Primary coil of an ideal transformer. I need an equation.

  • Thread starter Sarak
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Homework Statement


The primary coil of an ideal transformer is connected to a 120-V source and draws 1.0 A. The secondary coil has 600 turns and supplies an output current of 5.0 A to run an electrical device.
A) What is the voltage across the secondary coil?
B) How many turns are in the primary coil?
C) If the maximum power allowed by the device (before it is destroyed) is 210 W, what is the maximum input current to this transformer?


Homework Equations


V2=V1I1/I2
N1=N2I2/I1
and ??? I=P/V ????


The Attempt at a Solution


I found part A) to be V2=24, and part B) to be N1=3000. The only equation I can think of for part C) is I=P/V. P is given to be 210 W. I tried plugging V1 and V2 in that equation and haven't gotten the right answer. Can someone given me an idea how to go about this? I can't find anything in the text.
 

Answers and Replies

  • #2
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Maybe you won't see this reply or no more need help but let me reply.

Let's find R(load resistance) from first,

R=V/I = 24/5 ohm

and

Vs= Is*R = Is * 24/5 Volt

also

Ps=Pp= 210 watt

lets use these,

Ps= Vs*Is = Is * 24/5 * Is = 210 watt

Is^2 = 210 * 5/24 A

Is = 6.61 A

now let's find input current. From

Ip/Is = Ns/Np

and

Ip=Ns/Np * Is = 600/3000 * 6.61 = 1,322 A

I suppose approx. correct.
 

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