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 580 mA. If the primary (input) coil of the transformer has 840 turns, then how many turns are there on the secondary (output) coil

Answer 1

Answer:

35

Explanation:

We are given that

Initial voltage,$V_1=120 V$

Final voltage, $V_2=5 V$

Number of tuns in primary coil of the transformer, $N_p=840$

Rms current, $I_{rms}=580mA=580\times 10^{-3} A$

$1 mA=10^{-3} A$

We have to find the number of turns  are there on the secondary coil.

We know that

$\frac{N_s}{N_p}=\frac{V_2}{V_1}$

Using the formula

$\frac{N_s}{840}=\frac{5}{120}$

$N_s=\frac{5}{120}\times 840=35$

Hence, there are  number of turns on the secondary coil=35