Video Transcript
A circuit is powered by a battery with a terminal voltage of 2.5 volts. The circuit has a resistance of 3.5 ohms and the battery has an internal resistance of 0.65 ohms. What is the electromotive force of the battery? Give your answer to one decimal place.
Because this question asks us about a circuit, a good place to start is by drawing a circuit diagram. We’re told that this circuit is powered by a battery. So we’ll start with the circuit symbol for a battery. The question tells us that this battery powers a circuit and that this circuit has a resistance of 3.5 ohms. Now, we’re not actually told any of the components in this circuit. However, because we know its resistance, we can actually represent it in our circuit diagram just with a single 3.5-ohm resistor.
Looking back at the question, we’re also told that the battery has a terminal voltage of 2.5 volts. At this point, we need to recall that the terminal voltage of a battery is the potential difference between its positive and negative terminals when it’s connected to a circuit and charge is flowing. So in this case, the terminal voltage of our battery is the potential difference between two points located on either side of the battery. The final piece of information that we’re given in this question is that the battery has an internal resistance of 0.65 ohms. Now, in order to use this piece of information, we need to remember a bit about how batteries work.
We can recall that the role of a battery in a circuit is to provide a potential difference. In this sense, a battery is a lot like a cell, which we can see has a very similar circuit symbol. Now we use cells all the time when we’re drawing circuit diagrams and talking about how circuits work. However, usually, when we’re talking about a cell, we’re talking about an ideal version of a cell. This means that we assume the only thing a cell does in a circuit is provide a potential difference. This means that, usually, we work under the assumption that a cell doesn’t have any resistance. And this makes life much easier when we’re trying to calculate how circuits behave.
However, batteries are different. And usually, when we’re talking about batteries, we’re talking about real batteries. In real life, a battery does provide a potential difference. However, it also has some resistance, simply because the materials that we make batteries from aren’t perfect conductors. This resistance is called the internal resistance of the battery. And it has some important effects on how the battery behaves in a circuit.
So, whenever we see this phrase, internal resistance, we need to remember that we’re not dealing with an ideal cell or an ideal battery that doesn’t have any resistance. We’re talking about a real battery that does have some resistance. In order to account for this resistance, it can be useful to make it clear in our circuit diagram. And one way that we can do this is by drawing the battery in our circuit diagram as if it were an ideal cell connected in series to a fixed resistor.
In fact, a battery with some internal resistance is exactly equivalent to an ideal cell connected to a fixed resistor where the potential difference of this cell is equal to the electromotive force of the battery, denoted by the Greek letter 𝜀. And the resistance of this resistor is equal to the internal resistance of the battery, which is commonly denoted with a lowercase 𝑟. And at this point, it can be useful just to remind ourselves that even though electromotive force has force in the name, it’s actually just a potential difference, not a force. We could define electromotive force as being the potential difference across a battery when no charge is flowing. In other words, it’s the potential difference that we would measure across a battery’s terminals when it’s not connected to a circuit.
We actually find that as soon as a battery is connected to a circuit and charge starts to flow, the potential difference that we measure across its terminals decreases. This effect is due to the internal resistance of the battery. It causes a reduction in the total amount of potential difference that can be supplied to the circuit by the battery. We call this reduction in potential difference the lost volts of the battery. We can express all this using a simple equation. 𝑉 𝑇, that’s the terminal voltage of the battery measured when it is connected to a circuit and charge is flowing, is equal to 𝜀, the electromotive force of the battery, minus 𝑉 𝐿, the lost volts of the battery.
So the potential difference across a battery’s terminals when it’s connected to a circuit, in other words, the amount of potential difference that a battery supplies to a circuit, is equal to the electromotive force minus some amount of potential difference which is lost due to its internal resistance.
Now, in this question, we’ve been asked to find the electromotive force 𝜀. So let’s rearrange this equation to make 𝜀 the subject. If we add 𝑉 𝐿 to both sides of the equation, then we can see on the right-hand side of the equation, we have minus 𝑉 𝐿 plus 𝑉 𝐿. So these two terms cancel out, leaving us with this expression. Swapping the left- and right-hand sides around gives us 𝜀 equals 𝑉 𝐿 plus 𝑉 𝑇.
Now, in this question, we’ve actually been given the terminal voltage 𝑉 𝑇. We’re told in the question that that’s 2.5 volts. So, in order to find 𝜀, we need to work out the lost volts 𝑉 𝐿. Now, as we’ve said, the lost volts is the reduction in the potential difference provided by the battery due to its internal resistance. To help us understand this, remember that we’ve said a battery is equivalent to an ideal cell connected to a resistor. So we can imagine that inside any battery, there’s a cell which provides a certain potential difference known as the electromotive force and a resistor that opposes current and reduces the potential difference that we measure across the terminals of the battery. We can think of this resistor as using up some of the voltage.
So if we want to know the amount of potential difference that’s used up across this resistor, we’re in fact just looking for the potential difference across this resistor. And we can find this using Ohm’s law. This 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. We can apply Ohm’s law to the resistor that we’re using to represent the internal resistance of the battery. Doing so, we can say the potential difference across this resistor — that’s the lost volts 𝑉 𝐿 — is equal to the current in this resistor — since we don’t know what this is, we can just call it 𝐼 — multiplied by its resistance, which is the internal resistance of the battery represented by lowercase 𝑟.
Now, at this point, we run into another problem. We’re told in the question that the internal resistance of the battery is 0.65 ohms. However, we don’t know what the current in the battery is. We can notice that the circuit we’re dealing with is a single loop with components connected in series. This means that the current in the circuit, which we can draw going in the conventional direction from positive to negative, must be the same at every point in the circuit. So the current in the battery is the same as the current in the rest of the circuit, which is represented by this 3.5-ohm resistor.
We can actually work out the current by applying Ohm’s law to this resistor. This is because we know its resistance and because it’s connected directly to a battery with a terminal voltage of 2.5 volts. We know that the potential difference across this resistor must be 2.5 volts. Since we know these two variables, we can solve for the current 𝐼. So let’s start by making 𝐼 the subject of this equation. Dividing both sides of the equation by 𝑅 gives us 𝑉 over 𝑅 equals 𝐼. And then swapping around the left- and right-hand sides of the equation, we have 𝐼 equals 𝑉 over 𝑅.
Now, as we’ve noticed already, the voltage across this resistor is simply equal to the terminal voltage of the battery. So we can represent that 𝑉 𝑇. And 𝑅 in this equation is simply the resistance of the resistor. Note that we’re not using the internal resistance of the battery here, since at this point we’re not considering what’s happening inside the battery. We’re just applying Ohm’s law to the 3.5-ohm resistor in our circuit diagram.
We’re told in the question that the terminal voltage is 2.5 volts. And we’re told that the circuit we’re representing with this resistor has a resistance of 3.5 ohms. Notice that because we’ve given both of these quantities in SI units, we don’t need to do any unit conversions. Our answer will just be given using the SI unit for current. Using a calculator, 2.5 divided by 3.5 is 0.714 and so on. And the SI unit for current is the ampere or amp.
So we’ve now found the current in this part of the circuit. And because the entire circuit is a single-series loop, we know that this current is the same as the current inside the battery. So we can substitute this value for current into the equation that we’ve written for the lost volts. Doing so, we have 0.714 amps multiplied by the internal resistance of the battery, which we’re told is 0.65 ohms. Once again, both of these quantities are given in SI units, which means we don’t have to do any unit conversions. Evaluating this expression with a calculator gives us a value of 0.464 and a few more decimal places. And this is given in the SI units for potential difference, volts.
So we’ve now found the lost volts of the battery. We can say that because of its internal resistance, the battery loses 0.464 volts of potential difference when it’s connected to this circuit and that this leaves it with a terminal voltage of 2.5 volts. We’re now ready to calculate the electromotive force of the battery just by substituting these two values into this equation. So, 𝜀 is equal to 0.464 volts plus 2.5 volts. And adding both of these values together tells us that the electromotive force of the battery is 2.964 volts. The final thing to do is just round our answer to one decimal place. And 2.964 volts to one decimal place is 3.0 volts.
So, we’ve calculated that if a circuit with a resistance of 3.5 ohms is powered by a battery with a terminal voltage of 2.5 volts and an internal resistance of 0.65 ohms, then the electromotive force of the battery must be 3.0 volts.