imax?
Denominator on RHS will be 2 and this 2 will make 10 on LHS = 20A.
As given value of we is Vmax. So must be divided by under root 2.
Do same for I because we need Imax?
Denominator on RHS will be 2 and this 2 will make 10 on LHS = 20A.
Prashant said:
1 decade ago
How 20 ampere? Can anyone explain please?
Charan said:
1 decade ago
Can anyone explain with more detail?
Abhi said:
1 decade ago
(1000/500)*10
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