0.5V across 1.5m tungsten wire w/cross-sect. area 0.6 mm^2 - Find wire current

AI Thread Summary
A homework problem involves calculating the current through a 1.5-meter tungsten wire with a cross-sectional area of 0.6 mm² and a voltage of 0.5 volts. The resistivity of tungsten is given as 5.6E-8 ohm meters. The initial calculation for current yielded an incorrect value due to not converting the area from mm² to m², which is necessary for accurate results. After correcting the area conversion, the current was found to be approximately 3.57 amps. The discussion highlights the importance of unit conversion in physics calculations.
student_fun
Messages
8
Reaction score
0

Homework Statement


0.5 volts is maintained across a 1.5 meter tungsten wire that has a cross-sectional area of 0.6 mm2. What is the current in the wire?


Homework Equations


Resistivity of tungsten: 5.6E-8 ohm meters.

Current Density: J = \sigma \cdot \frac{V}{l}, J = \frac{I}{A}
Definition of resistivity: \rho = \frac{1}{\sigma}

Variable meanings in equations used:
I: current in amps, A: cross-sectional area of wire in m2, \sigma: material conductivity, V: volts, l: length of extruded cross-sectional area (in this case, a wire) in meters, \rho: resistivity in ohm meters

The Attempt at a Solution



Using the relations and data available it seems to make sense to find the current in the wire by relating the known tungsten resistivity to the two definitions of current density in which most variables are known except for the I (current) that I'm after. Doing this yields:

\frac{I}{A} = \sigma\frac{V}{l}
\sigma = \frac{1}{\rho}
\frac{I}{A} = \frac{1}{\rho} \cdot \frac{V}{l}
I = A\left[\frac{1}{\rho} \cdot \frac{V}{l}\right]

Then I plug and chug:

I = 0.6\left[\frac{1}{5.6E-8} \cdot \frac{0.5}{1.5}\right] = 3571428.571 Amps

However this answer is not valid. Can anyone see something I am doing that is outright wrong? Invalid logic somewhere?

Any hints to get me on the right track would be greatly appreciated.

Thank you for your time.
 
Physics news on Phys.org
Your answer has the right procedure but the area is in mm^2 you need to convert it to m^2. So you are off by a factor of 10^6 meaning 3.57 amps.
Below is an easier way to go about the problem and attain the same answer.

resistance = (resistivity*length)/cross sectional area in m^2

Current = voltage/ resistance
 
Nevermind, lol, solved my own problem. Found out that I didn't convert millimeters squared to meters squared. Got it right now :)
 
Thanks for the reply fball558 :) It's nice to get other views while your pourin over pages and pages of physics :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top