# Thread:

1. ## ratio of currents

I solved this first in that this appears to be an inversion.

$\frac{\frac{V^2}{R_2}}{\frac{V^2}{R_1}}<br /> \Rightarrow<br /> \frac{I_1}{I_2}<br /> \Rightarrow<br /> \frac{[1+60(.005)]}{[1+60(.004)]}<br /> \times<br /> \frac{[1+10(.004)]}{[1+10(.005)]}<br /> \Rightarrow<br /> \frac{(1.3)(1.04)}{(1.24)(1.05)} = 1.0384<br />$

I still don't understand what these temperature coefficients mean. or why the have the $C^{-1}$

appreciate any comments...

2. Here's how I read it:

the resistance of a resistor is being modeled as

$R=R_{0}(1+T\alpha),$ where $T$ is temperature in degrees C, and alpha is what is called the "temperature coefficient of resistance". $R_{0}$ is some initial resistance. It would, technically, be the resistance when the temperature is $0^{\circ}\text{C}.$ The resistance of a resistor changes with temperature, and here is one model for how it changes. The units of $\alpha$ must be inverse degrees C in order to cancel out the units of $T$. Otherwise, you wouldn't have units of Ohms on both sides. Hence the $C^{-1}.$ It would, perhaps, be better to write, for example, $\alpha_{1}=0.004(^{\circ}\text{C})^{-1}.$

Does this explanation clear things up?

3. Originally Posted by bigwave
I solved this first in that this appears to be an inversion.

$\frac{\frac{V^2}{R_2}}{\frac{V^2}{R_1}}<br /> \Rightarrow<br /> \frac{I_1}{I_2}<br /> \Rightarrow<br /> \frac{[1+60(.005)]}{[1+60(.004)]}<br /> \times<br /> \frac{[1+10(.004)]}{[1+10(.005)]}<br /> \Rightarrow<br /> \frac{(1.3)(1.04)}{(1.24)(1.05)} = 1.0384<br />$

I still don't understand what these temperature coefficients mean. or why the have the $C^{-1}$

appreciate any comments...
The resistors are in parallel so the potential drop across them is the same, in which case the current ratio is the reciprocal of the power ratio.

CB

4. thanks for the help, I see how this works now, I am finding EE a very interesting application of math.
was wondering if MHF considered an EE catagory... probabaly not enough potential subscribers tho.

5. You might be surprised how many people on MHF can do at least the basics of EE. Certainly digital logic and sophomore-level circuits are well within the capabilities of folks here. Also, signal analysis with Z transforms, Laplace transforms, etc., are all covered here. Maybe not antenna design!

Keep posting questions. I'm finding them interesting, at least. The worst thing that could happen would be that no one knows how to help!

6. ## will definitly keep posting the EE problems

I will definitly keep posting the EE problems.

Its a brave new world for me

I did some study with electronics about 20 years ago. but the math was not the focus.
just building kits and stuff..

now I don't see how it could even be understood at all without math

regards.

r

7. Ok, have a good one.
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