Linking charge, current and time

1. Current is the rate of flow of charge, so the charge that passes a point depends on both the current and the time for which it flows.
2. The relationship is $charge\; flow=current\times time$.
3. In symbols this is $Q=It$, where $Q$ is the charge in coulombs ($C$), $I$ is the current in amperes ($A$) and $t$ is the time in seconds ($s$).
4. You need to learn this equation and be able to use it without being given it.
5. A current of $1A$ means that $1C$ of charge passes a point every second, so $1A=1C/s$.
6. In the same way, one coulomb is the charge that passes a point when a current of $1A$ flows for $1s$, so $1C=1As$.
7. A larger current moves more charge each second, so for a fixed time, doubling the current doubles the charge that flows.
8. Leaving the same current flowing for longer also moves more charge, so for a fixed current, doubling the time doubles the charge.
9. For a steady current, charge flow is directly proportional to time, so a graph of charge against time is a straight line through the origin with a gradient equal to the current.
10. Dividing both sides of $Q=It$ by $t$ gives $I=\frac{Q}{t}$, which finds the current.
11. Dividing both sides of $Q=It$ by $I$ gives $t=\frac{Q}{I}$, which finds the time.

Using the right units

1. The equation only gives the right answer when charge is in coulombs, current is in amperes and time is in seconds.
2. Convert minutes to seconds by multiplying by $60$, so $5min=5\times 60=300s$.
3. Convert hours to seconds by multiplying by $3600$, so $2h=2\times 3600=7200s$.
4. Convert milliamperes to amperes by dividing by $1000$, so $250mA=0.25A$.
5. Convert kilocoulombs to coulombs by multiplying by $1000$, so $3.6kC=3600C$.
6. Give the final answer with its unit, rounded to the same number of significant figures as the least precise value you were given.