(II) (a) What is the potential difference between points a and...

Question

(II) (a) What is the potential difference between points a and d in Fig. 26–55 (similar to Fig. 26–12, Example 26–8), and (b) what is the terminal voltage of each battery?

Accepted Answer

Hi everyone. Let's take a look at this practice problem dealing with circuits. This problem says a physics student designs the circuit as shown below. If the student quickly measures the potential difference between points P and M in the diagram, what value should they attain? Well, the question we're given the circuit diagram that was mentioned in the problem. We're also given four possible choices as our answers. For choice A, we have 82 volts for choice B, we have 19 volts. For choice C, we have 63 volts and for choice D, we have 65 volts. Now we asked to calculate the potential difference between points M and P. So the easiest way to do that would be measure the voltage across the 33 ohm resistor in the top branch of the circuit in the diagram. In order to do that, I need to figure out what the current I A is because that's the current that's going through that resistor. And we'll need that information in order to calculate the voltage using OHMS law. So in order to calculate that current, we're actually going to have to use K cost rules. So we're gonna have to use both our junction rule and our loop rule. So we're actually going to start off with our junction rule. And we're going to look at the junction labeled point P in the diagram. Now we call for Kerkhoff junction rule. That is the sum of the incoming currents has to be equal to the sum of the outgoing currents. So the incoming currents is just the current I see in the diagram that's could be equal to the two outgoing currents, which are I A plus IB. Now, we also need to look at Kirk's loop role and we're actually going to use two loops. So our first loop is gonna be the upper half branch of our circuit. We have that loop going in a counterclockwise direction. We call that loop one. Our other loop is going to go around the outer edge of the circle. And we're gonna be going in a clockwise direction around that loop and we'll call that loop too. So we need to write down the um loop rule for both of those loops. So starting for loop one, we're going to start at point M in our diagram which is just to the left of one of the batteries in the circuit. And so as I go around that loop going in a counterclockwise direction, the first element I'm gonna hit is our battery which is 65 volts. And that's a voltage gain sense of going from the negative terminal to the positive terminal. The next element that I hit is a one ohm resistor. And I'm going across that resistor in the same direction that the current is flowing through that resistor. So this could be a voltage drop. So I'll have minus and recall your um formula for the voltage drop across the resistor. That's just gonna be IC multiplied by 10. So is the current multiplied by the resistance? The next element that I come to is a 35 ohm resistor. Again, we're going in the same direction as the current that's flowing through it, which is IC. So we again have a voltage drop. So we'll have minus IC multiplied by 35 ohms and then continuing around the next element that I come to is the 33 ohm resistor. And again, we're going to be going to cross that resistor in the same direction that the current is flowing. And so it'll be again, a voltage drop. So we'll have minus and this time we have the current I a multiplied by 33 ohms. And if I continue around my loop, I get back to my starting point. So all of this has to be equal to zero, I can combine the two IC terms just to simplify things a little bit. And so doing that, I'll get 65 volts minus I see multiplied by and here I'll have one ohm plus 35 ohms which will give me 36 ohms. And then I'll have minus I a multiplied by 33 ohms and all that has to be equal to zero. So that's our equation for loop one. Now, moving on to loop two, here we're gonna start at point V in the diagram, which is just to the right of another one of the batteries in the circuit. And here we're going to be go going around in a clockwise direction. So the first element that I'm going to come to is the battery, that battery has a voltage of 85 volts. And I'm going from the negative terminal to the positive terminal when I go across that battery. So it'll be a positive 85 volts. The next element I'm gonna come to is another one ohm resistor. And here I'm going across that resistor in the same direction that the current is flowing. So it's gonna be a voltage drop. So I'll have minus IB multiplied by 10. The next element I come to is the 22 ohm resistor. And again, I'm going to cross that resistor in the same direction it's current. So it's gonna be a voltage drop. So I'll have minus IB multiplied by 22 ohms. And if I continue on around the next element that I want to come to is the 33 ohm resistor. But here I'm going to cross that resistor in the opposite direction that the current is flowing. So that means I'm gonna have a voltage gain. So I have a plus I a multiplied by 33 ohms. And if I continue around my loop, I'd get back to my starting point. So all of this has to be equal to zero. I can get simplified by combining the two IB terms. And so I have 85 volts minus IB multiplied by. And here I'll have one ohm plus 22 ohms, which will give me 23 ohms don't have plus I a multiplied by 33 ohms and all that has to be equal to zero. That's my equation for the second loop. So at this point, I have three equations with three unknowns and I need to solve them simultaneously. Now I'm ultimately looking for the current I A and so what I want to do is take my equations from loop one and solve it or IC, which will give me that in terms of I A for loop two, I'll solve it for IB which will give me my current in terms of I A again. And then plug both of those back into my junction rule um equation to get me just an equation that has only I A as the variable. So from loop one, we can see that in order to solve this for IC, I'll need to add IC multiplied by 36 ohms to both sides and then divide both sides by 36 ohms. When I do that I'll get IC is equal to 65 volts minus I a multiplied by 33 ohms and all of that is divided by 36 homes. Now, for Luke two, we can solve this for IB. And in order to do that, I will need to add IB multiplied by 23 ohms to both sides and then divide both sides by 23 ohms. That would give me IB is equal to 85 volts plus I a multiplied by 33 ohms. That whole quantity is divided by 23 ohms. So now I have IC and IB written in terms of IA I can plug this into our junction rule equation. So looking at our junction rule equation. So from P we'll see that we have for IC, we'll just go ahead and plug in our value that we just calculated. We'll have these 65 volts minus I a multiplied by 33 ohms divided by 36 homes. And that's gonna be equal to, we have I A plus IB but the place where IB will put the expression that we just solved for which is the 85 volts plus I A multiplied by 33 ohms. That's gonna be divided by 23 ohms. So at this point I need to solve for I A. So I'll just want to collect all those terms on one side of the equation. So I'll move all those terms to the right hand side and move everything else to the left hand side. So on the left hand side, I'm gonna have 65 volts divided by 36 ohms. They don't have a minus 85 volts divided by 23 ohms. And that's gonna be equal to. And here I'm going to factor out an I A out of all the terms on the left hand side have I a multiplying the quantity of one plus 33 ohms divided by 23 owns and then we'll have a plus. Again, there are three ohms divided by 36 homes. So now the software I A I'll just divide over um the quantity that is multiplying I A and so I'll have I A is equal to the quantity of 65 divided by 36. And they'll have volts divided by ohms which gives me units of AMPS. They'll have minus 85 divided by 23. Again, I'll have volts divided by ohms which give me units of AMPS. And that whole quantity is divided by the quantity of one plus 33 divided by 23 plus 33 divided by 36. And here the units of ohms and the denominator actually cancel out. So now I can actually take that and plug it into my calculator. And when I do that I get I A is equal to minus 0.564. Yeah. And here we just rounded to three significant figures. The minus sign there indicates that the current is actually going in the opposite direction that was indicated in our circuit diagram. Now, the question actually wanted to know the potential difference between points M and point P. So we'll have VM minus BP and that's gonna be equal to minus I a multiplied by the 33 ohm resistor. And so here we include the minus sign out in front because we're gonna start at point P and end up at point M. And in that case, we're going across resistor in the same direction as current. So it would technically be a voltage drop. However, when we actually plug in our value for our current, we'll have VM minus VP is equal to minus and then that's multiplied by the minus 0.564 S and that's multiplied by the 33 ohms. So the two negative signs will actually cancel out. And so our overall potential difference will actually be positive. And so we'll have VM minus VP, it's equal to 19 vaults. And here have just rounded to two significant figures. So if I come up to my answer choices, look at them, the correct answer corresponds to answer choice B. So just a quick recap of what we've done here. In order to find the potential difference between points M and point P, we first had to find the current running through the 33 ohm resistor. And in order to do that, we actually had to use KKO rules, we wrote one equation for the junction rule and then two equations for two different loops. We then had to solve them simultaneously to get the current I A that was labeled in our diagram. And once we did that, we just had to multiply that by the resistance in order to get the potential difference between the two points. So I hope that this has been useful and I'll see you in the next video.

(II) (a) What is the potential difference between points a and d in Fig. 26–55 (similar to Fig. 26–12, Example 26–8), and (b) what is the terminal voltage of each battery?

Key Concepts

Potential Difference

Terminal Voltage

Kirchhoff's Voltage Law

Watch next