The 5.00-V battery in Fig. E26.28 is removed from the circuit and...

Question

The 5.00-V battery in Fig. E26.28 is removed from the circuit and replaced by a 15.00-V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure.  Find (a) the current in each branch.

Accepted Answer

Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The following circuit shown in the figure is built during a lab activity, calculate the current flowing through each of the branches ABC D and EF OK. So below our problem, they have provided us the circuit that they created during the lab activity in red, we have a resistor repre represented by a jagged line for 1.00 ohms. We also have a battery represented in red for a 5.00 volt battery. We also have a resistor on the right for 1.50 ohms. And this is all on the path branch for A B. And then for branch CD, we have a battery in blue for 7.50 volts and then a resistor for 0.50 ohms. And then we have another resistor in black for 2.00 ohms. And then finally, for branch ef we just have a resistor for 5.00 ohms. OK. We're also given some multiple choice answers. Let's read them off to see what our final answer set might be. A is for I A is equal to 1.20 D is equal to 1.30 years. IEF is equal to 0.10 years. And then for B it's I is equal to Sorry IAB is equal to 2.40. PI I CD is equal to 2.60 here. And IEF is equal to 0.20 years. C is I A is equal to 3.20 AM ID is equal to 2.60 pi and I ef is equal to 0.60. And finally D is I A is equal to 3.20 D is equal to 1.40 years and IEF is equal to 1.80 Pi Fantastic. So scrolling back up to look at our problem in our circuit diagram really quick. So first off, let us make the following assumptions. I one is to the left through the A B branch I two is to the right through the CD branch and I three is to the left through the EF branch. So the following figure which I've gone a gone ahead and drawn up a more detailed diagram. So to the right of the diagram that was provided to us in the problem is the diagram that I went ahead and you know, put it together for everyone to understand. So there's going in the counterclockwise direction. So for branch A B in a blue arrow, it's going to in the, to the left in the left direction. So it goes through the negative end of the battery towards the red resistor. And for loop one. So loop one focuses on branch A B and CD and Loop two focuses on branches CD and EF and for the EF branch, it's going to the left and then for the branch CD, it's going to the right. OK. So we'll go more in depth as we go on. But that's a quick little summary of what's going on here and a little visualization to what's going on. So this helps us visualize our assumptions that we just made. So to start solving this problem, let us start by using Kirchhoff's loop rule to solve for loop one by going around the loop in the counterclockwise direction. So going in the counterclockwise direction using Kirchhoff's loop rule, we can state that five point 00 volts since it's a five volt battery represented in red and since it's in the positive direction, it's positive, OK. Minus our resistor which is 1.00 arms multiplied by I one where eye represents the current just in case we're confused plus seven point 50 faults minus since it's going in the minus direction, as you see following our pathway here minus zero point 50 alms multiplied by I two minus 2.00 ohms multiplied by I two. Mine is 1.50. Arms multiplied by I won is equal to zero. Ok. Oh, that was a long one. So now we need to simplify this. And we can say when we simplify it, we can write 2.50 ohms multiplied by I one plus 2.50 ohms multiplied by I two is equal to 12.50 faults. So we can simplify this again to write I one plus I two. So this is simplifying and isolating I one and I two to just one side. So I one plus I two is equal to 5.00 here. So let's make a quick note here off to the side in blue that I the current in here is equal to the vaults D which is the charge divided by the resistance, which is in Ohms. OK. So we can state that our I one plus I two is equal to 5.00 pi we can call that equation one. So now we can start solving for our loop two. So using our Kirchhoff's loop rules again to solve for loop two following the same steps as we did before making sure we're following our loop direction as indicated by our arrows in our diagram. In this, in this case, it's going through. And then since it's going to the, to the right. It's gonna be negative. Ok. It's gonna be negative sir. Let's get cracking. So negative 7.50 volts plus 5.00. Ohms multiplied by I three plus two point 00. Alms multiplied by I two plus zero point 50. Alms multiplied by I two is equal to zero. So simplifying like we did before 2.50 ohms multiplied by I two plus 5.00. Ohms multiplied by I three is equal to 7.50. Oh OK. So doing like we did for equation one, let's isolate I two and I 3 to 1 side of the problem. So when we do that, we get I two plus I all right. So I two M so I two plus two multiplied by I three is equal to 3.00 here. And this will be equation two. So at this stage, we can apply the junction rule to point D to help us solve for equation three. So equation three, using the junction rule, we can state that I two is equal to I one plus I three. Now we must solve the system of three equations with three unknown values for I one I two and I three. So using equation three, we can write that I three is equal to I two minus I one. So now we can plug in the value for I three into equation two. So then we can write that I two plus two multiplied by I two minus I one is equal to 3.00 here. OK. So then we can finally solve for equation for. So equation four is negative two multiplied by I one plus three, multiplied by I two. Is he equal to three Aunt Piers? So now we can rewrite equation one to state that I one is equal to five PS minus I too. So now we can simplify this to solve for I two using some algebra. So I two is equal to 13.00 PS divided by 5.00 which is equal to 2.60 here. So now we can plug in the values for I one and I two into equation. OK. Wait, hold up. So now we can plug in the values I want and I two into equation three. But here, let's pause here for a second here. So we solve for I two. So now we need to plug in the value for I two into equation one. So yeah, we don't want to forget I one now. So we need to back back up a second and solve for I one. So I one is equal to 5.00 minus I two which is equal to 5.00 S minus 2.60 AMP here, which is equal to 2.40 amp. Years. Now, going back to what? Since I was skipping ahead for a brief second. I apologize. So now we can plug in the values for I one and I two into equation three. So now we know the values for I one and I two which let's quickly box those in green and these are part of our answers. So we don't wanna lose track of them. So plugging them back into equation three. So I three is equal to I two minus I one. So we can plug in our known values. So I three is equal to 2.60 minus all. Don't forget your unit. So 2.60 here minus 2.40 is equal to 0.20. And Piers, which is our final answer. Hooray, we did it. OK. So therefore the currents flowing through each branches. So for IAB, it's 2.40 years and then for I CD, it's 2.60 amp here. And then for IEF is equal to 0.20 amp here. And let's write that really quick. So IAB is 2.40 here. CD is equal to 2.6. Ers and IEF is zero point 20 and Piers. Awesome. So looking at our multiple choice answers, the correct answer has to be the letter B and letter B states that IAB is equal to 2.40 pi and I CD is equal to 2.60 pi and IEF is equal to 0.20. Pi thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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The 5.00-V battery in Fig. E26.28 is removed from the circuit and replaced by a 15.00-V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure. Find (a) the current in each branch.

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