Video Transcript
A battery with an electromotive force of 4.50 volts is connected to a circuit with a resistance of 2.75 ohms. The current in the circuit is 1.36 amps. What is the internal resistance of the battery? Give your answer to two decimal places.
Since this question involves a circuit, let’s start by drawing a circuit diagram. We’re told that the circuit involves a battery. So we can start with the circuit symbol for a battery. We’re then told that this battery is connected to a circuit with a resistance of 2.75 ohms. Now, we haven’t actually been told any of the components in this circuit. However, since we know the resistance of the entire thing, we can just treat the rest of the circuit as if it were a single 2.75-ohm resistor. The question also tells us that the current in the circuit is 1.36 amps. So we can label this current in our circuit diagram, noting that because the circuit is a single loop, the current will be the same at every point in the circuit.
We can also note that we’ve drawn our arrows pointing in the direction of conventional current, which flows from the positive terminal of the battery to the negative terminal. This direction is the opposite direction to the flow of electrons. Although the direction that we say current is going doesn’t actually change how we solve this question. The final piece of information that we’re given in this question is that the electromotive force of the battery is 4.50 volts. Now at this point, it might be tempting to label our battery with this potential difference. However, it’s important to remember that the electromotive force of a battery is not the same as its potential difference when it’s in a circuit. So we’ll avoid labeling our battery for now.
The question is asking us to calculate the internal resistance of the battery. In order to do this, we need to think carefully about how batteries work. We can recall that the role of a battery is to provide a potential difference to a circuit. And this potential difference is what creates a current in the circuit. In this sense, a battery is a lot like a cell, which has a very similar circuit symbol. However, there are some important differences between these two components.
A cell, or ideal cell, is a kind of theoretical component that we commonly use in circuit diagrams. An ideal cell provides a potential difference in a circuit. And crucially, we treat it as if it doesn’t have any resistance. It’s important to note that while ideal cells are really common in circuit diagrams, they don’t actually exist in real life. This is where batteries come in. A battery is a real device, and it converts chemical energy into electrical energy in order to basically perform the role of a cell in a real circuit. So a battery provides a potential difference. But unlike an ideal cell, it does have some electrical resistance. And we call this its internal resistance.
So, because a battery is basically the same as an ideal cell but with resistance, we can actually treat a battery as if it were an ideal cell connected to a fixed resistor. Specifically, the resistance of this resistor would be equal to the internal resistance of the battery, which we usually denote with a lowercase 𝑟. And the potential difference of this cell would be equal to the electromotive force of the battery, denoted by the Greek letter 𝜀. And we’ve been told that, in this case, the electromotive force of the battery is 4.50 volts. Since the battery in our question is equivalent to this combination of a cell and a resistor, we can substitute these components for the battery in our circuit diagram.
So, we’ve now shown that the description of the circuit in the question is equivalent to this circuit diagram. What we need to work out then is the resistance of the resistor that we’ve drawn here, which represents the internal resistance of the battery. We can do this by using Ohm’s law, which is expressed by the equation 𝑉 equals 𝐼𝑅. This tells us that the potential difference across a component 𝑉 is equal to the current in that component, 𝐼, multiplied by the resistance of that component, 𝑅. We can apply Ohm’s law to any resistor or group of resistors in a circuit.
Now, there are at least a couple of ways of approaching this problem. But what we’re going to do is apply Ohm’s law to both of the resistors in a circuit at the same time. In this case, Ohm’s law tells us that the potential difference across both resistors is equal to the current through both resistors multiplied by the resistance of both resistors. So for this specific case, the potential difference we’re dealing with is actually the electromotive force 𝜀 of the battery, since this provides a potential difference across both resistors in our circuit diagram. The current is the same at all points in the circuit, so we can just call this 𝐼. And the resistance 𝑅 that we’re dealing with is the combined resistance of the circuit and the internal resistance of the battery. We can call this total resistance 𝑅 tot.
We can recall that when we have several resistors connected in series, the total resistance 𝑅 tot is equal to the sum of the individual resistances of all the resistors that are connected in series. So, for three or more resistors, the general equation we use to work out the total resistance of series resistors looks like this. But in this case, since we only have two resistors connected in series, we can simplify the equation a bit like this. In this case, the two resistances that we’re adding together are the internal resistance of the cell, represented by lowercase 𝑟, and the resistance of the rest of the circuit, which we’ve been told is 2.75 ohms. For now, let’s call this resistance capital 𝑅. This means that the total resistance of both of the resistors in our circuit is equal to capital 𝑅 plus lowercase 𝑟.
We can substitute this into our statement of Ohm’s law to give us 𝜀 equals 𝐼 times capital 𝑅 plus lowercase 𝑟. So using Ohm’s law and the rule for series resistors, we’ve now obtained an equation that contains all of the known quantities in the problem, namely, the electromotive force, the current in the circuit, and the resistance of the rest of the circuit and the unknown quantity that we’re looking for, which is the internal resistance of the battery. All we need to do then is rearrange this equation to make lowercase 𝑟 the subject and then substitute in the known values.
The first step in rearranging this equation is to divide both sides of the equation by 𝐼. And doing this, we can see that a factor of 𝐼 in the numerator and the denominator on the right-hand side will cancel out, leaving us with 𝜀 over 𝐼 equals capital 𝑅 plus lowercase 𝑟. At this point, we can get rid of the parentheses and then subtract capital 𝑅 from both sides of the equation to give us 𝜀 over 𝐼 minus capital 𝑅 equals lowercase 𝑟. Finally, we’ll just swap the left- and right-hand sides of the equation round. All that’s left to do now is substitute in the values of the 𝜀, 𝐼, and capital 𝑅. So the value of 𝜀, that’s the electromotive force of the battery, is 4.50 volts. And the value of 𝐼, that’s the current in the circuit, is 1.36 amps. We then need to subtract capital 𝑅, that’s the resistance of the rest of the circuit, which is 2.75 ohms.
And at this point, we can note that because all the quantities use SI units, we don’t need to perform any unit conversions. Because this calculation will tell us a resistance, it will be given in ohms. Evaluating this fraction first, 4.50 volts divided by 1.36 amps gives us 3.309 and then some more decimal places, and this quantity is measured in ohms. Then, subtracting 2.75 ohms from this value gives us a result of 0.559 and so on ohms. The only thing left to do now is round this value to two decimal places as specified in the question, giving us a final answer for the internal resistance of the battery of 0.56 ohms.
If a battery with an electromotive force of 4.50 volts is connected to a circuit with a resistance of 2.75 ohms and the current in the circuit is 1.36 amps, then we know that the internal resistance of the battery must be 0.56 ohms.
