Question

keeping the potential difference constant, the resistance of a circuit is doubled .by what factor does the current change in the circuit

Answer 1

According to ohm's law
R =v/I...............(1)    ( R=resistance, V=voltage , I =current )

let I' be the current when Resistance become double
2R = V/I'   ............(2)                           (V is constant)

dividing equation (2) by (1) 

R/2R= (V/I)/V/I'
1/2=I'/I
I=2I'
I/2=I'
Current through conductor become half. 
 

Answer 2

Explanation:

The Ohm's law gives the relationship between the potential difference, current and the resistance of the circuit. Mathematically, it is given by :

$V=IR$

V = potential difference

I = current

R = resistance

If the resistance of the circuit is doubled, R' = 2R

$I=\dfrac{V}{R}$

$I'=\dfrac{V}{R'}$

$I'=\dfrac{V}{2R}$

$I'=\dfrac{1}{2}\dfrac{V}{R}$

$I'=\dfrac{1}{2}I$

So, the current will becomes the half of the initial current. Hence, this is the required solution.