An AC adapter for a telephone answering machine uses a transformer to reduce the

Q: An AC adapter for a telephone answering machine uses a transformer to reduce the

as we know Ns/Np = Vs/Vp thus Ns = NpVs/Vp = 720* 5/120 = 30 turns for ideal transformer VpIp = VsIs thus power drawn from the electric outlet = VsIs = 5 * 0.380 = 1.9W power on output of tranformer = (100 - 10.5) % of input power = 89.5% of input power thus 1.9W = 0.895 * input power input power = 1.9/0.895 = 2.123 W

ID: 1902738 • Letter: A

Question

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. The RMS current delivered to the answering machine is 400 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 10.5 percent of the input power is dissipated as heat in the coils and in the iron core of the

Explanation / Answer

as we know Ns/Np = Vs/Vp thus Ns = NpVs/Vp = 720* 5/120 = 30 turns

for ideal transformer VpIp = VsIs

thus power drawn from the electric outlet = VsIs = 5 * 0.380 = 1.9W

power on output of tranformer = (100 - 10.5) % of input power = 89.5% of input power

thus 1.9W = 0.895 * input power

input power = 1.9/0.895 = 2.123 W

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An AC adapter for a telephone answering machine uses a transformer to reduce the

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