Video Transcript
The diagram shows a circuit consisting of a battery and a resistor. The current through the circuit is 50 milliamps. Over a period of 1.5 hours, how much charge flows past point 𝑃 in the circuit?
Okay, so we’ve got a circuit consisting of a battery and a resistor. Here’s the battery. Here’s the resistor. We’ve been given the current in the circuit. And we’ve been given the amount of time for which we’re observing the circuit. We need to find out the amount of charge that flows past point 𝑃 in the circuit. So the first thing we can do is to observe our diagram, the diagram we’ve been given, and to notice that actually it’s a fairly simple circuit.
It doesn’t have for example any junctions at which the current can split. It’s just one single loop. This means that the current at any point in the circuit is the same as the current at any other point in the circuit. So in other words, the current at point 𝑃 is the same as the current anywhere else in the circuit. And this simplifies things for us massively, because now we’re just trying to work out the amount of charge that flows past any point in the circuit, not necessarily just point 𝑃.
So we’ve been given two quantities. We’ve got the current in the circuit, which we’ll call 𝐼, and the time for which the current flows. We’ll call this 𝑡. We’re asked to find out the charge, which we’ll call 𝑄. So we need to find a relationship that links together the current 𝐼, the time 𝑡, and the charge flowing 𝑄. To find this relationship, we can recall that current is the rate of flow of charge. In other words, a current is the amount of charge that flows per unit time. We can write that in symbols as 𝐼 is equal to 𝑄 over 𝑡 or the current is equal to the charge flowing divided by the time taken for that charge to flow.
So this is the relationship that we’re looking for. But let’s quickly discuss the units of each one of these quantities. The standard unit for current is the amp. For charge, it’s the coulomb. And for time, it’s the second. What this means is that if we have a current in amps and a time in seconds, then we find out our charge in coulombs. We should always try and give our final answer in standard units unless we’re told otherwise in the question. In this case, we’ve just been told to find the charge. We’ve not been told anything about the units.
So we need to find the charge in the standard units. That is, we need to give our final answer in coulombs. But in order to do that, we need to have the current in amps and the time in seconds. Currently, haha currently, we don’t have that. We’ve got the current in milliamps. And we’ve got the time in hours. So we’ve got a bit of converting to do. Let’s start with the current. Well one milliamp is defined as one thousandth of an amp. A milli anything is a thousandth of that thing. And therefore, 50 milliamps is 50 thousandths of an amp.
We can also write that like this. And that simplifies down to 0.05 amps. So going back to our current on the left-hand side of the screen, we can replace the 50 milliamps with 0.05 amps. At this point, we can move on to our time. We can convert from hours to seconds. Well we know that every hour has 60 minutes in it and every minute has 60 seconds in it. Using this information, we can work out how many seconds there are in one hour. Well one hour has 60 minutes and each minute has 60 seconds in it. So one hour has 60 times 60 seconds in it. In other words, one hour has 3600 seconds in it.
However, the time that we observing the circuit for is not one hour. It’s 1.5 hours. So the number of seconds in 1.5 hours is simply 1.5 times 3600. And this ends up being 5400 seconds. So going back to the left-hand side of the screen once again, we can replace the time with 5400 seconds. At this point, we’ve got both the current and the time in their standard units. So we’re ready to calculate the charge flowing past point 𝑃 or any point in the circuit in coulombs.
So let’s take our equation and rearrange it so that we’re solving for the charge 𝑄. To do this, we can multiply both sides of the equation by the time 𝑡. The time cancels on the right-hand side, leaving us with 𝑡𝐼 is equal to 𝑄. So all that’s left now is to substitute in the values. We have 5400, which is the time in seconds, multiplied by 0.05, which is the current in amps, is equal to 𝑄. And hence, our final answer is that the charge flowing past 𝑃 in 1.5 hours when the circuit has a current of 50 milliamps is 270 coulombs.
