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Question Video: Using Kirchoff’s Second Law to Calculate Voltage in a Circuit with Multiple Loops Physics

Find the decrease in potential across the resistor in the circuit shown. The batteries powering the circuit each have a terminal voltage of 2.5 V.

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Video Transcript

Find the decrease in potential across the resistor in the circuit shown. The batteries powering the circuit each have a terminal voltage of 2.5 volts.

Okay, so in this question, we have a circuit, which contains four components: a resistor and one, two, three batteries. The question tells us that the batteries powering the circuit each have a terminal voltage of 2.5 volts. And this is also shown on the diagram. Now, what the question wants us to do is find the decrease in potential across the resistor in the circuit. And we will label the potential difference across the resistor 𝑉 R.

Now, usually, when we’re working out the potential difference across a resistor, we use Ohm’s law, which tells us that the potential difference across a component in a circuit is equal to the current through that component multiplied by the resistance of that component. However, in this case, we don’t know the current through the resistor or even the resistance of the resistor. So we can’t use Ohm’s law to work out the answer to this question.

Instead, we can use a different law, namely, Kirchhoff’s second law, which tells us that the sum of the potential differences across each component in a loop in a circuit is zero. So this means that if we had 𝑁 components in a loop in a circuit and we added together the potential difference across each of these components, it would equal zero. We can use Kirchhoff’s second law to analyze the circuit in this question.

Our first step when using Kirchhoff’s second law is to identify the loops in the circuit. We can see that there is one loop that just contains the top two batteries. Then, there is a second loop containing the lower two batteries and the resistor and a third outer loop that contains the top and bottom batteries and the resistor. Now, remember that to answer this question, we need to calculate the potential difference across the resistor 𝑉 R. If we apply Kirchhoff’s law to one of the loops that contains the resistor, loop 2 or loop 3, then we will get an expression very similar to the one we had before containing 𝑉 R. We can then use that expression to calculate 𝑉 R itself.

Before we start, we should note that in Kirchhoff’s law, direction matters. Let’s say we had a battery with positive terminal on the left and negative terminal on the right. If we were to go across this from negative to positive, this would represent an increase in potential. So this potential difference would be positive in our calculations. However, if we were to go across the battery from positive terminal to negative terminal, this would be a decrease in potential. And therefore this potential difference would be negative in our calculations.

Something similar happens when we consider a resistor. We know that there is a decrease in potential across a resistor. However, we don’t know which direction this decrease is in. This means that we must take special care when we use Kirchhoff’s second law on a loop that contains a resistor.

Now, let’s go ahead and apply Kirchhoff’s second law to loop 3 of this circuit, which contains the upper and lower batteries and the resistor. As we said before, the direction we go around this loop matters, especially when dealing with resistors. To make things as simple as possible for ourselves, we will go counterclockwise around this loop. And by doing this, we make sure that when we go across each of the batteries, we’re going from the negative terminal to the positive terminal, meaning that each of these batteries will represent a positive value in our calculations. Similarly, we will take 𝑉 R to be positive in this counterclockwise direction.

Okay, now we’re ready to work our way around the loop and apply Kirchhoff’s second law to it. We will start at the positive terminal of the upper battery. And working our way around the loop in a counterclockwise direction, we reach the negative terminal of the lower battery. We are going across this battery from negative terminal to positive terminal. So this is a positive value when we use it in Kirchhoff’s second law. Continuing around the circuit, we reach the resistor. And we have said that 𝑉 R is positive in the direction we are traveling around this loop. So 𝑉 R is added to our expression. Continuing further around the loop, we reach the negative terminal of the upper battery. Once again, we are traveling from the negative terminal to the positive terminal of the battery. So this is positive and is, therefore, added to our expression.

We have now completed the loop. So we know that these three potential differences sum to zero volts. So Kirchhoff’s second law has given us an expression with one unknown, and that’s the potential difference across the resistor in the circuit. All we have to do to find out this potential difference is to rearrange the equation to make 𝑉 R the subject. The first thing we can note is that 2.5 volts plus 2.5 volts is just 5.0 volts. So this expression simplifies to 5.0 volts plus 𝑉 R is equal to zero volts. Next, we subtract 5.0 volts from both sides of the equation. And we see that five volts minus five volts is just zero volts, so these terms cancel. And zero volts minus five volts is just equal to negative five volts. And this gives us our value for the potential difference across the resistor.

However, the question asks for the decrease in potential, and this is just equal to five volts. So the decrease in potential across the resistor in the circuit shown is five volts. And we’ve worked this out by applying Kirchhoff’s second law to one of the loops in the circuit. However, we should also note that we could’ve applied this to the other loop in the circuit that also contains the resistor, the other loop being loop number 2.

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