# Video: Finding the Charge Flowing through a Point in a Circuit

The diagram shows a circuit consisting of a battery and a resistor. The current through the circuit is 2.0 A. Over a period of 45 seconds, how much charge flows past point P in the circuit?

02:09

### Video Transcript

The diagram shows a circuit consisting of a battery and a resistor. The current through the circuit is 2.0 amps. Over a period of 45 seconds, how much charge flows past point P in the circuit?

Taking a look at our circuit, we see the battery, point P, and then also our resistor. Notice that this symbol, the resistor being represented by a box, is just one way to show it. Another way is by using these jagged lines like this. But in either case, what the resistor does is the same. It resists the flow of current. We’re told that the current 𝐼, moving through the circuit, is 2.0 amperes and that this current runs for a time of 45 seconds. Over that time, we want to figure out just how much electric charge flows past the point P in the circuit.

As a side note, notice that the location of point P in this circuit isn’t important. That’s because the current is the same everywhere all throughout the circuit. So regardless of what point we picked, the charge that flows past that point, whatever it is, would be the same over this time interval. To figure out just how much charge does flow past a point in the circuit, let’s recall a relationship between charge, time, and current.

The current, represented with the letter capital 𝐼, in a circuit is equal to the amount of charge that passes a given point in the circuit divided by the time it took for the charge to pass. In our case though, we don’t want to solve for current. But we do want to solve for the charge 𝑄.

To get that, let’s rearrange this equation. We’ll do this algebraically by multiplying both sides of the equation by the time interval 𝑡. Having done that, the 𝑡 on the right-hand side of the equation cancels out since it appears above the top and the bottom of this fraction. We have then an expression for the charge 𝑄. It’s equal to 𝐼 times 𝑡.

Now, 𝐼 and 𝑡, current and time, are both given to us in the problem statement. 𝐼 is equal to 2.0 amperes. And 𝑡 is equal to 45 seconds. Multiplying these two numbers together, we get a result of 90. And the unit is coulombs. And recall that coulombs is the unit of electric charge. That then is our answer.

Over 45 seconds with a current of 2.0 amps running through the circuit, 90 coulombs of charge flows past point P.