Video Transcript
Find the potential drop across the resistor in the circuit shown. The batteries powering the circuit each have a terminal voltage of 2.5 volts.
In our circuit, we see this resistor across which we want to know the potential drop. And we also see three batteries, each with a terminal voltage of 2.5 volts. One way we might think of solving for the potential drop across the resistor is to use Ohm’s law. Applied to our resistor, this law says that the potential drop across the resistor equals the current in it multiplied by its resistance. The trouble is we don’t know neither can we solve for either the current in the resistor or its resistance.
Ohm’s law then will not help us answer this particular question. What will be helpful to us though is another law called Kirchhoff’s voltage law. In words, this law says that in a closed circuit loop, the sum of all changes in potential is zero. In our given circuit, we can identify three separate closed loops. There’s one loop here through what we could call the upper part of our circuit. A second closed loop in the circuit is here. And a third closed loop is a loop that goes all the way around the perimeter.
Since we want to solve for the potential drop across this resistor, we’ll want to choose one of the loops that includes that resistor. The one we choose doesn’t matter; we’ll get the same answer either way. But just to pick one, let’s choose the outer loop in blue.
Kirchhoff’s voltage law tells us that if we travel in one complete circuit around this loop, then the sum of all changes in potential around that circuit is zero. To travel around this loop, we’ll want to pick a direction, either clockwise or counterclockwise. Our answer doesn’t depend on our choice; either direction will work within Kirchhoff’s voltage law. Just to choose a direction, let’s say that we move around this loop counterclockwise.
Picking any point in this loop as a start point, let’s say that we start here and we travel in a counterclockwise direction until we reach this 2.5-volt battery. We notice that the positive terminal of this battery faces left and the negative terminal faces to the right. Therefore, as we cross over this battery from positive to negative, this contributes a negative potential difference. That is, in this direction, the battery contributes negative 2.5 volts of potential.
Continuing on in our circuit, we arrive at our unknown resistor. We’ve seen that we can’t use Ohm’s law to determine the potential drop across this component. Related to this, we don’t know the direction in which current crosses this component during normal operation of our circuit. That is, charge could flow normally through this resistor from top to bottom or from bottom to top. Depending on that current direction, as we move through the resistor from bottom to top in our circuit, we will either add or subtract potential.
Let’s do this. Let’s call the magnitude of potential difference across our resistor 𝑉 sub 𝑅. As we’ve seen, since we don’t know which way current points in this circuit, we don’t know whether we’re adding or subtracting 𝑉 sub 𝑅 as we move around our loop. We can then include that uncertainty in our voltage law equation. Once we’ve crossed over the resistor in our circuit moving in a counterclockwise direction, we have negative 2.5 volts plus or minus the voltage across the resistor.
Let’s now continue on in our counterclockwise loop. And since this loop does travel around the perimeter of our circuit, we follow this upper branch and encounter another 2.5-volt battery. Unlike before, as we cross this battery, now we’re moving from the negative to the positive terminal. This battery, therefore, contributes positive 2.5 volts to our overall potential difference. Once we’ve passed through this component, we then complete our circuit loop without passing through any other circuit components. By Kirchhoff’s voltage law, we can write that the sum of these potential differences is equal to zero volts.
Notice that on the left-hand side of our equation, we have negative 2.5 volts and positive 2.5 volts. These potential differences will effectively cancel one another out so that our remaining equation reads plus or minus 𝑉 sub 𝑅 equals zero volts. We can now see that it doesn’t matter whether we use a positive or a negative sign in front of 𝑉 sub 𝑅 because this potential drop is equal to zero volts. The potential drop across the unknown resistor in our circuit is zero volts.
And note that we would have found the same result if we had moved around this outer loop in a clockwise direction rather than counterclockwise or if we had used the other loop in our circuit that included this resistor. And lastly, even once we established which particular closed loop we would work with and which direction we would travel, we were able to begin at any point within that closed loop. And as long as we completed one circuit, our answer would be the same.
