Question

A cylindrical rod of length L is connected across a fixed potential difference, creating a current I through the rod. What would be the current if the length of the rod were doubled?
2
1/4
1/2
4

Answer 1

Answer:

The value of current through the rod becomes half

$i' = \frac{i}{2}$

Explanation:

As per Ohm's law we know that the current through a resistor is given as

$i = \frac{V}{R}$

here we know that

$R = \rho \frac{L}{A}$

here we know that the length of the cylinder is L and area is A so the value of current through the rod is given as

$i = \frac{V A}{\rho L}$

now we have change the length of the conductor to twice of initial value and rest all parameters will remain the same

so we will have

$i' = \frac{VA}{\rho (2L)}$

now from above two equations we have

$\frac{i}{i'} = 2$

so new current will become

$i' = \frac{i}{2}$