The circuit shown in Fig. E25.33

contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction)

Video transcript

Hey everyone in this problem. We have two non ideal power sources that are used to power the circuit shown, were asked to determine the magnitude and direction of the current in the circuit. Okay. All right. So when we're asked to find the magnitude and direction of the current, let's recall that we have Kirchoff slaw for voltage, which tells us that the sum of the voltage in this circuit that is a loop is going to be equal to zero. Now recall also that voltage is going to be equal to I The current times are the resistance according to law. So that gives us a way to relate this equation to the current I that we're looking for. Alright, so let's use this law and let's just go around a loop in our circuit. Now, we're going to assume that the current is traveling in this direction. K Counterclockwise, we're gonna do our calculation and then we're gonna see if our assumptions, right, okay. If the current is positive then it is indeed traveling in that direction. If the current is negative then it's actually traveling in the opposite direction. Okay, Alright, so let's start here. The first thing we're gonna pass when we're going counterclockwise, is this power source of 9.5 volts. Okay, we're going through the power source from the negative to the positive. And so this is going to be a positive 9.5V. Now we're taking the sum of all the voltages or all the potentials. Okay, so next thing we're gonna run into when we're going counterclockwise, is this resistor? Okay, It has 1.80m of resistance. And recall again homes law potential or voltage? Is I the current times are the resistance so we can write this as I the current times 1.8 times. OK. And you'll notice I put a negative. Okay. We're assuming that we're traveling in the direction of the current. Okay, so every time we meet a resistor that's going to be a negative value of negative potential. Okay. Alright. So we keep moving along we hit this other resistor 6.5. So again we're subtracting because we're in the direction of the current And then we have the current I times the resistance of 6.5. Os I times r gives us V. Moving along counterclockwise, another resistor 5.5 M So we have I times 5.5 ohms keep moving along another resistor of 1.2 homes so minus I times 1.2 OEMs. And we're running out of room here. So I'm just gonna write down below. And then the next thing and the last thing that we hit is this second power source. Okay, now we're going to this power source from the positive to the negative. Okay, so this is actually going to give us a negative voltage. So we have negative 4.5V And all of these things added together. Give us zero according to Kirchoff Slaw. Alright. So you'll notice in this equation that we have all of these eyes that are unknown and we know that the current through each of these resistors is going to be the same. So we can factor these eyes and solve for them because that's the only unknown in this equation. Okay, so we have 9.5V -4.5V. That's gonna give us 5V. Okay, we're gonna move everything else to the right hand side. So we're gonna have the current I times all of these resistance is 1.80ms plus 6. ohms plus 5. ohms plus 1.2. So this is 5V is equal to the current I times 15 homes if we add these all together. So when we divide we get that eye is going to be five volts divided by 15 OEMs. And when we're looking at units volt divided by almas. Gonna give us amp or ampere We have 5/15. Okay, both are divisible by five. We can simplify this to 1/3. Okay, okay. Up here. Okay, Alright. And what you'll see is that our current was positive in the end, that means that the direction of the current that we chose was the correct direction. Okay, we made the right assumption. So that means that the current is going to be traveling counterclockwise with a magnitude of 0.33 amps. So we have answer choice a 0.33 amps. Counterclockwise for the current in this circuit. Thanks everyone for watching. I hope this video helped you in the next one.