# Transformer and the brightness of bulbs

1. Oct 13, 2016

### Biker

1. The problem statement, all variables and given/known data
The brightness of the bulb A in the following picture equals:
A) half of the bulb B brightness
B) equal to bulb B brightness
C) Twice as bright as bulb B
D) four times as bright as bulb B

2. Relevant equations
Ohms law
Conservation of energy

3. The attempt at a solution
Now you can find the voltage of Bulb A by using faradays law and it will equal to 60 volts
Then you can find that $I_{bulb A} = 4 I_{bulb B}$
Using $I_s V_s = I_p V_p$
So you will choose D, That is the answer for the question. But that doesn't make sense. (It is an old exam problem ._.)

But my thinking was, Both of these bulbs use the same energy per second so they should glow with the same brightness, right?

2. Oct 13, 2016

### CWatters

Ok so perhaps I'm a bit rusty but how did you do that without knowing the resistance of the bulb?

I got answer D by noting that the winding ratio steps up the current by a factor of four. Without needing to make your first step of finding the voltage.

Power = I^2R and I assume R is the same unknown resistance for both bulbs.

3. Oct 13, 2016

### Biker

Yes it steps up the current by a factor of 4 but is the current the only thing that brightness depend on? Or power?
Because if it depends on power the brightness will be the same which seems more intuitive to me.

Using this equation:
$\frac{V_s}{N_s} = \frac{V_p}{N_p}$

4. Oct 13, 2016

### CWatters

The power isn't the same in each bulb.

In one it's I2R and in the other it's 4I2R

5. Oct 13, 2016

### CWatters

They don't dissipate the same energy per second (aka power).

If the transformer is ideal then yes the power going into the transformer = power coming out. However...

Yes the power dissipated in bulb B A does equal the power coming out.

But the power dissipated in bulb A B doesn't equal the power going into the transformer. That's because the voltage on A B isn't the same as the transformer primary voltage Vp.

Last edited: Oct 14, 2016
6. Oct 13, 2016

### Biker

Yep sorry my bad, I guess I haven't focused a bit on this.

So the voltage in bulb A is $0.25(V_s -V_{bulb B})$ that is why the voltage isnt the same

Just one last question, Why did you say $4 I^2_b R$ instead of $16I^2_b R$ ?

7. Oct 14, 2016

### CWatters

Deleted for reason stated in next post.

8. Oct 14, 2016

### CWatters

I made a mistake in post 5 and 7. I got Bulb A and B around the wrong way around. See strike outs and corrections.

9. Oct 14, 2016

### Biker

But secondary voltage is equal to $4I_b R$ while the voltage of bulb B is equal to $I_b R$
Using KVL
$V_{source} - V_b = V_p$
Then using Faraday's law you get
$0.25 V_p = V_{secondary}$
$0.25 (V_{source} - V_b) = V_{secondary}$

10. Oct 14, 2016

### CWatters

Looks like I was correcting my earlier post while you were typing.

Anyway..

Yes that's correct.

You are correct I forgot to square the current when I did power = I^2R

So none of the answers are correct.

11. Oct 14, 2016

### Biker

Thank you as always CWatters, Much appreciated :D

12. Oct 14, 2016

### CWatters

No problems. Sorry I didn't pick up my errors sooner. Been a bit distracted.