Calculating the potential difference across a lamp

[Goob]

Homework Statement

The potential divider XZ has a potential of 3.0 V across it. When the contact is at the position Y, the resistance of XY equals the resistance of YZ which equals 12Ω. The resistance of the lamp is 4.0Ω.

Picture of the relevant circuit:
http://www.mediafire.com/view/w8te8yb3zwy4qiu/Screen%20Shot%202014-01-11%20at%207.46.44%20pm.png

Heres a link - can't get the embed to work :/
http://www.mediafire.com/view/w8te8yb3zwy4qiu/Screen Shot 2014-01-11 at 7.46.44 pm.png

Homework Equations

V=IR

1/R1 + 1/R2 = 1/RT

R1 + R2 = RT

The Attempt at a Solution

So far I've worked out the resistance across the lamp from:

1/4 + 1/12 = 3Ω,

Then worked out the total resistance from:

3Ω + 12Ω = 15Ω.

And then worked out the total current from:

I=V/R = 3/15 =0.2A

At this point I know that if I can find the current across the lamp, the pd across the lamp will be easy to find... but this is where I get confused.

Thanks to whoever may answer!

 

[gneill]

Homework Statement

The potential divider XZ has a potential of 3.0 V across it. When the contact is at the position Y, the resistance of XY equals the resistance of YZ which equals 12Ω. The resistance of the lamp is 4.0Ω.

Picture of the relevant circuit:
http://www.mediafire.com/view/w8te8yb3zwy4qiu/Screen%20Shot%202014-01-11%20at%207.46.44%20pm.png

Heres a link - can't get the embed to work :/
http://www.mediafire.com/view/w8te8yb3zwy4qiu/Screen Shot 2014-01-11 at 7.46.44 pm.png

Homework Equations

V=IR

1/R1 + 1/R2 = 1/RT

R1 + R2 = RT

The Attempt at a Solution

So far I've worked out the resistance across the lamp from:

1/4 + 1/12 = 3Ω,

Then worked out the total resistance from:

3Ω + 12Ω = 15Ω.

And then worked out the total current from:

I=V/R = 3/15 =0.2A

At this point I know that if I can find the current across the lamp, the pd across the lamp will be easy to find... but this is where I get confused.

Thanks to whoever may answer!

Hi Goob, Welcome to Physics Forums.

Note that the lamp and the YZ resistance are in parallel, forming the total of 3 Ω that you calculated. What can you say about the potential across parallel components?

 

[Goob]

Hi Goob, Welcome to Physics Forums.

Note that the lamp and the YZ resistance are in parallel, forming the total of 3 Ω that you calculated. What can you say about the potential across parallel components?

I know that in parallel the current adds up, and the voltage in constant,
and in series that the current is constant but the voltage adds up,

so if I calculate the pd across YZ it would be the same as the pd across the bulb.

But if I try to do this I still encounter my original problem - I'm going wrong somewhere :(


edit:

Think I've solved it,

The whole circuit has a resistance of 15Ω, and the parallel component has a total resistance of 3Ω, 1/5 of the total resistance of the circuit. If I just divide the total pd across the circuit by 5, would I get the answer for the pd across the bulb?

This would get me a pd of 0.6V.

 

[gneill]

I know that in parallel the current adds up, and the voltage in constant,
and in series that the current is constant but the voltage adds up,

so if I calculate the pd across YZ it would be the same as the pd across the bulb.

But if I try to do this I still encounter my original problem - I'm going wrong somewhere :(

The potentials across the lamp and resistance YZ are the same Because those components are in parallel. So if you "replace" those components with the 3 Ω equivalent resistance that you calculated, you need only find the potential across that 3 Ω resistance in order to know what the potential across the lamp is, right?

 

[Goob]

The potentials across the lamp and resistance YZ are the same Because those components are in parallel. So if you "replace" those components with the 3 Ω equivalent resistance that you calculated, you need only find the potential across that 3 Ω resistance in order to know what the potential across the lamp is, right?

Yeap, I just solved it along those lines,

Thanks a lot for your help!