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?

 

Your definition of resistivity doesn't make sense to me. 5 E-7 m whats? Milliohms per meter^3?

 

the resistivity is 5x10^-7 ohm per meter

 

the resistivity is 5x10^-7 ohm per meter

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.

Please, please, please get in the habit of carrying your units through all of your work. Had you done this, you would have known that the resistivity number had to be Ωm and not Ω/m because, had it been Ω/m (and, therefore, had the width and thickness been extraneous information), your resistance calculation would not have come out to be Ω and you would have known to look closer. Similarly, had it actually been Ω/m, your units would not have worked out the way that you did it and, again, you would have known to look closer. As it is, you happen to do it correctly for the data as presented, but since you didn't sanity check your work, you really just got lucky.

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

 

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.

 

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:

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