The potential difference across a resistor increases by a factor of 4. How does the current change? (Ohm's law: V = IR)
A it increases by a factor of 4
B it decreases by a factor of 4
C it increases by a factor of 2
D it decreases by a factor of 2

Correct option is C it decreases by a factor of 2

A - it increases by a factor of 4

Explanation:

Ohm's law is V = IR where V = potential difference, I = current, and R = resistance. Assuming that resistance stays constant (since the question mentioned no change in resistance), we only need to look at "V" and "I" in the equation V = IR.

"V" and "I" are directly proportional. In other words, whatever change happens to "V" will also occur to "I". Since "V" increases by a factor of 4, then "I" will also increase by a factor of 4.

You could also approach this problem by plugging in a change of 4 for the potential difference in the equation: V = I*R.

If we plug in 4V for V, we now have:

4V = I*R

Since the left side increases by a factor of 4, then the right side (only looking at current in this case) will also need to increase by a factor of 4 in order to have the right side stay equal to the left. Therefore, current will also increase by a factor of 4.

reserve is the answer