An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 8.00 V. The RMS current delivered to the answering machine is 417 mA. If the primary (input) coil of the transformer has 1080 turns, then how many turns are there on the secondary (output) coil?
What is the power drawn from the electric outlet, if the transformer is assumed to be ideal?
What is the power drawn by the transformer, if 10.7 percent of the input power is dissipated as heat by the eddy currents in the iron core?
a) Voltage at the out put is 8 V and current is 417 mA. Hence
power consumed by the machine is = V*I = 3.336 W.
So the power drawn in case of ideal transformer will be the same as
3.336 W
b) If the efficiency of the transformer is 89.3% (100-10.7) then
the input power will be 3.7357 W (approx)
c) To get the number of turns in the secondary let us use Ns/Np =
Vs/Vp
Hence Ns = 1080*8/120 = 72 turns.
An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage...