Question

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 6.00 V. The RMS current delivered to the answering machine is 600 mA. If the primary (input) coil of the transformer has 360 turns, then how many turns are there on the secondary (output) coil? Tries 0/20

What is the power drawn from the electric outlet, if the transformer is assumed to be ideal? Tries 0/20

What is the power drawn by the transformer, if 11.5 percent of the input power is dissipated as heat in the coils and in the iron core of the transformer?

Answer 1

Ns/Np = Vs/Vp
Ns = (6/120) * 360 = 18 turns

P = Pp = Ps = Vs * Is (an ideal transformer)
= 6 * 0.600 = 3.6 watt

11.5 percent of the input power is dissipated,
then the power from output coil =(100 - 11.5) % = 88.5% of input power.
---> Po = 88.5% Pi
3.6 = 0.885 Pi --->

Pi = 3.6/0.885 = 4.067watt

I hope it was helpful.. please rate me positive..