How do you prove experimentally the ratio `V//I` is a constant for a given conductor ?

Explanation

Ohm's law : The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.
Verification :
Aim : To verify Ohm's law or to show that `(V)/(I)=`constant for a conductor.
Materials required :
6V Battery eliminator, 0 to 1A Ammeter, 0-6V volt meter, copper wires, 50 cm manganin coll, Rheostal, switch and 3V LED, etc.

Procedure :
1. Complete the circuit as shown in figure. knob should be adjusted to 4.5V at battery eliminator.
2. Using Rheostat change the potential difference between two ends of manganin wire from OV to 4.5V (maximum).
3. By using Rheostat adjust the potential difference 1V between two ends of manganin wire .
4. Now observe the electric current through Ammeter in the circuit and note down in the following table.

5. Using Rheostat change the potential difference with difference value upto 4.5V and note down the current value (I) in the table.
6. Take atleast five values of and 1 and note down in the table.
7. Find `(V)/(I) for each set of values.
8. we notice that `(V)/(I)` is a constant
`ValphaIrArr(V)/(I)="constant"`
This constant is known as resistance of the conductor, denoted by R.
`rArr(V)/(I)=R.`
`therefore` Ohm's law is verified.
II. Aim : To show `V//I` is not constant in LED type conductors.
. The long terminal of LED is attached to positive terminal of the battery. short one to negative terminal of the battery.
. With the help of Rheostat change the potential difference (up to 3V) and note down current (I) and potential difference (V) in each instace in the Table.
. Calculate the `V//I` value.
. You come to know `V//I` is not contant.