Video Transcript
A battery has an electromotive force of 2.50 volts. The terminal voltage of the battery is 2.42 volts when it is connected to a circuit, and there is a current of 435 milliamperes in the circuit. What is the internal resistance of the battery? Give your answer to three decimal places.
Let’s say that this is our battery and that it’s connected to this electrical circuit. When the current in this circuit is 435 milliamperes, the terminal voltage of the battery, that is, the potential difference across the positive and negative terminals of the battery, is 2.42 volts. Our problem statement tells us that the battery has an electromotive force, which is also a potential difference, of 2.50 volts. This value is different from the terminal voltage because of the internal resistance of the battery. We can think of a battery as a cell, a supplier of potential difference, that has its own resistor built in. This is the internal resistance of the battery.
The battery generates an emf, an electromotive force, which is then diminished by the battery’s internal resistor. We can quantify that diminishment using Ohm’s law. This law says that the potential difference across the component in a circuit, such as a resistor, is equal to the current through that component multiplied by its resistance. Ohm’s law tells us then that the decrease in potential across the internal resistance of our battery is equal to the current through the battery multiplied by the battery’s internal resistance.
Often, internal resistance is represented by a lowercase 𝑟. If we call the potential difference across the internal resistor of our battery 𝑉 sub r, then by Ohm’s law, that’s equal to the current through the battery multiplied by the internal resistance, lowercase 𝑟. If we subtract 𝑉 sub r from the emf generated by a battery, then we get the terminal voltage of the battery. We’ve called this terminal voltage 𝑉. And we see that it’s 2.42 volts. Let’s now plug in everything we know about this equation. The terminal voltage 𝑉 is 2.42 volts, the emf is 2.50 volts, and 𝑉 sub r is equal to 𝐼 times 𝑟. And lastly, we know the value of 𝐼; it’s 435 milliamperes.
Knowing all this, we want to solve this equation for 𝑟, the internal resistance. First, we subtract 2.50 volts from each side so that positive 2.50 volts minus 2.50 volts on the right equals zero. Then on the left, negative 2.50 volts plus 2.42 volts equals negative 0.08 volts. We’re going to get to a point of dividing both sides of this remaining equation by our current.
But before we do, let’s convert the units of this current from milliamperes into amperes. 1000 milliamperes is equal to one ampere. So that means to convert milliamperes to amperes, we’ll divide by 1000. 435 divided by 1000 equals 0.435. That’s our current in units of amperes. Now we divide by our current so that negative 0.435 amperes cancels out on the right. And on the left, notice that the negative sign in the numerator will cancel with the negative sign in the denominator.
We arrived then at this simplified expression for the internal resistance 𝑟. Our numerator in units of volts divided by a denominator in units of amperes will give an answer in units of ohms. To three decimal places, 𝑟 is 0.184 ohms. This is the internal resistance of the battery.
