Question Video: Calculating Potential Difference in a Combination Circuit | Nagwa Question Video: Calculating Potential Difference in a Combination Circuit | Nagwa

Question Video: Calculating Potential Difference in a Combination Circuit Physics • Third Year of Secondary School

The circuit shown has one resistor with an unknown potential drop and some other resistors with known potential drops. What is the potential drop across the unknown resistor? What is the potential difference between the points 𝐴 and 𝐵?

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Video Transcript

The circuit shown has one resistor with an unknown potential drop and some other resistors with known potential drops. What is the potential drop across the unknown resistor? What is the potential difference between the points 𝐴 and 𝐵?

Let’s start off with the first part of our question, which asks us about the potential drop across this unknown resistor in our circuit. We note that all the other resistors in the circuit have a potential drop across them indicated. This tells us how much electrical potential drops as charge crosses each one of these resistors.

To begin solving for the potential drop across the unknown resistor, let’s clear some space the top of our screen and let’s recall what is known as Kirchhoff’s voltage law. This law tells us that within any complete current loop, that is, within any closed circuit, the sum of all voltages across that loop is zero. In the circuit shown in our diagram, we have two such loops. The first loop completes the interior part of our circuit like this, and the second loop goes all the way around the outside through point 𝐴 like this. We could use either one of these two loops along with Kirchhoff’s voltage law to solve for the unknown potential drop across our resistor.

In fact, since this resistor here has a voltage drop equal to this resistor here, that is the two resistors in the parallel part of our circuit, whether we analyze the circuit by looking at the inner loop or the outer one, the process will be the same. Just to pick one, let’s choose the interior loop, the one that goes through point 𝐵 in our circuit. Let’s say we start by passing through our cell giving us 18 volts of potential difference. We then come to this first resistor, where, while crossing over it, we lose 5.5 volts. That makes our running tally of voltage 18 volts minus 5.5 volts. And then we follow this loop around until we reach this resistor. The voltage drop experienced here is 2.5 volts. And then, we follow this loop until we reach our unknown resistor.

Let’s call the potential drop experienced across this resistor 𝑉 sub 𝑢 because it’s unknown. We then reach this resistor with a 4.5-volt potential drop and, after passing through, come back to complete this loop of our circuit. By Kirchhoff’s voltage law, we can say that all of this is equal to zero, that is, zero volts.

We can now use this equation to solve for 𝑉 sub 𝑢, our unknown voltage drop. Let’s first notice that negative 5.5 minus 2.5 is negative 8.0 and then negative 8.0 minus 4.5 is equal to negative 12.5. If we now add 𝑉 sub 𝑢 to both sides of this equation so that 𝑉 sub 𝑢 minus 𝑉 sub 𝑢 equals zero on the left, we find that 18 volts minus 12.5 volts equals 𝑉 sub 𝑢. And that means 𝑉 sub 𝑢 is 5.5 volts. This is the potential drop across the unknown resistor in our circuit.

Moving on to the next part of our question, we want to know what is the potential difference between the points 𝐴 and 𝐵 in the circuit.

Notice that point 𝐴 and point 𝐵 in our circuit are not separated by any circuit components. The only thing that comes between these points is a segment of wire, which we assume to have zero resistance. That means that this amount of wire will not have any potential drop across it. And that means that whatever the electrical potential of the circuit at point 𝐵, it will have that same potential at point 𝐴. The potential difference between these two points then is zero.

Since the units of potential difference are volts, we’ll report our answer in those units. The potential difference between the points 𝐴 and 𝐵 in our circuit is zero volts.

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