What is the resistance of a non-ohmic resistor in v-i graph

[chetan:)]

What is the resistance of a non-ohmic resistor (lamp) in v-i graph given a voltage?
I thought it would be the gradiant for specific voltages.
But apparentlly it is the ratio of the specific voltage/current which makes sense too.
So what is the correct out of the two?

 

[cragar]

is voltage on the x-axis and current on the y or the other way around .
if voltage is on the x-axis then the slope of that line would be 1/R
if current is on the x-axis then the slope of that line would be R
using I=V/R

 

[dulrich]

For a non-ohmic resistor, the ratio V/I is not necessarily the same as the slope of the curve.

If y = x^2, then x = 3 implies y = 9 so y/x = 3, but dy/dx = 6.

Since resistance is defined via Ohm's law the resistance is the ratio, not the slope.

 

[berkeman]

For a non-ohmic resistor, the ratio V/I is not necessarily the same as the slope of the curve.

If y = x^2, then x = 3 implies y = 9 so y/x = 3, but dy/dx = 6.

Since resistance is defined via Ohm's law the resistance is the ratio, not the slope.

I'm not sure I'm understanding which one you are promoting, but the impedance is definitely the slope of the line:

Z = dV/dI

This is used all the time in circuit analysis.

 

[dulrich]

I was arguing on the opposite side. I'm not an expert on electronics, so I defer to you. Is there a difference between impedance and resistance? I was thinking in terms of DC voltage.

 

[berkeman]

I was arguing on the opposite side. I'm not an expert on electronics, so I defer to you. Is there a difference between impedance and resistance? I was thinking in terms of DC voltage.

If all the components are resistors, then impedance = resistance.

If some of the components are non-liinear (like diodes for example), then impedance = Z = dV/dI, and it is still real.

If some of the coponents are reactive (inductors and capacitors), then you get a complex Z = dV/dI, with real and imaginary (in-phase and quadrature-phase) components.