We will apply the concept related to the current change given in the same problem. We will divide both currents into two states: the new current and the old current. As the current is the change of the load in a certain time, we will have that the old current is,
[tex]I_{old} = \frac{dq}{dt}[/tex]
If it takes 5 times more time, then we will have the new current is,
[tex]I_{new} = \frac{dq}{5dt}[/tex]
[tex]I_{new} = \frac{1}{5}(\frac{dq}{dt})[/tex]
Replacing the given value of the old current we will have to,
[tex]I_{new} = \frac{I_{old}}{5}[/tex]
Therefore the new current will be [tex]\frac{1}{5}[/tex] the old current.
