Question

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 4.00 V. The RMS current delivered to the answering machine is 660 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 17.5 percent of the input power is dissipated as heat in the coils and in the iron core of the transformer?

Answer 1

Part A.

We know that In transformers:

Ns/Np = Vs/Vp

Ns = number of secondary turns = Np*(Vs/Vp)

Ns = 600*(4/120)

Ns = 20 turns

Part B.

Power drawn from electric outlet (secondary) will be

Ps = Vs*Is

Ps = 4*660*10^-3 = 2.64 W

Part C.

power drawn by the transformer will be,

given that 17.5% of input power is dissipated

Which means power in output coil will be

Ps = Pp - 17.5% of Pp

Ps = Pp - 0.175*Pp

Pp = Ps/(1 - 0.175)

Pp = 2.64/(1 - 0.175)

Pp = 3.2 W

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