The 10.00-V battery in Fig. E26.28 is removed from the circuit an...

Question

The 10.00-V battery in Fig. E26.28 is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure.  Find (a) the current in each branch.

Accepted Answer

Welcome back. Everyone in this problem. A student investigating how this current through a circuit depends on the polarity of a battery built the circuit shown below, we want to determine the amount of current that goes on each branch of the circuit. For our answer choices. A says the current in the upper branch is 0.671 S. The middle branch, 0.177 pires and the lower branch, 0.848 S B says it's 1.005 S 0.266 S and 1.27 S respectively. C says it's 2.01 S 0.531 S and 2.541 S respectively. And D says it's 2.68 S 0.708 pires and 3.388 pires respectively. Now, if we want to, we want to find the amount of current that goes on each branch. First, let's try to label it on our diagram here. So we want the current on each branch. Let's assume that these currents, the current in the upper branch is I one current in the lower branch in the middle branch is I two and in the lower branch is I three. And let's assume that I one and I two going in the same direction while I three is going in the opposite direction. I remember we can assume that because the sign will tell us what direction they're going in. OK. Now, once we've made this assumption here, there are two rules if we can recall, that can help us Kirkoff loop rule and the junction rule. Now recall that Kirkoff loop rule tells us that the sum of all the potential differences around the loop is equal to zero. So here, if we try to make two loops order for a circuit, here, we can have one loop. OK going in the counterclockwise direction for our nine volt current source. So let's call that loop one. And then we can have another loop going in the counter clockwise direction for our seven volt current, our seventh volt voltage source. Let's call that loop two. So with these two loops, we could come up with two algebraic equations from Kirk Off's loop root. So let's let's go ahead and do that. So by Kirkoff loop, by Kos loop rule for loop one or in loop one rather let me rewrite that by alert Kirkoff loop in loop one, eh then starting from our voltage source, we have our nine volts going in the same direction counterclockwise. So we have that gonna be positive. So we'll have nine volts. We have our current passing through our one ohm resistor going in the same direction as the loop. So that's minus one ohm. Multiplied by that current I one. So one I, one or just I, one really? But I'm going to leave it as one ohm, multiplied by I one. Just to make sure we're seeing that, then we come around and we're at our six ohm resistor where the current is going in the opposite direction to the loop. So that's gonna be plus six ohms multiplied by that current I two. OK. And then we come around and here we are at our 14 ohm resistor where the loop is going in the same direction as the current. So that will be minus 14 ohms multiplied by I one. And we're saying all of that is going to be equal to zero. We can simplify here because we have one I one minus one I one minus 14 I one. So that's minus 15 I one. So as a matter of fact here, well, let me keep it in black. So that's nine volts. OK. Minus 15 I one plus six I two is gonna be equal to zero. So what that tells us then is that 15 I one minus six I two are going to is going to be equal to nine votes. So from our first loop, we have an equation, let's call that equation one. No no equation. OK. So we have an equation. We have the first equation here from Kirkoff loop rule. Let's see if we can, we can find an equation from our second loop. OK. So from loop two, OK. Sorry. In loop two in loop two here starting from our going or, or starting from our current source. OK. Notice that our current source. OK? Or, or yeah, we or we could start from our six ohm resistor rather. So our six ohm resistor is going in the direction of the current. So let me put that in red. Actually, this is loop two in red. Then we'll have negative six ohms multiplied by current in the middle branch. When we come around. Notice that in our loop, our current is again going in the same direction as the loop. So that's minus one ohm multiplied by I one. That's a potential difference there or I three. Sorry, I three. OK. Now notice that our seven volts or, or voltage source is going against the direction of our loop. OK. So no, that means we're going to have going to have it. As now here, the current is flowing in the direction of the, the uh our voltage source. So that's plus several volts. OK. Then we go to our six ohm resistor in the same direction as the loop that has a minus six ohms multiplied by I three and all of that equals zero. Let me just put this in a circuit to show that it's different from our equation. OK. So now we can simplify loop two, right? Because here we have like terms negative six I two. Uh oh sorry, I mean negative I three minus six I three. So you can write that down as negative six I two minus seven I three plus seven volts equals zero. Then we can bring our current or, or, or terms with our currents in there to the uh to the, to the right hand side and the suture equation on our own, which will then leave us with six I two plus seven I three being equal to seven volts. So loop two gives us a second equation. But now if you notice when we compare both, we have three unknowns, two equations. So can we get another equation somehow? Well, we also have Kirkoff junction rule and Kirkoff junction rule tells us that the current entering a junction will be equal to the current that's leaving the junction. So let's see if we can use keras junction rule at or junction here where all these three currents meet. Let's call that junction F OK. Well, at Jong Shan F and let me put this information right here. OK. Add junction F buy Kirkoff junction rule. OK. We can say then that a current going into the junction I one plus I two, this is going to be equal to the current leaving the junction, which is I three. OK. So I one plus I two equals I three. And that is our third equation. So now that we have all three equations, all right, we have all three equations here. Let me highlight them. That means no, we can use them to figure out the coins in the upper middle and lower branch. First, let's use equation three to eliminate our third current in equation two. So we could replace I three in equation two with I one plus I two to eventually help us to get our equation in terms of I one and I two. When we do that, then we will be able to compare equation one to equation two because both of them will have the same terms. Let's actually do it. Let me stop talking. Let's let's do something. OK? So now substituting equations, substituting, don't tell anyone I didn't spell that properly. OK? Substituting equation three into equation two. OK? Then now we can rewrite the equation too as six I two plus seven multiplied by I two plus I three. Which when we expand that, that's gonna be six I two. Sorry. Uh I one plus I two. My bad I one plus I two, which we can expand that to be six I two plus seven I one plus seven I two and six I two plus seven I two equals 13 I two. So now our equation will be seven I one. OK plus 13 I two equals seven volts. Eh No, here now that we have it in terms of I one N I two, you can compare it to I one. So by comparison, OK, when we compare it to I one, our equation one, remember that equation one was 15, I two M I one minus six, I two equals nine volts. So that's 15. I won minus, let me put that properly here. 15, I won minus six I two equals nine volts. E and now when we compare them, notice that we could solve it simultaneously by trying to eliminate one of the variables. So uh let's say that we want to eliminate I one. OK, then we can multiply this equation by 15 and multiply equation one by seven and then we can subtract. So when we do that, OK, then down here we're going to have uh 715 multiplied by seven, which is 105. So that's 105 I one plus 13 multiplied by 15. That's 195 I to equals 105 votes equals 15, multiplied by seven. That equals 105. So that's 105 I one seven multiplied by negative six I two. That's negative uh 42 I two. OK. And that is gonna be equal to nine multiplied by seven, which is 63 volts. Now we can subtract one equation from the other and no 105 minus 105 Cancers I 1195 minus negative 42. That equals 237. Ok. That 237 I two and 105 minus 63. Ok, 105 minus 63. Let's let's do a little extra math here. So that's uh 45 minus two. That's actually 42. Hm. Yeah, 42 votes. Now here let me come over to the next side to save some space. This tells us then that I too is gonna be equal to 200 sorry 42 volts. OK? Divided by 237 ohms. Because remember this was still in terms of Ohms and that means I two is gonna be equal to 0.177 pairs. So we have a value for the current in the middle branch. Now let's use our first equation to solve for I two for I one because we know it's expressed in terms of I one and I two. So using equation on one to solve for I two or to solve for I one. OK. Then know by that equation we know that it was 15 I one minus six I two. So that's 15 I one minus six, multiplied by 0.177 empires. And all of that is going to be equal to our nine volts. So now let's multiply six by 0.177 S which is equal to 1.062. So no, that means 15 1 minus 1.062 uh is volts because that was six ohms multiplied by 0.177 M pairs that equals nine volts. Therefore, 15, I one equals nine plus 1.062 which is 10.062 volts. Therefore, I one is gonna be equal to 10.062 volts divided by 15. We, we divide both sides by 15. OK, Divided by 15 ohms which is equal to 0.671 empires. So now we have values for it too and we have it for I one. So finally, we can solve for I three. I remember that equation three tells us I three equals I one plus I two. So therefore, from equation three, from equation three, that means I three is going to be, let me put that in red. Actually, I three is going to be equal to I 10.671 pairs plus I 20.177 pairs which equals 0.848 pairs. So that would be the value of I three. So the current in the upper branch I one is 0.671 S in the middle branch, 0.177 pires and in the lower branch, 0.848 S OK. Which when we look back on our answer choices that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

The 10.00-V battery in Fig. E26.28 is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. Find (a) the current in each branch.

Key Concepts

Ohm's Law

Circuit Analysis

Polarity of Voltage Sources

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