Power of transformer

1. Jul 5, 2015

steve oliver

when i compare the brightness of the light bulb in the primary transformer and the secondary transformer (step up transformer) the secondary light bulb is brighter . teacher say because of v^2/r after step up the v increase and so the power. however if i use i^2r after step up i decrease power should be decreased also. whu would i get a different result in different calculating methods

2. Jul 5, 2015

CWatters

How are you measuring or calculating i?

3. Jul 5, 2015

Staff: Mentor

I don't blame you for being confused. Your transformer raises the voltage and decreases the current, relative to what? Relative to the primary side voltage and current. But what happens on the secondary side also affects the primary side. Considering the two sides independently causes the confusion.

Look up transformers on Wikipedia. https://en.m.wikipedia.org/wiki/Transformer. In the ideal tranfpsformer equations , you can see that you can replace the transformer and the load resistance R, by a single resistance Rp=a*a*Rs, where a is the number of primary side turns divided by the number of secondary side turns, Rs is the load resistance of the light bulb. Rp is the apparent resistance of the light seen from the primary side.

Now you can analyze your circuit using V=I*Rp as if the transformer was not there. If you do that, then the seeming cotraditictions and confusion should go away.

If that is not clear, I can try to draw the pictures for you, when I get access to a real computer.

4. Jul 5, 2015

CWatters

The point is that the current through the load does not decrease in this situation. The transformer increases the voltage and the load determines the current according to I = V/R. V has been increased (relative to what it was when the load was connected to the primary) so I increases. Hence my question..

5. Jul 6, 2015

steve oliver

i get it now thank you

6. Jul 6, 2015

steve oliver

Then in the following question how should i calculate.
Two identical light bubls x and y are connected respectively to the primary and secondary coils of an ideal step downtransformer.The turn ratio of the transformer is 2:1.Calculate the power dissipated in x and y.
method 1: i^2r x:y=1:4
method 2: v^2/r x:y=4:1
method 3 :vi x:y=1:1
why would be 3 different result?also shouldnt the power in primary =power in secondary?

7. Jul 7, 2015

CWatters

ix = V/R
iy = 0.5V/R

So using W = i2R you get

Wx = (V/R)2 R = V2/R and
Wy = (0.5V/R)2 R = 0.25V2/R
Ratio is 4:1

Vx = V
Vy = V/2

Using V2/R ...

Wx = V2/R
Wy = (V/2)2/R = 0.25V2/R
Ratio is 4:1

Vx = V
Vy = 0.5V

Ix = V/R
Iy = 0.5V/R

Using W = IV

Wx = V * V/R = V2/R
Wy = 0.5V * 0.5V = 0.25V2
Ratio = 4:1

So they are all the same ratio 4:1

8. Jul 7, 2015

CWatters

You are confusing two different things..

a) Power in vs Power out of the transformer - yes it's the same
with
b) Power dissipated in the bulb when connected to the primary vs Power dissipated in the bulb when connected to the secondary - no it's not the same because the bulb draws more power when connected to the primary.

9. Jul 7, 2015

steve oliver

But shouldnt iy have double current that of ix?

10. Jul 7, 2015

CWatters

x and y are not the same circuit so Ix and Iy are not related by the transformer equation...

Circuit "x" has the bulb connected to the primary. In that situation there is nothing connected to the secondary of the transformer so the secondary current is zero. Therefore the current in the primary is also zero. The current in the bulb is V/R.

Circuit "y" has the bulb connected to the secondary. The current in the secondary is 0.5V/R (It's 0.5V because it's a 2:1 step down transformer). The current in the primary is equal to half the secondary current or (0.5V/R) / 2.

11. Jul 7, 2015

CWatters

I will try and do a drawing later. Have to go out for a few hours.

12. Jul 7, 2015

CWatters

13. Jul 9, 2015

steve oliver

then what if the light bulb in primary coil is connected with the light bulb in secondary coil. i mean they are in the same circuit.

14. Jul 9, 2015

CWatters

Then the current drawn from the source V is just the sum of the two (kirchhoff's current law).

Itotal = Ix + Iin = (V/R) + (0.25V/R) = 1.25V/R

Everything else is the same.