 # Transformer examples

Now we know

V_1/V_2=(Number\ \turns\ \1)/(Number\ \turns\ \2)

and

V_1xxI_1=V_2xxI_2

We can now work out many problems.

Example 1

A transformer converts 40 volts of alternating current to a higher voltage. The current in the primary coil is 12 amps and the secondary side is 4 amps.

Question A – What is the secondary voltage?

So if we write out all the formulae:

V_1/V_2=(Number\ \turns\ \1)/(Number\ \turns\ \2).......1

V_1xxI_1=V_2xxI_2........2

using formula 2 we have:

40xx12=V_2xx4

therefore

V_2=(40xx12)/4=120\ \v\o\l\t\s

There are 120 volts produced in the secondary side.

Question B – If there are 100 primary coils, how many secondary coils are there?

Using formula 1 we have:

40/120=100/(Numer\ \turns\ \2)

Turns\ \2=(100xx120)/40

Turns\ \2=100xx3

Turns\ \2=300

There are 300 turns on the secondary side.

Example 2

A transformer is 100% efficient. It has 200 turns on the primary coil and an input current of 0.5amps. If the secondary coil has 3,000 turns what is the output current?

We know

V_1/V_2=N_1/N_2

and also

V_1\ \I_1=V_2\ \I_2

Therfore

V_1/V_2=I_2/I_1

and therefore

I_2/I_1=N_1/N_2

Using this last formula we get

I_2/0.5amps=(200\ \turns)/(3000\ \turns)

I_2=(200xx0.5)/3000

I_2=0.033dot3\ \amps

The output current is 0.033dot3\ \amps

Example 3

An AC power supply is connected to a transformer. The AC power supply is 12 volts and 2.5 amps. The secondary coil gives a reading of 4 volts.

i.  Calculate the power input to the transformer.

We know Power = Village Idiot (see mnemonic)

P=VI

Therefore

P=12xx2.5=30\ \w\a\t\t\s

ii.  Calculate the current in the secondary coil.

We know power cannot be lost or gained

Voltage 1xxCurrent 1 = Primary power = Secondary power = Voltage 2xx Current 2

V_1\ \I_1=V_2\ \I_2

12xx2.5=4xxI_2

I_2=(12xx2.5)/4

I_2=7.5\ \amps

iii.  The primary coil has 15 turns; how many turns does the secondary coil have?

We know

V_1/V_2=N_1/N_2

12/4=15/N_2

N_2=(15xx4)/12

N_2=5\ \turns