Power of Transformer: Primary & Secondary Light Bulb

In summary: The power dissipated in the bulbs will still be different because their resistances are different. But the total power dissipated in both bulbs will still be equal to the power drawn from the source.
  • #1
steve oliver
12
0
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
 
Physics news on Phys.org
  • #2
steve oliver said:
however if i use i^2r after step up i decrease power

How are you measuring or calculating i?
 
  • #3
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
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..

CWatters said:
How are you measuring or calculating i?
 
  • #5
i get it now thank you
 
  • #6
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 shouldn't the power in primary =power in secondary?
 
  • #7
You made mistakes somewhere...

steve oliver said:
method 1: i^2r x:y=1:4

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

steve oliver said:
method 2: v^2/r x:y=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

steve oliver said:
method 3 :vi x:y=1:1

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
steve oliver said:
also shouldn't the power in primary =power in secondary?

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
But shouldn't iy have double current that of ix?
 
  • #10
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
I will try and do a drawing later. Have to go out for a few hours.
 
  • #12
Bulb XnY.jpg
 
  • #13
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
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.
 

1. What is a power transformer?

A power transformer is a device that is used to transfer electrical energy from one circuit to another through electromagnetic induction. It is composed of two or more coils of wire, known as primary and secondary windings, that are wrapped around a core of iron or ferrite.

2. How does a power transformer work?

A power transformer works on the principle of electromagnetic induction. When an alternating current (AC) flows through the primary winding, it creates a changing magnetic field which induces a voltage in the secondary winding. This voltage is then used to power the secondary circuit, such as a light bulb.

3. What is the role of the primary winding in a power transformer?

The primary winding in a power transformer is responsible for receiving the incoming alternating current (AC) and converting it into a changing magnetic field. This changing magnetic field then induces a voltage in the secondary winding, which is used to power the secondary circuit.

4. How does the secondary light bulb in a power transformer work?

The secondary light bulb in a power transformer works by receiving the induced voltage from the changing magnetic field in the primary winding. This voltage then powers the light bulb, causing it to light up. The brightness of the light bulb can be controlled by adjusting the number of turns in the secondary winding.

5. What are the advantages of using a power transformer in a circuit?

There are several advantages of using a power transformer in a circuit. These include: efficient transfer of electrical energy, ability to step up or step down the voltage, isolation between the primary and secondary circuit, and the ability to control the voltage and current in the secondary circuit. Additionally, power transformers are reliable, cost-effective, and have a long lifespan.

Similar threads

  • Electrical Engineering
Replies
8
Views
916
Replies
6
Views
2K
  • Electromagnetism
Replies
3
Views
725
  • Mechanics
Replies
2
Views
3K
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
182
  • Electromagnetism
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
924
Back
Top