Question

Ohm’s Law in terms of the applied potential. But we can re-express the potential as the integral of the electric field along some path. We can also re-express electrical resistance in terms of conductivity, length, and cross-section. Use these relationships to write Ohm’s Law in terms of the electric field and the current density.  What physical parameter does a linear fit to the slope of the resistor current graph represent? Why is a similar graphical analysis of the diode current not appropriate?

Answer 1

The ohms law can be written in form of applied potential as

Let V = applied potential

I = current through the resistor

R = resistor of the circuit

Hence the ohms law can be written as

V = I R

Now the electric field E = V/d

Where d = distance at which the electric field applied

Hence V = E d

Now the resistance of the circuit can be written as

= resistivity of the resistor

d = length of the resistor or the length at which the electric field applied

A = crossection area of the resistor

Now the ohms law can be written as

V = I R

Where J = current density = I/A

Now the coductivity of the resistor is reciprocal of the resistivity

i.e

This is the expression for the ohms law in term of electric field and coductivity.

Now from the resistor you can see that

The resistance depends on the length of the resistor , resistivity of the resistor , and the crossection area of the resistor

4) diode current is exponential in nature and the current through the resistor is linear with the applied voltage

Hence diode current first 0 A for some voltage but as the voltage across the diode increases the current through the diode increase suddenly after the applied voltage cross the knee voltage or break down voltage .

But in the simple resistor Tere is no knee or breakdown voltage hence the current is linear with the applied voltage