# Video: Finding the Current through a Point in a Circuit

The diagram shows a circuit consisting of a battery and a light-emitting diode (LED). Over a period of 25 seconds, a charge of 50 coulombs passes through point 𝑃 in the circuit. What is the current in the circuit during this period?

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### Video Transcript

The diagram shows a circuit consisting of a battery and a light-emitting diode, LED. Over a period of 25 seconds, a charge of 50 coulombs passes through point 𝑃 in the circuit. What is the current in the circuit during this period?

Taking a look at our diagram, we see it’s an electrical circuit. This circuit is powered by a battery, which sends a current 𝐼 through the loop counterclockwise. The current passes through a light-emitting diode or LED and every other point in this loop including point 𝑃. In the problem statement, we’re told that, over a time period of 25 seconds, a certain amount of charge, 50 coulombs, passes through point 𝑃 in the circuit.

Now let’s think about that for a second, a certain amount of charge passing through a point over a certain period of time. Thinking of that, we can recollect the definition for electric current. It’s charge divided by time. In other words, we’ve been given the information for the right-hand side of this equation, charge and time. And based on that, we want to solve for the electric current in the circuit.

Going back over to our diagram for a second, notice that this entire electric circuit is just one closed loop. That tells us that the current 𝐼 in the circuit is the same everywhere. And this means that the current at point 𝑃, which is technically where we’ve been counting the passage of charge and the passage of time, will be the same as the current at any other point in the loop.

So let’s go ahead and figure out what that current 𝐼 is. We’re told that the time 𝑡 is 25 seconds and that the charge passing a certain point in the circuit, point 𝑃, is 50 coulombs. Before we calculate this fraction though, take a look at the units in it. We have coulombs per second. If we take that ratio, one coulomb over one second, then that ratio is equal to one ampere of current. This means we can write the current in our circuit this way. It’s equal to 50 divided by 25 amperes. 50 divided by 25 is two. And that gives us our answer.

The current in the circuit over this 25-second time interval is two amps.