In the circuit shown in Fig. E26.31 the batteries have negligible...

Question

In the circuit shown in Fig. E26.31 the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 15.0 V.

(b) What will the ammeter read when the switch is closed?

Accepted Answer

Hi, everyone in this particular practice problem, we are asked to determine the meter reading when the 18 ohm resistor is replaced with a lower sister shunt When we have an experimental circuit which is shown in the circuit below. And we have all meters and all batteries to be ideal. And when we have the 1800m resistor to be plucked out, creating a gap, default meter will actually display 12 fault. And that is pretty much the only information that we have. So we want to pretty much start from there. Okay. So first, what we want to do is to actually determine the current going through the 10 ohm resistor using Homes Law. So we want to recall using OEM slaw that current will equal to the fold divided by the resistance. And in this case, we are given that when the 18 home here is plucked out, the four m will display 12 fold, 12 fold is going to be actually on the stand um here. So we have 12 fold and 10 and therefore our eye is going to be 1.2 amp Just like. So, so the total current in the circuit is 1.2 app. Next, we want to realize that the 40 member sister and The 41 resistor here and this two resistors in series are going through parallel to one another, especially when this is plucked out. So I'm gonna find the equivalent resistance for this side of the loop. Okay. So first, I want to find the equivalent resistance in series. So I'm just going to call that our series and recall that the resistance in series will just be the sum of the two resistance. So this will essentially just be 39 just like so okay. And then we want to find the equivalent resistor resistance for this whole thing. So in parallel, so are equally one over R equivalent is going to be one over R series Plus 1/40 So one over R E Q, it's going to equal 1/39 plus 1/40 which actually will, it corresponds to Um 79 over So actually in first thing that our EQ will then be 15 60 oh Over 79, which will then be 19.7. Okay. After we find the equivalent resistance for this right, set off for sister, what we want to find is the fault ege. So using home slow again, we know that V will equals two I multiply by our. So in this case, the eye we know is still 1.2 M while the resistance on the right side, right side is 19.7. So this is going to be 1.2 M Multiplied by 19.7. And that will corresponds to a v value of 23. volt, just like. So okay. Now that we know the potential there, we wanna use care chops law for the 10 Omer sister and the R E Q on all the right side, in order for us to actually determine the E okay. So using Carrot Slaw law, we will have this equation right here. So E minus the I multiplied by first, this R which is stan and then we will also have minus I multiplied by our key, Which is going to be dis overall heating, which is 19.7 which will equals to zero. We have actually calculated IR and I are a key and this will then be E minus, this is going to be 12 fold here because recall that this is essentially just I multiplied by R which is going to be this voltage value potential value here which is 12 fold minus. This is going to actually be this fee value that we just calculated 23.7 fold equals to zero. And therefore the E will equals to 35.7 fold just like so Okay. So considering a loop containing an e of 35. fold or a potential of 35.7 fold here I'm gonna this value here is 35.7 fold. And at the same time, we will have a 52 fold power sources. On this side, we can conclude that between 35.7 and 50. uh and 52 the 52 fold is actually going to be um more important in this case. So because of that, we can conclude that the 52 fold is actually going to be controlling the electric current going through the circuit. So the 52 fold will drive a clockwise current going through the right most loop here, which is the one that we're interested in because we're asked what the AM meter reading is. So in this right, most current here, right, most loop there will be a clockwise current going through the loop. Okay. I'm going to draw the current going. This is going to be essentially the eye going clockwise just like. So, okay. So since the 35.7 fold of the value that we just found here does not really influence the circuit will consider a loop only going through the 52 fold, going to the 25 to 14 To the meter. And remember that the 18-fold is going to be replaced with a lower system shunt. So that is essentially kind of negligible. So we will consider this loop right here going clockwise. Okay. So again, we want to use Care Shelves Law on Tier Shelves law on the right most Luke. So doing this, we have this is going to be E but essentially that is just going to be the 52 fold right here minus I multiplied by 25 minus I multiplied by 14. That will equals to zero because the 1800 is replaced with a low resistance shunt. And from here, we can actually kind of group this together. So 52 fold is actually going to equal to I multiplied by 25 M plus 14 M. So DI is essentially just going to be 52-fold divided by 39, which will be 1.3 Or 1.29 am just like so and that will be the answer to this particular practice problem. So that will correspond to option B that we have here. And that will be all for this particular practice problem. If you guys have any sort of confusion on this, please make sure to check out our other lessons, videos on similar topics, especially on rule of law and also just how to sum up their resistance is in series and in parallel and almost law as well. And that will be all for this video. Thank you.

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In the circuit shown in Fig. E26.31 the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 15.0 V.
(b) What will the ammeter read when the switch is closed?

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