Electronics - Inductors - Discussion

84. 

What is the maximum source current for a transformer rated at 10 kVA, 1000/500 60 Hz?

[A]. 2 A
[B]. 5 A
[C]. 10 A
[D]. 20 A

Answer: Option D

Explanation:

No answer description available for this question.

Abhi said: (Mar 1, 2011)  
(1000/500)*10

Charan said: (Dec 19, 2014)  
Can anyone explain with more detail?

Prashant said: (Sep 6, 2015)  
How 20 ampere? Can anyone explain please?

Nisarkhan said: (Jan 16, 2016)  
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.

Kiran V said: (Jan 25, 2017)  
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.

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