Finding the potential difference across two wires

AI Thread Summary
The discussion revolves around calculating the potential difference across two copper wires connecting a lamp to a power supply. The user initially miscalculated the potential difference, assuming it to be 12V for each wire without considering their resistance. After receiving guidance, they correctly calculated the resistance of the wires based on their length and cross-sectional area. Using the formula V = IR, they derived a potential difference of 3.06 x 10^-5V across each wire. The conversation emphasizes the importance of accurately determining wire resistance in electrical calculations.
SamS90
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Hello everyone, as you can tell I'm new and just wondering if I could get some help with this question as I'm not sure where exactly I've gone wrong:

Homework Statement



Two wires are used to connect a lamp to a power supply of negligible internal resistance. The potential difference across the lamp is 12V and its power is 36W. Calculate the potential difference across each wire. The two wires in question are made of copper and have a resistivity of 1.7 x 10^-8 Ohm meter.


Homework Equations



Not 100% sure but:

V=W/Q, R=V/I, I=P/V, E= I (R + r)

And I'm not sure about an others.

The Attempt at a Solution



I = 36/12, I = 3
R = 12/3, R = 4
E = 3 (4 + 1.7x10^-8) = 12V

Therefore the potential difference across each wire is 12V

I know I've gone wrong some where and am hoping someone can put me right.

Many thanks,
Sam
 
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SamS90 said:

The Attempt at a Solution



I = 36/12, I = 3
Good. You found the current, which you'll need.
R = 12/3, R = 4
That's the resistance of the lamp. OK, but not needed.
E = 3 (4 + 1.7x10^-8) = 12V
:confused:

Hint: What's the resistance of each wire? How long are they?
 
Thanks for the help. :smile:

Using 0.6m as length and 0.001m as the cross sectional area I have derived an answer of 3.06 x 10^-5 after first findinng the resistance of both wires (as you stated) then using the resistance found in the equation V = IR to find the potential difference.

Thanks again.
 
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