# What happens to resistance if we double current intensity?

problem: what happens to resistance if we double current intensity ?

my attempt:

ohm's law tells us $$R = { V \over I }$$ so : $${R \over 2} = { V \over 2I }$$

so resistance should decrease to half if we double the current intensity , however my textbook says that the resistance should stay the same and i doubt that and i wanted to know if this is a typo or i have mistaken

Resistance is a propety of electrical components, not a consequence of the active circuit - a 100-ohm resistor has a resistance of 100 ohms, whether it's hooked up to a AAA battery or the power grid. In Ohm's Law, the quantity $$R$$ is typically fixed. So, when you double the current running through an element, the voltage across that element also doubles, leaving you with the same resistance for the element.
What you've done is answered the question while keeping the quantity $$V$$ fixed. The result you've found shows that for a power source of a specific, set voltage, an element with resistance $$2R$$ will draw half the current ($$I/2$$) of an element with resistance $$R$$ that draws current $$I$$. Just keep in mind that it's the value for $$R$$ which is usually a constant in Ohm's law, while $$V$$ and $$I$$ are inversely related to maintain that fixed resistance value.