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A Few RC Circuits/Electricity Problems

by cse63146
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Feb9-08, 05:05 PM
P: 452
1. The problem statement, all variables and given/known data

1. What electric field strength is needed to create a 5.0 A current in a 2.0-mm-diameter iron wire?

2. You need to design a 1.0 A fuse that "blows" if the current exceeds 1.0 A. The fuse material in your stockroom melts at a current density of 500 A/cm^2. What diameter wire of this material will do the job?

3.A car battery is rated at 90 Ahr, meaning that it can supply a 90 A current for 1 hr before being completely discharged. If you leave your headlights on until the battery is completely dead, how much charge leaves the battery? Express your answer in Coulombs to three significant figures

4. The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q = 4t - t^2 , where t is in seconds and t>=0.

Which is an expression for the current in the wire at time t.

2. Relevant equations

E = [tex]\frac{I}{(\sigma)(\pi)r^2}[/tex]

J = [tex]\frac{I}{(\pi)(r^2)}[/tex]

Q = [tex]\frac{I}{t}[/tex]

3. The attempt at a solution

1. I know [tex]\sigma[/tex] is 10^7 so

E = [tex]\frac{5}{(10^7)(\pi)(0.001^2)}[/tex] = 0.159 N/c

2. 500x10^-2=5 A/m^2

r = [tex]\sqrt{\frac{I}{(\pi)J}}[/tex] = [tex]\sqrt{\frac{1}{(\pi)5}}[/tex] = 0.252 m and to get diameter I multiply it by two to get 0.504. The question asks me to express it in mm so I got 504mm.

3. 1 hour = 3600 seconds

Q = [tex]\frac{I}{t}[/tex] = Q = [tex]\frac{90}{3600}[/tex] = 0.025 C

4. Current is the derivative of charge so it would be the third equation.

Did I make a math (or any other kind) error?
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Feb11-08, 10:28 AM
P: 452
#1 and #4 are correct, but I cannot get the correct answers for #2 and #3. Does anyone know why?
Feb11-08, 10:38 AM
P: 288
If you have 500 amps for every square CENTImeter, should you have a lot less per square meter, or, say, a LOT more?

#3, assuming you did it right, has only 2 significant figures

it's always the easy ones ^_^

Feb11-08, 10:48 AM
P: 452
A Few RC Circuits/Electricity Problems

so 1 Square Centimeter = 0.0001 Square Meters therefore 500 A per Square Centimeters = 0.05 A per Square Meters

r = [tex]\sqrt{\frac{I}{(\pi)J}}[/tex] = [tex]\sqrt{\frac{1}{(\pi)0.05}}[/tex] = 5.046 m = 5046 mm

and to change 0.025 from 2 significant digits to 3, I'll need to use scientific notation so 0.025 = 2.50 x 10^-2
Feb11-08, 11:02 AM
P: 288
ehh, you kinda missed my point for #2, here's a hint

Make sure you can get that to work right. If I can fix x things into a square centimeter, I'd better be able to fit a helluva lot more into a square meter, which is notably bigger than a cm^2
Feb11-08, 11:21 AM
P: 452
Ah, now I get it thanks. Let's give this another try:

r = [tex]\sqrt{\frac{I}{(\pi)J}}[/tex] = [tex]\sqrt{\frac{1}{(\pi)(5000000)}}[/tex] = 0.000252 m and to get the diameter I multiply it by two to get 5.05 x 10^-4 m = 0.505 mm.

And I also figured out why #3 is wrong. It's not Q = I/t, it's Q = I*t so:

Q = I*t = (90)(3600) = 324000 C

Just tried them, and it said I was correct. Thank you blochwave.

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