Video Transcript
Which of the following statements correctly relates the electromotive force ε of a battery to the current 𝐼 through the battery, the terminal voltage 𝑉 of the battery, and the internal resistance 𝑟 of the battery? (A) ε equals 𝑉 minus 𝐼𝑟, (B) ε equals 𝑉𝑟 plus 𝐼, (C) ε equals 𝑉 plus 𝐼𝑟, (D) 𝑉 equals ε𝐼𝑟, or (E) 𝑉 equals ε𝑟 plus 𝐼.
So, in this question, we’ve been given a bunch of different variables that relate to a battery. These are the electromotive force represented by the symbol ε, the current represented by the symbol 𝐼, the terminal voltage represented by the symbol 𝑉, and the internal resistance, which is represented by a lowercase 𝑟. We need to decide which of these expressions gives the correct relationship between these four variables. So, to start with, let’s remind ourselves of how batteries behave.
A battery is a device which converts chemical energy into electrical energy, and we can use a battery to provide a potential difference to a circuit. In this sense, a battery is a lot like an ideal cell, which we often use in circuit diagrams. However, there are some important differences between the two. Now, a cell or ideal cell is a kind of theoretical component. It provides a potential difference, and we assume that it doesn’t have any resistance. A battery, on the other hand, is a real device. It provides a potential difference. However, it does have some resistance. This means that when we’re dealing with a battery, we can’t really describe its behavior accurately by treating it like a cell.
In fact, because it has some resistance, a battery actually behaves like an ideal cell that’s connected to a fixed resistor in series. When we represent a battery in this way, this resistor represents the internal resistance of the battery, which we can denote with a lowercase 𝑟. This ideal cell represents the source of the electromotive force of the battery, which we denote with ε. Now, as well as the electromotive force and the internal resistance of the battery, this question asks us to consider the current through the battery. And in order for there to be a current, the battery must be connected to a circuit.
So let’s imagine that our battery is just connected in a simple series circuit with a resistor. With the battery connected in this way, we would find that charge starts to flow. In other words, there would be a current in the circuit. Let’s label this current 𝐼, noting that here we’re showing current going in the direction of conventional current, that is, from the positive to the negative terminal of the battery.
Now, since we’ve shown that a battery is effectively the same as an ideal cell and a resistor connected in series, we can draw these components inside our battery to show how it works. At this point, we’ve labeled three of the given quantities in our diagram. These are ε, the electromotive force of the battery; lowercase 𝑟, the internal resistance of the battery; and 𝐼, the current through the battery. The remaining quantity that’s mentioned in the question is the terminal voltage 𝑉. At this point, we need to remember that the terminal voltage is the voltage or potential difference between the positive and negative terminals of the battery when it’s connected to a circuit and charge is flowing. So we can label this quantity on our diagram like this.
Okay, so now we have these quantities represented and labeled in a circuit diagram, we need to figure out a relationship between all four of them. And we can do this by thinking about the effect of the internal resistance of the battery. Now, we know that connecting a battery to a circuit like this causes charge to flow. In other words, it creates a current. Importantly, this current isn’t only present in the circuit itself; it’s also present inside the battery. Now, we’ve shown that a battery behaves just like a cell and a resistor connected in series, which means that the current 𝐼 is effectively passing through a resistor here. Whenever there’s a current through a resistor, we see that there’s a potential difference across that resistor.
We can think of resistors as using up some of the potential difference which is supplied to the circuit. So, in the case of a battery, we can think of the electromotive force, which we’re representing with this cell, as being the maximum potential difference that we could ever measure across the battery. However, as soon as we connect the battery to a circuit and charge starts to flow, the internal resistance of the battery acts to effectively use up some of the potential difference and decrease the amount of potential difference that’s supplied to the rest of the circuit.
Now, we can calculate the exact size of this reduction in potential difference that’s caused by the battery’s internal resistance using Ohm’s law, which is represented by the equation 𝑉 equals 𝐼 times 𝑅. Ohm’s law 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. Let’s apply Ohm’s law to the resistor that we’re using to represent the internal resistance of the battery. So, in this case, 𝑉 will be the potential difference across this resistor. Because this is effectively the amount of potential difference which is lost due to the battery’s internal resistance, it’s often called the lost volts of the battery, which we can represent with the symbol 𝑉 sub L.
Next, the current 𝐼 in this case will be the current in this resistor, which is the same as the current in the rest of this circuit, and we’re just calling this 𝐼. And 𝑅 in this case will be the resistance of this resistor, which we’re representing with a lowercase 𝑟. So we’ve used Ohm’s law to show that the lost volts in the battery are equal to the current multiplied by the internal resistance. But how does this help us find a relationship between the four variables which we’re given in the question?
Well, if we consider the fact that the battery behaves just like an ideal cell connected to a resistor, then we know that the potential difference across the entire battery is equal to the potential difference provided by this cell minus the amount of potential difference that’s lost or used up by this resistor. In other words, the terminal voltage 𝑉 is equal to the electromotive force ε minus the lost volts 𝑉 sub L. And since we’ve shown that the lost volts are equal to 𝐼 times lowercase 𝑟, we can substitute 𝐼 times lowercase 𝑟 in place of 𝑉 sub L in this expression. This gives us 𝑉 equals ε minus 𝐼 times lowercase 𝑟. The terminal voltage of a battery is equal to the electromotive force of that battery minus the current through the battery times its internal resistance.
Now, as it stands, this expression doesn’t exactly match any of the answer options which we’ve been given, but we can fix this by rearranging it. If we add 𝐼𝑟 to both sides of this expression, we obtain 𝑉 plus 𝐼𝑟 equals ε. And then, swapping the left- and right-hand sides of this equation round, we obtain ε equals 𝑉 plus 𝐼𝑟, which we can see matches option (C). So this is our final answer. If we have a battery that has electromotive force ε, terminal voltage 𝑉, and internal resistance 𝑟 that’s experiencing a current 𝐼, then these four quantities are related by the equation ε equals 𝑉 plus 𝐼𝑟.
