Find the emfs ε_1 and ε_2 in the circuit of Fig. E26.26, and find...

Question

Find the emfs ε_1 and ε_2 in the circuit of Fig. E26.26, and find the potential difference of point b relative to point a.

Accepted Answer

Welcome back. Everyone. In this problem, we want to calculate the electromotive forces E one and E two and the electrical potential defense. VF minus V in the diagram shown below for our answer choices. A says E one is nine volts, three, E two is 3.5 volts and VF minus V is negative 6.5 volts. B says it's 12 volts, seven volts and 4.5 volts respectively. C says it's 14.5 volts, 6.5 volts and negative eight volts respectively. And D says it's 15 volts, 12 volts and 10 volts respectively. Now, if we want to find the electromotive forces E one and E two, we're going to need our current in all the branches to figure that out. Once we have the current, then we'll be able to find the voltages. Now, do we know anything that can help us to find voltages and currents in a circuit? Well, how about Kirk's loop rule? And Kirkoff junction rule recall that we can use both of them, we can use the junction rule to find currents and the loop rule to find potential differences. Recall that Kirk's junction rule tells us that the current entering a junction must be the same as the current leaving a junction. So by Kos junction rule, right, if we need, we need that current, that's in the middle branch, then by Kos junction rule that tells us then that at junction F so let me put that there. So at junction F OK, then at junction, if the current entering the junction which is 500 millions or 0.5 pires must be equal. OK? To the current leaving the junction which is our 250 million pi 0.25 pi that goes to the upper branch and the current in the middle branch, let's call that current I, OK. So if that's the case, then that means our current eye, I is gonna be equal to 0.5 amperes minus 0.25 amperes which is 0.25 amperes. So in other words here, we know that the current in our middle branch is 0.25 empires. So now that we have that current, OK? Now that we have that current, then we should be able to figure out our electromotive force E one and E two. And we can do that by using loop. Now here, if we think about it for our first loop, for our first loop here, we can look at E one, the loop that E one is in. So we could go in this direction. That's uh let me put that in black here, we could go in a counterclockwise direction for loop one and I'm not going to tell any lie. Sometimes I, I kind of mess up. I don't remember which one is clockwise or counterclockwise. So what helps me is the face of a clock? OK. Remember that? A clock because that's where clockwise comes from. Remember that if it's going in the direction of the face of the clock, it's clockwise, it's going the other way, it's counterclockwise. So now I remember so we can go in the counter clockwise direction to for loop one by ko loop route. OK. So if we do that, let me put that in red here. OK? So by ko loop rule, don't let me, OK, let me put that here properly. So buy Caro looper for loop one, we can find something that's going to include E one. That's our purpose for no. So let's try to include E one somewhere. So by ko loop rule for loop one going counterclockwise, OK, then our loop will say that we'll have our 10 volt source going in the direction of the current. So that's 10 volts. OK? Minus 0.25 multiplied by two ohms. That's a potential difference across a two ohm resistor. The loop is going in the same direction as the current. OK. So that's minus zero point or rather let me write, let me write it like this two ohms multiplied by 0.25 umpires. Next we come down to our eight ohm resistor. Now, notice we have the current going opposite to the direction of the loop. So it will be positive here. So we'll have plus 0.25 amperes multiplied by eight ohms. Ok. Next, we have our two arm resistor. Again, the current is going in the opposite direction. So that's plus two ohms multiplied by 0.2 file for the potential difference there. OK. Next, we have E one and it's going uh the, the current is going in the same direction as E one. So we have minus E one here. OK. And then we go to our 12 ohm resistor where the loop is going in the direction of the current again. So that's minus 12 ohms multiplied by 0.25 pairs. OK. And all of that is going to be equal to zero. No, what's pretty good here is that E one is our only unknown. So we should be able to solve for E one because we have the values for the rest of the potential differences and emfs. So by doing that here, let's write it a bit further. So that means we're going to have 10 words minus two, multiplied by 0.25 which is 0.5 volts. OK. Plus 0.25 multiplied by eight. which is that, that's basically a quarter of eight, which is uh two volts OK. So that's plus two volts, two multiplied by 0.25. Again, that's plus 0.5 volts minus E one K minus 12 multiplied by 0.25. That's a quarter of 12, which is three volts and all of that equals zero. If we combine those 10 minus 0.5 is 9.5 plus two, that's 11.5 plus 0.5 that's 12 minus three. That's nine. So nine volts minus E one equals zero, which means that E one is going to be equal to nine volts. So we have a value for E one. Next, let's see if we can use Kirk Off's loop rule to find a value for E two. Now, when we look at E two, here, we could come up with another loop, we could come up with a loop that goes around the outer branches. In other words, we could have a loop going in this direction. OK? And we could call that loop too. This is loop two. So let's see if we can figure out how to solve using loop solving for E two. So by ko rule bar by carers, loop rule for loop two. OK. Here again, notice that we're starting at our 10 volt source. That's the one we know. So we have 10 words and then OK, we have um we come around here, we have our two arm resist. So that is going in the same direction. As we as our current here. Ok. So here that's uh our 250 milliamp current. So that's 10 volts minus 10 points minus two ohms multiplied by 0.25 empires. Ok. Next, we have our two ohm resistor at the bottom beside the beside tube that's going in the same direction as our 500. Well, or it has the current for 500 mi uh current is going in the same direction. So that's minus two ohms multiplied by 0.5 pairs. Next, we have the potential difference across our four ohm resistor. So that's minus four ohms multiplied by 0.5 pairs. OK. And oops, it looks like I forgot E two, I forgot E two. But remember E two is here. So we still have minus E two. My apologies. So we have minus E two here. OK. Don't forget that. Don't do like me. And then we come up to our 12 ohm resistor which has our 0.25 pi cos so that's minus 0.25 pairs multiplied by 12 ohms. And all of that, we come back to our 10 volt source. So all of that is gonna be equal to zero. Again, we're in a good situation here because we have only one unknown E two. So by the loophole, we can start to simplify. And then so for E two, so we're gonna have 10 volts minus two multiply by 0.25 0.0 0.5 volts. They're multiplying ohms by amperes. Again, we'll have two multiplied by 0.5. That's one volt four multiplied by 0.5. That's two volts. Ok. Minus E two minus a quarter of 12 adds three volts. So all of that will be equal to zero 10 minus 0.59 0.5 minus 18.5 minus 26.5 minus three. That's 3.5. So 3.5 minus E two equals zero, which means that E two is gonna be equal to 3.5 volts. So we have values for E one and the E two. Now let's move on to the last part of our problem and it asks us to find the electrical potential difference between VF and V. So let's focus on V on that branch a bit or middle branch from F to E. Now, if we think about it here, if we were to find a potential difference, we can look at all of the potential differences from E to F and that is going to be able to give us what the potential difference is going to be. OK? So going along that branch, going along that mid branch there, let's isolate it. OK? So solving for, put that in blue, solving for VF minus VE OK, if we think about that branch, OK, starting at E I'm starting at point E, we had um an eight arm resistor check. Let me come up just to make sure. So we have our eight arm resistor and then we have our two arm resistor. OK? And then we have our EMF E one, OK? And then it went to F OK. Our eight ohms two ohms E one which we don't draw is nine volts and then it goes to F. So if we think about it here, if we were to look at the voltages starting from F, then that means the voltage at F OK, we're gonna have the voltage at F coming this way plus nine volts. OK. Minus two multiplied by the current. We know the current in this branch was 0.25 pires. OK? So that's minus two multiplied by 0.25 pires, right minus eight ohms and design ohms multiplied by 0.25 pairs. OK? And all of that is going to be equal to V. Now here, remember we just want the difference between VF and V. So if we simplify, we should be able to make that the subject of our equation, even though we might not know VF and V, we just want their difference. That's what really matters, the potential difference. So that means here VF minus V is going to be equal to whatever we had here. Or by the way, let me solve what? Let me simplify what's on this side first. So we have nine volts two multiplied by 0.25 that's minus 0.5 volts. And we have eight multiplied by 0.25 0.2 words, all of that equals ve nine minus 0.5 minus two equals 6.5. OK? So 6.5 volts equals V. So that tells us then that the difference between VF and V OK is negative 6.5 volts. So that's the potential difference between the junctions fef and EVF and V. So our potential difference is negative 6.5 volts. E two is 3.5 volts and E one is nine volts answer choice. A then would be the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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Find the emfs ε_1 and ε_2 in the circuit of Fig. E26.26, and find the potential difference of point b relative to point a.

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