# simple resistance question

Discussion in 'Homework Help' started by fan_boy17, May 8, 2012.

1. ### fan_boy17 Thread Starter New Member

Apr 17, 2012
12
0
The rear screen heater for the new Nisota saloon car will comprise 8 resistive filaments bonded to the surface of the glass and connected together in parallel. Each filament will be made from a material of resistivity 5 × 10^-7 ohm.m with a width of 1.2 mm, a thickness of 0.1 mm and a length of 1.2 m. The car battery will have a capacity of 80 A.hr, will provide an electromotive force of 12 V, and will be of negligible internal resistance.

what would be the total resistance in the heater?

I get 0.625ohms. is this correct?

if it is, would the total current drawn from the battery just be
I=V/R=12/0.625?

Last edited: May 8, 2012
2. ### #12 Expert

Nov 30, 2010
18,076
9,678
Your definition of resistivity doesn't make sense to me. 5 E-7 m whats? Milliohms per meter^3?

3. ### fan_boy17 Thread Starter New Member

Apr 17, 2012
12
0
the resistivity is 5x10^-7 ohm per meter

4. ### WBahn Moderator

Mar 31, 2012
23,856
7,382
I would be highly, highly surprised at this. While Ω/m might be used to describe the resistance of wires made of this stuff, it won't be much of a heater if a meter of the stuff only has half a microohm of resistance.

It is much, much more likely that the bulk resistivity of the material is (5E-7)Ωm, which, by the way, is how you obviously used it to come up with 5Ω for each of the 8 lines to yield 0.625Ω overall.

Yes, the current drawn from the batter would then be I = V/R = 12V/0.625Ω = 19.2A.

Do you want the engineer that is designing the electrical system in your car to be lucky, or thorough and competent?

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,018
1,772
did you forget the length of the wires were 1.2 meters and if you assumed the resistivity was 5 ohms per meter, you need to account for that other 0.2 meters in your calculation. Eight 1.2 meter pieces in parallel is not 0.625 ohms as you have calculated. Eight 1 meter pieces in parallel is the 0.625 ohms that you have calculated.

6. ### WBahn Moderator

Mar 31, 2012
23,856
7,382
Who's assuming 5Ω/m?

The choices are a resistivity of 5E-7Ωm, or a "resistivity" of 5E-7Ω/m. The problem states that the given resistivity is a property of the material and bulk resistivity has units of resistance times length. The second "resistivity" would be a property not only of the material, but also of the shape of the material and is more properly referred to as "length resistance" or "wire resistance".

For a rectangular resistor that is WxTxL (connected on the two WxT ends) and bulk resistivity rho, the resistance is:

$
R = \rho \frac{L}{WT}
$

So, in this case, the resistance of each filament is:

$
R = 5 \times 10^{-7} \Omega m \frac{1.2m}{(1.2mm)(0.1mm)}
R = 5 \times 10^{-7} \Omega m \left(\frac{1.2m}{(1.2mm)(0.1mm)}\right)\left( \frac{10^3mm}{1m}\right)\left( \frac{10^3mm}{1m}\right)
R = \frac{(5)(10^{-7})(1.2)(10^3)(10^3)}{(1.2)(0.1)}\Omega \ = 5\Omega
$