Question Video: Identifying How the Current through a Point in a Circuit Changes Depending on the Paths through the Circuit | Nagwa Question Video: Identifying How the Current through a Point in a Circuit Changes Depending on the Paths through the Circuit | Nagwa

# Question Video: Identifying How the Current through a Point in a Circuit Changes Depending on the Paths through the Circuit Physics • Third Year of Secondary School

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Isabella sets up the circuit shown in the diagram. Each resistor has a resistance of 10 ฮฉ. The value of ๐ผโ is initially 3 A. What is the value of ๐ผ_total? If ๐โ is then closed, what happens to the value of ๐ผโ? With ๐โ still closed, is the resistance of the circuit lower than, higher than, or the same as it was when ๐โ was open? ๐โ and ๐โ are kept closed, but ๐โ is opened. Does the resistance in the circuit increase, decrease, or stay the same?

14:55

### Video Transcript

Isabella sets up the circuit shown in the diagram. Each resistor has a resistance of 10 ohms. The value of ๐ผ two is initially three amps. What is the value of ๐ผ total? If ๐ one is then closed, what happens to the value of ๐ผ two? With ๐ one still closed, is the resistance of the circuit lower than, higher than, or the same as it was when ๐ one was open? ๐ one and ๐ three are kept closed, but ๐ two is opened. Does the resistance in the circuit increase, decrease, or stay the same?

Okay, so there are lots of parts to this question, and the circuit itself is actually quite a complicated one. So the first thing that we can do is to label the diagram with the resistances of the resistors. Weโve been told in the question that each resistor has a resistance of 10 ohms. So we can label each resistor with 10 ohms.

Now the first sub question tells us that the value of ๐ผ two is initially three amps. In other words, the reading on this ammeter is three amps. What weโve been asked to do is to find out the value of ๐ผ total; thatโs the reading on this ammeter here. To do this, the best thing to do is to consider the flow of current in the circuit.

Now in our case, weโll consider conventional current. Thatโs the current that flows from the positive terminal of the cell to the negative terminal, so itโs flowing in this direction. Now this conventional current would travel through the ammeter that reads ๐ผ sub total and then reach this junction point here. At this junction, the current would potentially split. Normally, some would travel this way and the rest would travel this way.

Now the current traveling this way would just complete a loop and meet the current coming this way at this junction here, at which point those currents would be combined back into one final current that will go to the negative terminal of the cell. So is this whatโs happening in this circuit?

Well, in this situation no, because if we look carefully along this branch here, this switch is open. So because the circuit is not complete there, the current cannot flow along that branch. Instead, when the switch ๐ one is open, the only current that flows is down this way. Therefore, none of the current splits at this junction and so all of the current goes down this way. Hence, the current reading on the ammeter that reads ๐ผ total will be the same as the current reading on the ammeter that reads ๐ผ two.

Remember, this is only true when the switch ๐ one is open. However, the first part of the question is exactly this case. Weโve been told that the value of ๐ผ two is initially three amps. And initially, this circuit is in the orientation shown at the diagram. So initially, the value of ๐ผ total is going to be the same as ๐ผ two, and so our answer to the first part of the question is three amps.

Letโs move on to the next bit then. If ๐ one is then closed, what happens to the value of ๐ผ two? So letโs get rid of all the markings we made on the diagram for the previous part of the question. And letโs now consider what happens when ๐ one, the switch, is closed. Now if ๐ one is closed, then the argument that we used in the first part of the question cannot be true, because now that ๐ one is closed, there is a complete path for current to flow along this branch now as well.

So what happens to the conventional current? Well it starts out at the positive terminal of the cell, goes through ๐ผ total as usual, arrives at this junction, but this time it will split because there are two feasible paths for the current to go down. One is straight down this way, and the other is to the right as weโve already said.

Now what happens along this branch is a little bit complicated and weโll come back to that later. However, what we need to worry about this point in time is what happens to this current because this is the current thatโs going to flow through the ammeter that reads ๐ผ two. Now initially when the switch was open, all of the current was flowing this way. However, now that switch ๐ one has been closed, some current is flowing this way because some of it is also going this way.

Therefore, the current going this way is now less than what it was before. In other words, the current reading ๐ผ two is going to decrease. Now if we read the second subquestion carefully, then all weโve been asked to do is to find out what happens to the value of ๐ผ two. We donโt need to calculate it; we just need to say what happens to it. And the answer to that is that it decreases, so weโve got the answer to the second subquestion.

Letโs then move on to the third subquestion. With ๐ one still closed, is the resistance of the circuit lower than, higher than, or the same as it was when ๐ one was open? Okay, so letโs get rid of the markings of the diagram once again, and letโs now think about what weโve been asked to consider. What weโve been asked to find is what happens to the resistance of the circuit, the entire circuit, when the switch ๐ one is closed compared to the resistance of the circuit when the switch was open.

Now when the switch is open, this is an easy case to consider, because remember when the switch is open, there was no current flowing down this path. All of the current was flowing down this branch. And all we had was one loop of current going from the positive terminal of the cell around this ammeter through this resistor around this switch and back to the negative terminal of the cell. Now along this entire path, it was only passing through one resistor, this one.

And hence the resistance of the entire circuit in this situation was 10 ohms. So weโll write down in the bottom left of the screen the total resistance initially, which weโve called ๐ sub tot init, was 10 ohms, at which point we can go about working out the resistance of the circuit after this switch ๐ one was closed. Now this is a lot more complicated, so weโll need some space on the screen.

Letโs clear the first two subquestions. Okay, so now weโve cleared the space, we can work out the total resistance of the circuit when the switch ๐ one is closed. Notice that the question doesnโt necessarily need us to do this. Weโve only been asked to find out whether the resistance of the circuit is lower than, higher than, or the same as before.

However, the most intuitive way to understand whatโs happening to the circuit is to actually work out the resistance of the circuit. So letโs go about doing that. Okay, so this time when the switch ๐ one is closed, weโre going to have current flowing both this way along this branch and this way along this branch now. In this case, the two branches are in parallel. So letโs call this branch, the top branch, branch one. And this branch, the second branch, will be branch two.

Now we can break the circuit down into the two branches. And as weโll see later, weโll break down each branch into their components as well. So first of all, we already know the resistance of branch two. Weโve seen it before; itโs 10 ohms, because remember once again weโve assumed implicitly that the resistance of the ammeter is zero. And the only resistance that matters is the resistance of the resistor.

So letโs say that the resistance of branch two, which weโll call ๐ sub ๐ต sub two, the resistance of branch two is equal to 10 ohms. Now we can go about finding out the resistance of branch one. Now with branch one, itโs a little bit more complicated, because as current flows along branch one, it reaches this junction here. And the current once again splits. So some current will continue to travel to the right and some current will go down this way around this loop, at which point both of the currents will meet at this junction here and travel this way.

And remember, this is because the switch ๐ one is closed. So current actually can travel this way. So in the middle of branch one, weโve got two resistors that the current is going through: this resistor and this resistor. However, both of these resistors are in parallel. And once again, weโve assumed that there is no resistance for the ammeter. And of course the same is true for the switches because switches are just a piece of wire that complete the circuit.

Anyway, so the resistance of branch one is the resistance of 10 ohms and 10 ohms in parallel. So we need to recall the equation that gives us resistances in parallel. The equation in question tells us that if weโve got a certain number of resistances in parallel, then the total resistance ๐ is given by one over ๐ is equal to one over the first resistance plus one over the second resistance plus so on and so forth up to however many resistances we have.

Now in this case for branch one, weโve got two resistors. So the resistance of just branch one โ remember, weโre not calculating the full resistance of the circuit yet โ the resistance of just branch one is given by one divided by ๐ sub ๐ต sub one, the resistance of branch one, is equal to one divided by the resistance of the first resistor, which is 10 ohms, plus one divided by the resistance of the second resistor, which is also 10 ohms. And this is equal to two divided by 10.

And now if we find the reciprocal of both sides of this equation by flipping the fraction and the same for the right-hand side, we find that ๐ sub ๐ต sub one divided by one which is just ๐ sub ๐ต sub one is equal to 10 divided by two. And that is equal to five. And we know that unit is ohms. So we have the resistance of branch one now. As well as this, we already had the resistance of branch two so we can now work out the resistance of the whole circuit.

Now the whole circuit consists of branch one and branch two in parallel with each other. So once again we can use the parallel resistances formula to give us the resistance of the whole circuit. This time though, instead of applying it to this set of components here to find out the resistance of the branch, weโre applying it to the two branches, this one and this one, to work out the resistance of the entire circuit.

So we say that the resistance of the entire circuit is given by one divided by โ and weโll call the resistance of the entire circuit ๐ sub tot comma fin, because it stands for the total resistance at the finish which is when weโve closed the switch ๐ one. And remember, early we call the initial resistance ๐ sub tot comma init. So we say that this expression is given by one divided by the resistance of the first branch, which is five ohms as weโve worked out here, plus one over the resistance of the second branch, which is 10 ohms as weโve worked out here.

Now at this point, we can see that thereโs a common denominator of 10 because we can multiply both the top and bottom of this fraction by two, so we get two divided by 10 and then we can say two-tenths plus one-tenth is equal to three-tenths. Now that we have a single fraction on the right-hand side, we can find its reciprocal on both sides of the equation by flipping the fraction once again. We find that ๐ sub tot comma fin divided by one which is just ๐ sub tot comma fin is equal to 10 divided by three. And this is equal to 3.33 reoccurring ohms.

Now this is the resistance of the circuit after weโve closed the switch ๐ one. So the initial resistance of the circuit when switch ๐ one was open was 10 ohms. And when the switch was closed, it became 3.33 ohms. So the resistance of the circuit decreased. In other words, the answer to the third sub question is lower. The total resistance of the circuit is lower than when ๐ one was open, at which point then we can move on to the fourth subquestion.

This sub question says that ๐ one and ๐ three are kept closed, but ๐ two is opened. Does the resistance in the circuit increase, decrease, or stay the same? Okay, so once again we get rid of all the drawings on the diagram apart from the important stuff. And now we need to consider what happens to the circuit when the switch ๐ two is opened whilst ๐ one and ๐ three are still kept closed. In other words, switch ๐ two is no longer closed, itโs now open.

Letโs draw it this way. So letโs again consider what happens to the current. The current comes from the positive terminal then, goes this way, and reaches the first junction, at which point as usual some of it goes this way and some of it goes this way. Now when it reaches this junction, we would expect it to go this way and this way. However, because switch ๐ two is opened, it cannot go this way because there is no complete path for the current to go.

So instead, all of the current that was coming this way will end up going this way and then through switch ๐ one which, remember, is closed and back around the loop. And of course, all of this current would do the usual: go around the long way. So in this case, the current is going through only two resistors: this one and this one. But remember, these resistors are in parallel as well, because weโve got the first resistor on this branch and the second one on this branch.

So essentially, the difference between this question and the previous question is that switch ๐ two is open. So this set of components is no longer a set of components, itโs just this resistor here. Now before we closed switch ๐ two, we worked out the resistance was ๐ sub tot comma fin. Now fin might not have been the best word to use because this is not the final resistance of the circuit. Weโve still got one more to work out in this part of the question. However, weโll stick with it.

Weโll say that the resistance of the circuit was 3.33 ohms, so it was ๐ sub tot comma fin, and then we opened switch ๐ two at which point the resistance became ๐ sub tot comma END in capital letters to emphasize it actually is the final one. So we donโt need to consider this resistance of the circuit anymore. All we need to do is to consider the difference between the resistance before switch ๐ two was opened and after switch ๐ two is opened.

So what is the resistance after switch ๐ two was opened? Well once again letโs consider the resistances of branch one and branch two. Well, the resistance of branch one, ๐ sub ๐ต sub one, is just equal to 10 ohms because the current is only flowing through this resistor, which is 10 ohms. And so this entire part of the circuit is not relevant. And for branch two, once again itโs only 10 ohms cause thereโs only one resistor for it to flow through.

So ๐ sub ๐ต sub two is 10 ohms as well, once again. So whatโs the total resistance of the circuit? Well we use the formula for resistances in parallel once again. One divided by the total resistance is equal to one divided by the resistance of the first branch, which is 10 ohms, plus one divided by the resistance of the second branch, which is also 10 ohms. And one-tenth plus one-tenth is equal to two-tenths, at which one we have a single fraction on the right-hand side so we can find its reciprocal and the same with the left-hand side by flipping the fractions.

Doing this gives us ๐ sub tot comma end divided by one, which is the same as ๐ sub tot comma END, is equal to 10 divided by two, which is five. And letโs not forget the unit of ohms, at which one we say that ๐ sub tot comma end is equal to five ohms. So before switch ๐ two was opened, we had a resistance of the circuit of 3.33 ohms. After switch ๐ two was opened, it became five ohms. And remember, switches ๐ one and ๐ three were still kept closed. So we didnโt need to worry about them.

So whatโs happened to the resistance of the circuit after we opened switch ๐ two? Well itโs gone from 3.33 to five ohms; itโs increased. And so we say that when ๐ two is opened, the resistance of the circuit increases, at which point we found our final answer to the final part of the question.

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