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 570 mA. If the primary (input) coil of the transformer has 600 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 13.5 percent of the input power is dissipated as heat in the coils and in the iron core of the transformer?
Hello Samantha, in case of ideal transformer the power at the out put would be the same as that in the input. To get the power we need voltage and current. Voltage at the out put is 8 V and current is 570 mA. Hence power consumed by the machine is 4.560 W. So the power drawn in case of ideal transformer will be the same as 4.560 W If the efficiency of the transformer is 86.5% (100-13.5) then the input power will be 5.27 W (approx) To get the number of turns in the secondary let us use Ns/Np Vs/Vp Hence Ns 600*8/120 40 turns.
Ns/Np Vs/Vp Ns (8/120) * 600 40 turns P Pp Ps Vs * Is (an ideal transformer) 8 * 0.570 4.56 watt 13.5 percent of the input power is dissipated, then the power from output coil (100 - 13.5) % 86.5% of input power. ---Po 86.5% Pi 4.56 0.86 Pi ---Pi 4.56/0.86 5.30 watt
