How Does Efficiency Loss Affect Voltage and Current in a Transformer?

[michaelplease]

Homework Statement

A step-down transformer going from 240 V to 12 V is 95% efficient. The input current is 20 A.
a) Determine the primary turns: secondary turns ratio necessary in order to maintain a 12V output.
b) Determine the output current.

I have solved it as if I needed an secondary output of 12 * (100/95) = 12.6 V. But what I am wondering about is whether the 5% loss of power causes a 5% loss in secondary voltage, or a 5% loss in secondary current, or a loss in both that multiplies up to 5%, and why. Could someone please explain?

Homework Equations

P = IV
Np/Ns = Vp/Vs

The Attempt at a Solution

a) Vs = 12V * (100/95) = 12.6V
Np/Ns = Vp/Vs
= 240 / 12.6
= 19 : 1
b) Pp = IpVp
= 240(20)
= 4800W
Ps = Pp (0.95)
= 4560 W
Ps = IsVs
4560 = (12)(I)
I = 380 A

 

[Andrew Mason]

It would depend on how the transformer is built and how the losses occur. If the loss of energy is due to resistance of the windings, ie. I^2R losses, most of the losses will likely be in the secondary, where the current is going to be roughly 20 times higher than in the primary and the resistance 1/20 of that of the primary (1/20th no. of turns). If it is due to hysteresis losses or eddy currents in the core, I expect the loss would be distributed more or less equally over the primary and secondary.

You don't need to know this to calculate the secondary current. You just need to know the overall efficiency at a particular primary current and voltage.

You know the power in the secondary is .95 of the power in the primary. So if the voltage in the secondary is 12 V.:

$E_sI_s = .95(E_pI_p)$ from which you can work out the secondary current.

The emf generated in the secondary is determined by the primary voltage and the ratio of the number of turns of primary/secondary. Since you are not told what the source of the energy loss is, I think you have to stick to the 20:1 ratio of turns to produce 12 V. output.

AM