Transformer and the brightness of bulbs

[Biker]

Homework Statement

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

Homework Equations

Ohms law
Conservation of energy
Faraday's law

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?

 

[CWatters]

Now you can find the voltage of Bulb A by using faradays law and it will equal to 60 volts

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.

 

[Biker]

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

I get answer D by noting that the winding ratio steps up the current by a factor of

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:
$\frac{V_s}{N_s} = \frac{V_p}{N_p}$

 

[CWatters]

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.

The power isn't the same in each bulb.

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

 

[CWatters]

But my thinking was, Both of these bulbs use the same energy per second..

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 V_p.

 

[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 isn't the same

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

 

[CWatters]

Deleted for reason stated in next post.

 

[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.

 

[Biker]

I don't follow that. The voltage on bulb B is the same as the secondary voltage.

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
$0.25 V_p = V_{secondary}$
$0.25 (V_{source} - V_b) = V_{secondary}$

 

[CWatters]

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

Anyway..

So the voltage in bulb A is 0.25(Vs−V_bulbB) that is why the voltage isn't the same

Yes that's correct.

Just one last question, Why did you say 4I2bR4Ib2R 4 I^2_b R instead of 16I2bR16Ib2R 16I^2_b R ?

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

So none of the answers are correct.

 

[Biker]

Thank you as always CWatters, Much appreciated :D

 

[CWatters]

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