An ideal transformer steps 8V up to 2000V and the 4000 turn secondary coil carries 2 A.

a. Find the number of turns on the primary.
b. FInd the current value in the primary.
c. What is the power of this transformer?
For A and B, I am just checking my work. I do not know how to do C, though.
Thanks!

1- Vs = voltage in secondary coil
Vp= voltage in primary

Ns = number of turns in secondary coil
Np= number of turns in primary coil


using the formula : Vs/Vp = Ns/Np , Np = 4000/(2000/8) , so, Np=16 turns.


2- here, Is = current in the secondary,
Ip= current in the primary

using the formula, Vs * Is = Vp * Ip, Ip = (2000*2)/ 8, So, Ip = 500 A


3- to find the power, power input and output is calculated , power input= Vp * Ip , that is, P-in = 8* 500, so, power in = 4000 W or 4 kW

power output= Vs * Is , P-out = 2000 * 2 , so, power out= 4000 W, or 4 kW.

so,since power in and power out are equal, the transformer is an ideal one

Vp / Vs = Np / Ns = is / ip

8 / 2000 = Np / 4000 = 2 / ip

a.) ..... Np = 8(4000) / 2000 = 16

b.) ..... ip = 2(2000) / 8 = 500 A

c.) ..... Power = (Vp) (ip) = (Vs) (is)
...................... (8) (500) = (2000) (2) = 4000 W rating

An ideal transformer consumes zero power; it only transforms it's voltage
to current ratio.