Question

The charger for your electronic devices is a transformer. Suppose a $60\mathrm{Hz}$outlet voltage of $120\mathrm{V}$needs to be reduced to a device voltage of $3.0\mathrm{V}$. The side of the transformer attached to the electronic device has $60\mathrm{turns}$of wire. How many turns are on the side that plugs into the outlet?

Short answer

Turns on the side that plugs into the outlet is$2400\mathrm{turns}$.

Step-by-step solution

Formula for number of turns

The secondary coil is the side that is connected to the electronic gadget.

The primary coil is the one that is connected to the plug.

The primary coil must contain more coils than the secondary coil to scale down the voltage.

The number of turns is precisely proportional to the induced emf, according to Faraday's law.

$\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}=\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}$

Where${\mathrm{V}}_1$andlocalid="1648828624813" ${\mathrm{N}}_1$are the primary coil's emf and the number of turns.

While${\mathrm{V}}_2$and localid="1648828631790" ${\mathrm{N}}_2$are the secondary coil's emf and the number of turns.

${\mathrm{N}}_1=\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}{\mathrm{N}}_2$

Calculation of number of turns

As it stands,

${\mathrm{V}}_1=120\mathrm{V}$

${\mathrm{V}}_2=3\mathrm{V}$

${\mathrm{N}}_2=60$

Number of turns,

${\mathrm{N}}_1=\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}{\mathrm{N}}_2$

$=\left(\frac{120\mathrm{V}}{3\mathrm{V}}\right)\times 60\mathrm{turns}$

$=2400\mathrm{turns}$