Minimum output e.m.f. of the power source

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SUMMARY

The discussion centers on calculating the minimum output electromotive force (e.m.f.) of a power source in a circuit involving resistors. The participant initially calculated the resistance of a cable made of two parallel wires, resulting in a total resistance of 2 Ω. They then computed the voltage across the cable as 0.12V, leading to a total potential difference of 16.12V. However, they later realized that the wires should be considered in series, correcting their approach and confirming the correct answer.

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Janiceleong26
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Homework Statement


image.jpg


Homework Equations


V=IR
For resistors connected in parallel, 1/R =1/R1 + 1/R2 +...

The Attempt at a Solution


I first calculated the resistance of the the cable which consists of two wires,
1/R= 1/0.0050 + 1/0.0050
R = 2.5x10^-3 x 800
= 2 Ω
Then I calculate the voltage across the cable, V=0.060x2= 0.12V
So the total p.d would be 0.12V+16.0V=16.12V, and I chose A, but the correct answer is C. Why?
Please help, thanks
 
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Janiceleong26 said:

Homework Equations


V=IR
For resistors connected in parallel, 1/R =1/R1 + 1/R2 +...
One of the wires in the cable carries current toward the relay and one carries the current back toward the power supply, right? If you drew that as a circuit diagram, would that makes those wires parallel or series?
 
jbriggs444 said:
One of the wires in the cable carries current toward the relay and one carries the current back toward the power supply, right? If you drew that as a circuit diagram, would that makes those wires parallel or series?
Oh right! It's series. I misread the question. I've figured out the answer already. Thank you!
 

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