### Video Transcript

A circuit is powered by a battery with an electromotive force of 3.6 volts. The circuit has a resistance of 5.5 ohms and the battery has an internal resistance of 0.75 ohms. What is the terminal voltage of the battery? Give your answer to one decimal place.

Here, we are being asked a question about a circuit. So a good place to start is by drawing a circuit diagram. The first thing we’re told about the circuit is that it’s powered by a battery, so we can start by drawing the circuit symbol for battery. We’re also told that the circuit has a resistance of 5.5 ohms. Now we haven’t actually been told any of the components in the rest of the circuit. However, because we know its resistance, we can actually represent it in our circuit diagram with a single 5.5-ohm resistor.

The other information that we’ve been given about the circuit is that the battery has an electromotive force of 3.6 volts and an internal resistance of 0.75 ohms. Now in order to use these pieces of information, we need to think carefully about how batteries behave. We can start by remembering that the role of a battery in a circuit is to provide a potential difference. And it’s this potential difference that creates a current in the other components and enables them to do their job. In this sense, a battery is a lot like a cell, which has a really similar circuit symbol.

Now when we talk about a cell, we’re usually talking about an ideal version of a cell. So it does its job of providing a potential difference, but to make our lives easier when we’re looking at circuits, we assume that it doesn’t have any resistance. However, when we talk about a battery, we’re often talking about a real battery, not just an ideal component. Now just like an ideal cell, a real battery will provide a potential difference. However, it will also have some resistance because the materials that we make batteries from in real life are not perfect conductors.

We call this resistance the internal resistance of the battery, and it has some important effects on how batteries behave in circuits. So batteries don’t behave like ideal cells. In fact, a battery behaves just like an ideal cell that’s connected in series with a fixed resistor. We can actually say that a real battery is equivalent to an ideal cell connected to a fixed resistor, where the resistor has a resistance equal to the internal resistance of the battery, which we usually denote with a lowercase 𝑟. And the cell has a potential difference equal to the electromotive force of the battery, which we usually denote with a lowercase Greek letter 𝜀.

In this question, we’re told that the battery in the circuit has an electromotive force of 3.6 volts and an internal resistance of 0.75 ohms. This means that the battery in our circuit is exactly equivalent to a cell with a potential difference of 3.6 volts connected to a 0.75-ohm resistor. This means that we can substitute these components in place of the battery in our circuit diagram like this.

Doing this, we can see that we have simplified the circuit that was given in the question. The question describes a mystery circuit with a resistance of 5.5 ohms that’s powered by a battery with a certain electromotive force and internal resistance. We’ve shown that the circuit described in the question behaves in exactly the same way as a circuit consisting of a cell with a potential difference of 3.6 volts connected in series with two resistors, one with a resistance of 0.75 ohms and one with a resistance of 5.5 ohms.

So now that we have this nice, clear circuit diagram, let’s take another look at the question. We’re being asked to find the terminal voltage of the battery. We should recall that the terminal voltage of a battery refers to the potential difference between the terminals of the battery when it’s connected to a circuit and charge is flowing. Since the battery in our circuit diagram is represented by these components, the terminal voltage therefore refers to the potential difference between this point and this point. And we can label this terminal voltage 𝑉 sub 𝑇.

There are actually a few different ways that we could work out the terminal voltage here. The method we’re going to use relies on Ohm’s law, which is expressed by the equation 𝑉 equals 𝐼 times 𝑅. This equation tells us that the potential difference 𝑉 across a component is equal to the current 𝐼 in that component multiplied by the resistance 𝑅 of that component. And we can apply Ohm’s law to any single resistor or group of resistors in a circuit. We’re going to apply Ohm’s law to both of the resistors in our circuit diagram at the same time.

So in this case, 𝑉 will be the total potential difference across both resistors, which we’ll call 𝑉 sub tot, 𝐼 will be the current in both resistors. And here, we should notice that all of the components in our circuit diagram lie on a single loop, which means that the current is the same at all points in the circuit. So in this case, we’ll just call the current 𝐼. Finally, 𝑅 will be the combined resistance of both the resistors we’ve drawn in our circuit diagram. And we’ll call this 𝑅 sub tot.

Applying Ohm’s law in this way will enable us to calculate the current in the circuit. To do this, we first need to rearrange the equation to make 𝐼 the subject. Dividing both sides of the equation by 𝑅 sub tot gives us this expression, and then we can just swap around the left and right sides to give us 𝐼 equals 𝑉 sub tot over 𝑅 sub tot.

Now looking at our circuit diagram, we can see that 𝑉 tot, that is, the potential difference across both resistors, must be equal to the potential difference that’s provided by this cell. So 𝑉 tot in this case is equal to the electromotive force of the battery, which is 3.6 volts.

Next, we can calculate 𝑅 tot, the combined resistance of both resistors, by considering the fact that they’re connected in series. We can recall that whenever we have resistors connected in series, the total resistance is given by the sum of the individual resistances. So for example, if we had three resistors connected in series with resistances 𝑅 one, 𝑅 two, and 𝑅 three, the total resistance of this combination, 𝑅 tot, will be equal to 𝑅 one plus 𝑅 two plus 𝑅 three. However, since in this question we’re just dealing with two resistors, the equation we use looks like this. So the total resistance of these two resistors in our circuit diagram is 0.75 ohms plus 5.5 ohms, which is equal to 6.25 ohms.

We can now evaluate this fraction, noting that we’ve used SI units for both potential difference in the numerator and resistance in the denominator. This means that we don’t need to do any unit conversions, and we’ll obtain a result measured in the SI units for current. 3.6 divided by 6.25 is 0.576. And the SI unit for current is the ampere or amp.

Okay, now that we found the current in the circuit, let’s make a note of this result and clear some space on screen. Now that we know the current at all points in our circuit, we can use Ohm’s law to find the potential difference across any of the resistors. So what we’re going to do is find the potential difference across this resistor, which represents the internal resistance of the battery. The potential difference of this resistor is effectively the amount of potential difference that’s lost within the battery due to its internal resistance. This is also known as the lost volts of the battery, which we can represent with the symbol 𝑉 sub L.

Using Ohm’s law, we can say that 𝑉 sub L, that is, the potential difference across this resistor that we’ve drawn, is equal to the current in that resistor, which in this case we’re calling 𝐼, multiplied by its resistance, which in this case we’ve denoted lowercase 𝑟 to represent the internal resistance of the battery. Since we know the values of 𝐼 and lowercase 𝑟, we can substitute these values straight in to give us 0.576 amps multiplied by 0.75 ohms. And once again, we can notice that we’re using SI units for both of these quantities. That’s the amp for current and the ohm for resistance. So again, we don’t need to do any unit conversions.

Evaluating this expression with a calculator gives us a result of 0.432 volts. Here, we should note that because this quantity, the lost volts, depends on the current in the circuit, it’s not a fixed property of the battery. The size of the lost volts for any given battery depends both on the electromotive force of the battery and on the circuit that it’s connected to. But now that we found this value, it’s relatively easy to calculate the terminal voltage.

We can say that the terminal voltage 𝑉𝑇 is equal to the electromotive force 𝜀 minus the lost volts 𝑉𝐿. We’re told in the question that the electromotive force produced by the battery is 3.6 volts. And we just calculated that the lost volts is equal to 0.432 volts. Substituting these into our expression, we have that the terminal voltage 𝑉𝑇 is equal to 3.6 volts minus 0.432 volts, which gives us a result of 3.168 volts.

All that’s left to do now is round our answer to one decimal place as specified in the question. 3.168 volts rounded to one decimal place gives us a final answer of 3.2 volts. If a circuit with a resistance of 5.5 ohms is powered by a battery with an electromotive force of 3.6 volts and an internal resistance of 0.75 ohms, the terminal voltage of the battery must be 3.2 volts.