# A little transformer help?

#### Declan

Joined Jun 10, 2007
1
Hi, my first post, so please be kind!

Anyway, I'm kind of confused about transformers.

I know that they step up or step down the voltage depending on the ratio of the primary turns to secondary, and that they do the opposite to the current. And it conserves power (ignoring small losses).

But I always thought that a voltage across a resistance drives a current, which is Ohms Law. I thought that if you had 100V across a 20 Ohm resistor, there would be 5 Amps. Simple.

But this http://www.allaboutcircuits.com/vol_2/chpt_9/2.html says that on one side of the transformer, there is "High voltage, low current", and on the other side, there's "Low Voltage, High current". How is this possible? Doesn't this directly contradict Ohm's Law? What is the voltage across, the two poles of the transformer on one of the sides?

Thanks a lot!

#### thingmaker3

Joined May 16, 2005
5,084
Ohms Law is still in effect. One side of the transformer has a different impedance than the other.

Example: A transformer draws 100mA from a 120Vac wall outlet. The secondary provides 12Vac at 1Amp. Impedance of the primary is 1200 Ohms. Impedance of the secondary is 12 Ohms.

#### bloguetronica

Joined Apr 27, 2007
1,372
Hi, my first post, so please be kind!

Anyway, I'm kind of confused about transformers.

I know that they step up or step down the voltage depending on the ratio of the primary turns to secondary, and that they do the opposite to the current. And it conserves power (ignoring small losses).

But I always thought that a voltage across a resistance drives a current, which is Ohms Law. I thought that if you had 100V across a 20 Ohm resistor, there would be 5 Amps. Simple.

But this http://www.allaboutcircuits.com/vol_2/chpt_9/2.html says that on one side of the transformer, there is "High voltage, low current", and on the other side, there's "Low Voltage, High current". How is this possible? Doesn't this directly contradict Ohm's Law? What is the voltage across, the two poles of the transformer on one of the sides?

Thanks a lot!
Not necessarily. If we ignore the effects of indutance, and lets say, read the resistance of the primary winding, we would reading a higher resistance than the one read on the secondary. Thats because the primary has many turns with thin wire and the secondary lesser turns with ticker wire. But I think that is due to the nature of the transformer, since the primary has to be thinner, and the secondary has to be thicker, in a step-down transformer. So, the secondary has a smaller resistance.
But resistances aside, don't forget that the primary inducts current on the secondary. Being the emf per turn (voltage per turn) constant for a transformer, this dictates a direct relationship between the ratio between Vprimary and Vsecondary, and between the turns ratio. Since in an ideal transformer no energy is loss, and thus, we have no power loss, Pprimary = Psecondary. So what happens is that in the secondary we have a bigger current than in the primary, and that those two are related. So if we use a thicker wire in the primary, we have a bigger current in that winding, and thus we have to expect a bigger current in the secondary.

#### subtech

Joined Nov 21, 2006
123
Although I'll not fault anything that has been stated so far, let's go with a more simplistic view of the transformer. There are many applications for transformers of various designs, but in this case we are concerned with transferring energy from the high voltage or source side of the transformer to the low voltage or load side of the transformer. The main thing to remember is that you are transforming energy, or power (watts) from one side to the other.
Assume that the high voltage side of a transformer is energised at 240 volts and the low voltage side is rated at 12 volts. (that is a ratio of 20 to 1 by the way, remember that) Connected to the low voltage side of the transformer is a lamp which is a 60 watt load. With 12 volts applied to the lamp, the current draw would be 5 amperes. (E*I=P) We now know the quantities that are present on the low voltage, or secondary side of the transformer.
On the high voltage side the quantities will be different or course, but in proportion to the winding ratio of this transformer. (remember the 20 to 1)
Remember that the power in to the transformer has to equal the power out.
If that is true in this case, then the load on the transformer is 60 watts, the voltage on the primary side of the transformer is 240 volts then the current in the primary winding must be .25 amperes. Sure enough, .25*240=60
In keeping with your original post, notice that the current flow in the high voltage winding is less than that in the secondary winding. It is less by a factor of 20. There you have it. High voltage-low current. Or, if you prefer, low voltage-high current. Just remember, you transforming energy(watts) from a high voltage source to low voltage load. Watts in to the transformer must equal the watts out.
Hope this helps.

#### alim

Joined Dec 27, 2005
113
Hi I would state my response this way The power consumed by the secondary must be equal to the power supplied by the primary.What happens in the primary depends on what happens in the secondary.

#### recca02

Joined Apr 2, 2007
1,214
i will only be repeating the above replies but cud not stop myself from posting

a transformer steps up/down the voltage without adding/reducing the power
(neglecting very small losses).
so the power remains constant.
ohms law must be applied to the side where the load is connected.
on the other side the load is transformed as
z(1) = z(2)*sq{v(1)}/sq{v(2)}--------sine power remains constant.
now ohms law can be applied using the transformed value.

transformers are such kool devices but they get hot quickly 