The circuit shown in Fig. E25.30

contains two batteries, each with an emf and an internal resistance, and two resistors. Find (c) the potential difference Vac of point a with respect to point c.

Video transcript

Hey everyone in this problem. We have an experimental procedure that states that you assemble the circuit shown below using two non ideal batteries and three resistors were asked to determine the potential drop VPs across points P and S. Alright, so we're looking for the potential drop VPs and recall that this can be given by the potential at v minus the potential at S. Okay. Alright, so let's just consider moving from point P to point S. Okay. And we're going to do so in a counterclockwise direction like this. Alright, so going starting at point P. Ok, so we're gonna have the potential at point P. Okay, we're gonna run into this resistor of homes. Okay, so our potential here is positive. We're going counterclockwise. So the resisters we encounter those are going to be negative. Okay. And this is gonna be by the current Times the resistance which is 15 homes. Okay, remember that B is equal to I. R. Holmes law. Alright then we keep moving along and we run into another resistor. Okay, so again we're going to subtract and we have the potential V is going to be equal to I times the resistance are so we have I times 4.5% we keep moving along, we hit another resistor again subtracting I times The resistance 1.250ms. And then we're hitting this battery because it's not an ideal battery and we're moving through it from the positive to the negative end. Okay, so that's gonna be in the negative direction. So we're gonna have a negative five volts. Okay. And this all of this is going to get us to the potential at point S. So we go from point P. We have this initial potential at point P. We go through all of these resistors that are going to decrease our potential. Okay, we're going to end up with some potential at point S. Now we can move things around so that on the left hand side we have V P minus V. S. And on the right hand side we're gonna have 5V plus I times the sum of all of these resistance is OEMs plus 4.5 Ems plus 1.25 Ems. Okay, we can factor out that I because we know that the current is going to be the same through all of those resistors. Alright, so you'll notice on the left hand side, this is just VPs, that's what we're trying to find. We're trying to find VPs. So now we have an equation for V. P. S here, This is gonna be equal to 5V plus the current I times 20.75 bombs. Okay, now, in order to calculate this VPs, this potential drop across points P and S. We need to find the current. I All right, so how can we do that? Well, let's recall Geoff's law. Okay, for voltage where the sum of the potentials across the circuit is equal to zero. Okay, So we're gonna start up here. Okay with this 12 volt battery and we are going to take the same direction as before to be our direction of travel. So we're going in the counterclockwise direction. Mhm. Alright, so first thing we're going to run into is this battery, It has a potential of 12V and we're going through it from the negative to the positive direction. So that's gonna be positive potential. So we're going to have to give us a little bit more room but I don't want to get it so we can't see this diagram. So we have 12 volts there. Okay then we're going through this resistor that's drawn in red of 1.85 Okay, now because we're going this direction are resistors are going to be negative. Okay, we're going in the direction of the current And so we're gonna have I times the resistance which is 1.85. OK, recall that voltage is I times are so we have I times the resistance r. All right, let me keep going around this corner, we hit another resistor. 8.5 ohms. Again we're in the direction of the current. So we end up with I times 8.5 OEMs. Okay, carry along. Same thing 15. Um so subtract I times 15. OEMs keep going another resistor. I times 4.50ms okay we go up here to this blue resistor. Okay, which has a quantity of 1.25 ems so again minus I times 1.25. All of these quantities are negative because we're going in the direction of the current. Okay, and then the last thing we're going to go through this second battery in blue from positive to negative. Okay, so that's going to be a negative voltage or negative potential and it has a magnitude of five. Okay. Alright, so finishing off this equation -5V from that second battery and all of this is going to equal zero mm All right, so let's simplify. Okay, now you can see that we have these iVs in this equation. That's the only thing we don't know in the equation that's going to allow us to solve for I so we can get back to this potential drop here. VPs. Alright, so on the left hand side, Okay, we have 7V mm 7V. And on the right hand side we're gonna get the current I times the sum of all of these resistance is so 1.85 Homes plus 8.5 homes plus homes plus 4. plus 1.25. Okay, and just be careful with your signs here, it's a really easy place to mix things up. Okay, so just check your signs all the way through. Alright, so seven V is equal to I times 31.1 Os. Okay. And if we divide we are going to get that I the current is equal to 0.22508. Okay. The unit is going to BVV divided by um so we're going to end up with amps amperes. Okay. Alright. So there's our current I so now we can go back to our equation for the potential drop that we're looking for and use this value of I. Okay, so this is going to be equal to five volts plus 0. Amps Times 20.75 homes. If we work this out, This is going to give us a value of 9.67041. Okay. Alright. So that's gonna be our potential draw from point P. To point S. So we go back up to our answer choices and the question okay. We found that the potential drop VPs across points P and S. Was 9.67 volts Which is gonna be answer choice. A thanks everyone for watching. I hope this video helped you in the next one.