Question

1. An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 10.00 V. The RMS current delivered to the answering machine is 364 mA. If the primary (input) coil of the transformer has 960 turns, then how many turns are there on the secondary (output) coil?

2. What is the power drawn from the electric outlet, if the transformer is assumed to be ideal?

Answer 1

given,

voltage in secondary coil = Vs = 10 V

Voltage in primary coil = V p = 120 V

number of turns in primary coil = Np = 960 turns

number of turns in secondary coil

using formula,

`(N_s)/(N_p)=(V_s)/(V_p)`

`N_s=(V_s)/(V_p)* N_p`

`N_s=(10)/(120)* 960`

`N_s= 80 \turns`

b) Power drawn out from the transformer

P = I V

P =0.364 x 10

P = 3.64 Watt