Finding the potential difference across two wires

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SUMMARY

The discussion centers on calculating the potential difference across two copper wires connecting a lamp to a power supply, where the lamp has a potential difference of 12V and power of 36W. The user initially miscalculated the potential difference across the wires, incorrectly concluding it to be 12V. The correct approach involves determining the resistance of the wires using their resistivity of 1.7 x 10^-8 Ohm meter, length of 0.6m, and cross-sectional area of 0.001m², leading to a potential difference of 3.06 x 10^-5V across each wire.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of electrical power calculations (P = VI)
  • Familiarity with resistivity and resistance formulas
  • Basic concepts of electrical circuits
NEXT STEPS
  • Learn how to calculate resistance using resistivity, length, and cross-sectional area
  • Study the implications of wire gauge on electrical resistance
  • Explore the concept of voltage drop in electrical circuits
  • Investigate the effects of wire material on conductivity and resistance
USEFUL FOR

Students studying electrical engineering, electricians, and anyone interested in understanding the principles of electrical circuits and potential differences in wiring systems.

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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