Mammoth Memory

Transformer examples

Now we know 

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



using formula 2 we have:



`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


and also

`V_1\ \I_1=V_2\ \I_2`



and therefore


Using this last formula we get

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


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


`V_1\ \I_1=V_2\ \I_2`



`I_2=7.5\ \amps`


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


We know




`N_2=5\ \turns`

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