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In a real physical problem the answer to "is the relationship linear or nonlinear?" is non-linear > 99% of the time. That answer is not that helpful since we often don't have closed-form solutions to non-linear problems.

For most undergraduate courses, Resistance is assumed not to be a function of current, and Ohm's law can be used in its most boring form, I = V/R

2007-12-10 04:12:18 · answer #1 · answered by flyin520 3 · 0 0

In an equation for this relationship R=E/I the only way it could be nonlinear is for the components to change by squaring(or rather by an exponent) or by reduction (additions of radicals) as an example lets solve ONE
e=12 I=6 R=2
now that we have set R at 2 you can do any thing you want to to E and I as long as its the same and R=2 SO the graph of these changes will by these numbers be a straight line positive in direction with a slope of 2/1.......making it linear..

If the equation was in a format where the Numerator or denominator were squared a quick look at it would form a graph of a parabola looking up or down depending on which was squared. making the relationship nonlinear.

Merry Christmas ...from the E..

2007-12-10 10:55:35 · answer #2 · answered by Edesigner 6 · 0 1

It depends on whether resistance is constant or not. Power (P) = Current (I) times Voltage (V) but when you start varying the resistance (ie, the wire gets hot or you add a bunch more of it) you have to use Ohm's law to figure it out and it becomes nonlinear.

2007-12-10 10:47:00 · answer #3 · answered by Andrew 1 · 0 0

Always non-linear. But for some materials in the right range of currents/voltages one can get a very good linear relationship. Those are the ones that we use as resistors.

2007-12-10 10:42:39 · answer #4 · answered by Anonymous · 1 0

Ideally, Ohm's (V = IR) law makes it a linear relationship.

2007-12-10 11:01:33 · answer #5 · answered by navster15 1 · 1 0

It depends on the material
For good conductors V=RI

v= voltage, I current. It is ohm's law

2007-12-10 10:27:40 · answer #6 · answered by maussy 7 · 0 0

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