Ohm's law states that $V=IR$ for a simple circuit, where $V$ is the total voltage, $I$ is the amount of current running through the resistor and $R$ is the resistance of the resistor. If $I=40$ amperes when $R=3$ ohms, find the amount of current (in amperes) when the resistor has a resistance of $20$ ohms. Assume voltage remains constant.
This problem involves Ohm's law, which relates voltage, current, and resistance in a simple circuit. I need to find the amount of current $I$ when $R=20$ ohms, given that $I=40$ amperes when $R=3$ ohms. I also know that the voltage $V$ remains constant, so I can use that to relate the two scenarios. When $I=40$ amperes and $R=3$ ohms, the voltage is $V=IR=40\times 3=120$ volts. This means that when $R=20$ ohms, the voltage is still $V=120$ volts. Now I can use Ohm's law again to find $I$ when $R=20$ ohms and $V=120$ volts. I have $V=IR$, so I can solve for $I$ by dividing both sides by $R$. This gives me $I=\frac{V}{R}=\frac{120}{20}=6$ amperes. So the amount of current when the resistor has a resistance of $20$ ohms is $6$ amperes. 