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Change in potential energy across wires differing composition

  • Thread starter Mebmt
  • Start date
12
0
1. The problem statement, all variables and given/known data

9 V battery is connected to wire made up of three segments of different materials connected in a series: 10 cm copper, then 12 cm iron, then 18 cm tungsten. All wires are 0.26 mm diameter. Find potential difference across each wire.

Constants:
p Copper= 1.7 x 10-8
p Iron = 9.7 x 10 -8
p Tungsten = 5.6 x 10-8

26 mm diameter = 0.13 mm radius or 1.3 x 10-4 m

2. Relevant equations

R=pl/A

area of wire (1.3 x 10-4)(1.3 x 10-4) x pi = 5.31 x 10-8 meters squared

3. The attempt at a solution

Copper Resistance
R=pl/a
R= (1.7 x 10-8 ohm m)(.1m)/5.31 x 10-8 meters squared
R=0.032 ohm

Iron Resistance
R=pl/a
R=(9.7 x 10-8 ohm m)(.12m)/5.31 x 10-8 meters squared
R=0.22 ohm

Tungsten Resistance
R=pl/a
R=(5.6 x 10-8 ohm m)(0.18 m)/5.31 x 10-8 meters squared
R=0.19

R total series = r1 + r2 + r3
= .032 + .22 + .19
= 0.442 ohm


V copper = vtotal * Rcopper/R total
= 9 V * .032/.442
= 0.65 V

V silver = vtotal *Rsilver/R total
=9 V * .22/.442
=4.48 V

V Tungsten = vtotal * Rtungsten/R total
=9V *.387/.442
= 3.87 V

Volts all add up to 9 so 2nd Law is good but my answers are wrong somewhere...

Thanks,
Scott
 

PeterO

Homework Helper
2,425
46
1. The problem statement, all variables and given/known data

9 V battery is connected to wire made up of three segments of different materials connected in a series: 10 cm copper, then 12 cm iron, then 18 cm tungsten. All wires are 0.26 mm diameter. Find potential difference across each wire.

Constants:
p Copper= 1.7 x 10-8
p Iron = 9.7 x 10 -8
p Tungsten = 5.6 x 10-8

26 mm diameter = 0.13 mm radius or 1.3 x 10-4 m

2. Relevant equations

R=pl/A

area of wire (1.3 x 10-4)(1.3 x 10-4) x pi = 5.31 x 10-8 meters squared

3. The attempt at a solution

Copper Resistance
R=pl/a
R= (1.7 x 10-8 ohm m)(.1m)/5.31 x 10-8 meters squared
R=0.032 ohm

Iron Resistance
R=pl/a
R=(9.7 x 10-8 ohm m)(.12m)/5.31 x 10-8 meters squared
R=0.22 ohm

Tungsten Resistance
R=pl/a
R=(5.6 x 10-8 ohm m)(0.18 m)/5.31 x 10-8 meters squared
R=0.19

R total series = r1 + r2 + r3
= .032 + .22 + .19
= 0.442 ohm


V copper = vtotal * Rcopper/R total
= 9 V * .032/.442
= 0.65 V

V silver = vtotal *Rsilver/R total
=9 V * .22/.442
=4.48 V

V Tungsten = vtotal * Rtungsten/R total
=9V *.387/.442
= 3.87 V

Volts all add up to 9 so 2nd Law is good but my answers are wrong somewhere...

Thanks,
Scott
I like your reasoning - and your answers - so check that you have transcribed the figures correctly, and looked at the correct answers in you book.
 
12
0
Thanks Peter. The prof had changed the numbers on the website from those provided in the book. I had overlooked one of them.
 

PeterO

Homework Helper
2,425
46
Thanks Peter. The prof had changed the numbers on the website from those provided in the book. I had overlooked one of them.
Note: when I worked this out I ignored the diameter - since they were all the same, as well as the 10-8 on each of the resistivities, [they all had the same factor] since I knew at the end I was only going use the resistances in the form of fractions - where those values would cancel anyway.
 

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