The question says the potential difference is what is changing, which means we're solving for V.
It tells us that potential difference increases by a factor of two, which just means V doubles.
With this info, we can pick some numbers, plug it into Ohms law and see what happens.
Here's an example where I just picked random numbers that are easy to work with:
V=I*R 10=I*5 I=2 Lets increase the potential difference (V) by a factor of two and see what happens to current: V=I*R 20=I*5 (all I've done is double the potential difference from 10 to 20) I=4
When we increase V by a factor of 2, I increases by a factor of 2. We went from I=2 to I=4. We can increase V by factor of 2 again and see: V=I*R 40=I*5 I=8
Okay, current just increased by a factor of 2 again when we increased the potential difference by a factor of 2.
It's always good to check work with alternate numbers, so here's one more set: V=I*R 16=I*4 (remember, we know we're solving for V, so I'm just plugging in random numbers for I and R) I=4 Increase V by factor of 2: 32=I*4 I=8
So, when we increase V (the potential difference) by a factor of 2, I (current) always increases by a factor of 2 as well.
