University Physics 2 Exam 2 Solutions Fall 2025

Hi everybody, it's Dr. Shabbo and I'm going to talk through the solutions for the exam 2 in the fall of 2025 for university physics 2. All right, so the first question is what is the equivalent unit of farad? and farad remember is the unit for capacitance and capacitance is fundamentally how much charge something something stir stores per volt. and so charge per volt is per and the units of that are kms here per volt. So kms per volt is choice In the second question, we have done this in the lab. You measure equipotential lines. So these are lines of constant voltage. and you measure the distance between two lines at the location indicated. What's the approximate magnitude of the direction and direction of the average electric field at point So at point we do know that the electric field is going to point downhill and it's going to be towards lower potential and it's going to be perpendicular to the field lines at every point. And so we know at point it's going to be to the right. And so the choice is to the right. And then in terms of the magnitude an approximation for the electric field remember is that it can be written in terms of the volts per meter. and so that's going to be delta over delta And so here I'm going to take the delta between the two adjacent lines. That's 2 volts. And then over the And do want to put this in terms of meters. So 0.05 And so get 40. And could write that as volts per meter. Or unit of electric field is also newtons per km. So we are looking at choice for newtons per km. 40 newtons per kum towards the right. All right. So for number three here, this is RC circuit. And the RC circuit has the battery essentially the power supply has voltage of 5 and we're currently starting with an initially uncharged capacitor at t=0. And then we're going to close the switch. And want to find the total current through the battery after the switch has been closed for long time. So notice that that total is going to be equal to the sum of two currents. So one of the currents at this first junction, one of the currents is going to go through this loop with the capacitor at it. and I'm going to call that just I1 since it passes through R1. And then I'm also going to have current that passes through this branch. and then they are going to rejoin at this junction and com combine back into this total. So my total over here is the sum of I1 plus I2 and so we have to look at each of those currents separately. So here we have the branch I2. If we look at the branch I2 first, there is no capacitor in that branch and we simply have that delta across it is going to be 5 The resistance R2 is 1 kiloohm. And so we then can say delta across 2 is equal to I2 R2. and we end up getting the result that I2 equals 5 milliamps. So, we're going to hold on to that for second. And now we need to figure out what I1 is. I1 is going to be the current through the branch with the capacitor. So, this is an IC branch. and we're going to be charging the capacitor. So initially it starts uncharged and then we flip the switch and then the current flows for while and then it decays down. So there will be some current flow for while when first flip the switch. and can write the current through that capacitor branch which is I1 as function of is going to be equal to its IAX value which would actually be the same 5 millias* minus over tow. Now for this problem though we do know right that after long time what's going to happen is it drops to zero. So one goes to zero for long time. So when we add up to find total we see that total after very long time is simply going to be the I2 and we get choice Okay, let's go ahead on to number four. An isolated parallel plate capacitor is charged by connecting it to battery. Has dialectric material with dialectric constant cappa= 4 is inserted to fill the area between the plates. Once fully charged, it stores charge initial and then it stays connected to the battery. So staying connected to the battery means that my initial is going to be fixed. And so that's going to equal my Qf final no matter what do because that charge is stuck on that battery. and so now I'm going to remove the dialectric. So we have this equation for the capacitance of parallel plate. It's kappa epsilon over so initial is going to be equal to something and then that's going to just be equal to my 4 epsilon over And then final I'm removing that capacitor. So actually see that final is actually equal to initial divided by 4. So the final capacitance goes down goes down by factor of four because that's the dialectric. and want to know the new charge stored on the capacitor. And even though the capacitance went down, what that means is that the voltage will actually go up. So the new charge stored on the capacitor is going to be the same charge that originally stored. The voltage if go through the the process here, Q= CV. Then Vfinal is actually going to be equal to 4 initial. but it didn't ask about that. It asked about the charge on the capacitor plates. So the new charge stored remains unchanged because it has nowhere to go. but if it had asked about voltage, then that would be four times the voltage. Okay. Number five. have two point charges plus and minus 2 They are located distance apart. How much work must an external agent like me do in order to move them the increase the distance to 3D. So not drawn to scale obviously running out of room here. but what would be the work done by me to pull them apart? Well, first if we just do conceptual check, they are moving they want to be close together because they are attracted to each other. plus and minus. So they're attracted to each other. And then when pull them apart, I'm going making them go the opposite of the way they want to go. so I'm moving them opposite how they want to go. So the analogy is it's like lifting box in gravitational field. do positive work when lift that box, right? I'm moving them opposite how they want. So what expect is that am going to do positive work. So that's going to be conceptual check at the end that I'm moving them up so the way they want. I'm lifting the box in the gravitational field. Mathematically, we know that the work by an external agent is going to be equal to the change in the potential energy. And the potential energy of pair is going to be equal to here if do final minus initial ufal is going to be equal to q1 q2 over final minus q1 q2 / initial. And so if plug in my values here, get and then they have you know Q1 is positive Q2 is minus 2 and keep that minus sign and then final is 3D and then minus and then and then min - 2 all over my original And so when work through this what find is indeed do get positive number and the positive number is equal to 4/3 ^2 overd. and so if we see why that is can make this smaller to give myself little bit more space. then can say that this is equal to 2 ^2 that's negative over 3d and then minus negative. So this is now plus 2 ^2 / And if make those fractions work out, this becomes 6 over 3. And end up here with my 4/3 ^2 over All right. So moving on to number six. for number six we have this function for potential electric potential that varies with and And so we want to find the electric field in that region. And so we know that the electric field is equal to negative dell dot or sorry dell * and so we have essentially the gradient of the potential here. And so we can write that out as negative partial partial and that's in the ihat direction minus don't have component so I'm not going to use that one minus partial partial in the khat direction. When take partial derivative do it as if everything else in that equation is constant. So have minus and if I'm taking the partial derivative of that with respect to I'm treating as my constant. So get minus and then are my constants * 3 x^2 for that first partial derivative. And then the next term has no in it. So I'm going to get zero for the partial derivative with respect to of that term with bzz ^2 in it. So that's my ihat component. And then that's going to be minus my partial partial So now treat the terms as constants. So minus cubed partial partial partial is actually just one here. and then minus 2 and that's my hat component here. So then just look to see which one matches that and see that it is indeed the first one here that matches. Okay. So here we have the resistors and they are shown here. And key idea of this picture is that this is parallel configuration meaning that everything on this left side is all at the same positive voltage that you get from the positive side of the battery. and on the left side of the battery then everything that is being highlighted when this switch is closed everything that's highlighted here is also at the same voltage as the negative side of the battery. So fundamentally the voltage across this top branch the delta across BC at all times delta across BC equals the voltage of the battery whether or not the switch is open because what happens in that parallel branch branch doesn't impact the other parallel branch. in household, this is you being able to turn on light and plug in light or unplug light and then, you know, the TV still works because they're wired in parallel. They're both still getting the full amount. and so for this one here, we would see that resistor which is in that BC branch, is going to have the same power as before. Nothing changes in that parallel branch. nothing at all. not even the current through it, right? Because we would have that delta VBC BC equals the current through that BC branch times the resistance of that BC branch. haven't changed any of those. So therefore, the power hasn't changed either. It's key concept of circuits. in terms of thinking about the current through the circuit, the current through the total circuit does change. So, total, if we look here, and I'm going to use different color so you don't think I'm talking about the voltage anymore. But if look here at my total, my total does change because now my equivalent resistance of the circuit has gone down. I'm sending more current out from the battery. This makes total sense because plugged in lamp and my TV is still working, right? and so if I'm going to have my current through my BC branch, which is unchanged, now also might have current through sub And then my total current is going to be the sum of those. So if you looked at it through current lens, you would still ultimately have to see that it's unchanged because that voltage across BC is unchanged. Okay, let's go ahead and talk about number eight. So in number eight, wire of resistivity row must be replaced in circuit by wire of the same material. So that means that it's going to be the same row that is four times as long. If the total resistance is to remain as before, what be the what must be the radius of the new wire? So remember that the area it's the cross-sectional area. So it's pi 2. So when we use this equation that equals row over then initial right we want to have the the same amount here. So if have some initial initial then that's going to have to equal my row lfal final because the row is the same in both. So basically if my final equals 4 initial then can see that what get is if put in 4 initial for my final then get that my final cross everything else off is going to have to equal 4 * my initial. So need four * the area. and so 4 * the area equ= 2 * the radius in because of that squared term. So for number eight then we get the choice All right number nine the electric field in region is found to be zero. What can you say about the electric potential in this region? So this is key part of this equation which is that delta v= integral dot Now the key part here is the delta the fact that it's the change in potential. so what we see is that that means that my change in potential is going to be equal to zero when plug in zero electric field. It does not mean that have zero potential. It hasn't doesn't have to do with the slope. It means that have no change in my potential so that I'm essentially at that same energy. The analogy to gravitational field, I'm at the same height. It doesn't mean that I'm at the ground floor. It just means that I'm at the same height in my gravitational field. So have no change in potential. and so that choice is here that the electric potential is constant value everywhere. So it is not changing but it is not necessarily zero. All right. Number 10. resistor carries this time varying current that's given here and is positive constant. how much total energy is dissipated by the resistor over the time interval 0 to capital so couple ideas here. One is that power is equal to I^2 This comes from IV and V= IIR. and we know that power is simply equal to energy dissipated per time, right? Work delivered per time which is energy per time. And we can write that as du dt. So from that we see that we can get that the energy delivered is the integral of the power dt. And so then can plug in for this i^2 So would get my I'm going to square that time which was given in the problem. And I'm going to integrate that from 0 to capital And so when do that, see that I'm just integrating ^2. And so get here and then cubed over 3 * evaluated from 0 to capital And so doing so we then see where comes from for this choice. All right. For number 11, series RC circuit has the voltage as 10 volts. The capacitance is 2 microfarads. The resistance is 300 kiloohms. And at t=0, the capacitor is initially uncharged. So I'm going to be closing the switch and starting to charge this. So I'm going to make note that I'm in the charging phase here. What is the charge stored on the capacitor at = 1.2 seconds? so first of all, we can look at this and note that since I'm in the charging phase, right, my is going to be equal to maximum value. It's going to start at zero and then it's going to rise up, right? want my to increase because I'm charging. and so my increasing exponential mathematically is written as 1 minus to the minus over towo. So we're going to have to figure out what our tow is. Well, this is just simple RC circuit. So tow with just single resistor and single capacitor. So tow is just * which in this case we have the 300 kiloohms* 200 microfarads. so we get 300 kiloohms times my 200 micro farads. And so we actually end up with this one being 0.6 seconds for my time constant. So actually notice then if look at my time that was provided that this is actually equal to 2 tow. Okay. So that has eliminated or dealt with this portion of the exponential where over tow is going to be equal to 2 for us. And then in terms of what is the maximum max we can find from Q= CV. So the capacitance is fixed. So max is equal to * max. And the maximum voltage that the capacitor gets is the full thing of the battery once it's fully charged. It's actually getting that full 10 volts. So this max is 10 vol. And so the max is equal to which was 2 microfarads * 10 vol. and so we see that that's going to be 20 microucms for that maximum value. So now we can go ahead and just see which answer matches what we found. We found our max and then we have our increasing term where over tow is 2. So that gives us choice Okay, moving on. So number 12 here, this is just simple junction rule from Kirkoff's laws where have currents flowing into and out of the junction. And so know that the total current in has to equal the total current out. So if look at all of the currents heading in, have 4 amps, 1 amp, and 2 amps headed in. And can see that I'm going to end up having to have one coming out because three is less than that. So, so I'm going to put the unknown on the outside. and I'm going to have here 4 + 1 + 2. And then the current heading out is going to be 3 amps plus this unknown So, my currents headed out of the junction would be 3 + and if you put it on the wrong side, if you put assuming it was going in, you would get negative number and that would let you know that you picked the wrong direction. so then here we have that is simply equal to 4 amps and the direction because it needs to head to to the out of the junction, it needs to be away from the node. So we end up seeing that it is 4 amps out of node And then the last of the standalone problems number 13. we have Kirkoff loop rule question. where we're asked to find the equation for this outer loop. and so what will do is just like to go clockwise around the loop. It's not required. and generally start at the bottom left corner. So, am going to go around the loop sort of starting here. and I'm going to to see what pass. Now, before do that, am going to So, that's my KVL loop. So, that's going to be my KVL loop that I'm going to look at. But, want to label the plus and minuses of my resistor. So notice that R1 here has I1 going through it. So I1 is going towards the right and the current points downhill through my resistors from positive to negative voltage because I'm losing energy when go downhill and then I3 points to the left. So as I'm going to go here then again the current has to point downhill cuz current goes in the direction of electric field. and the electric field points towards lower potential. So go and make sure have those labels done and then now can traverse this loop. So first come through this first power supply and end up with + 28 vol here. and then I'm going to continue around my loop. pass this first resistor and end on that negative sign. I'm going downhill now. get negative I1 R1. and I'd like to kind of make the visual of what this looks like. So, let's say that we start here at some energy at point and then I'm going to go ahead and I'm going to go through this 28 volt. So, I'm going uphill by some amount. and in that and then I'm going to, you know, go around the loop and then hit this resistor. That's going to cause me to go downhill little. So that's now making me go downhill little and now I'm going to continue along. And so continue along and now I'm going uphill because I'm going to end on the positive through R3. So then get plus I3 R3. And so now that's me going up hill. Say let's say that hill here. You wouldn't really know the relative heights of the hills until you do the numbers. and then I'm going to continue along and then come here to this 7 volts. and then know I'm going to have to go back downhill. and come back around and so that min - 7 vol equals and will have ended at the same energy that had started at. And so that's the equal zero portion. look through which of these match my equation and get choice Okay, so let's go ahead and talk about these last long problems. And this first one is just bunch of capacitors with power supply that is 18 volts. and it asks bunch of questions including the equivalent capacitance and bunch of other rankings that really show that we we really just want to understand the entire circuit. So what I'm going to do is that I've already set up this table that want to fill in where can find the equivalent capacitance. can look up at this top line at the total circuit and then can go ahead and kind of do some logic mapping onto other parts of this table. So let's just go ahead and work through that. So for CEQ to solve for CEQ first know that C1 is in series with C2 which is in parallel with C3 do notice that my last step is parallel step and that's going to come in handy when start to analyze that table. So, first need to find C12. And for C12, because they're in series and they're capacitors, 1 / C12 equ= 1 / C1 + 1 / C2. so that's 1 12 + 1 over 4 = 1 3. So get the result that C12 = 3. And the units here, just like the others, are microfarads. So in my table, I'm going to put the microfarad unit. And then the C12 is three. And then what have here is with that C12 see that C12 is in parallel with C3. So for CQ can get that CQ is going to be C12 plus C3. So get my 3 plus my 6 which is going to be equal to my 9 microfarads. So my equivalent capacitance is my 9 microfarads. And go ahead and can see that that is in fact the first answer for this question. So over here then now I'm ready to analyze the rest of the circuit. and the next thing that I'm looking for is the voltage V2 across capacitor C2. So, I'm going to make box around that cell so that know that that's one of the values I'm looking for. don't always go straight there because know that sometimes need to take this kind of windy path through the table, but always start with the top line of the equivalent circuit. My total voltage in the circuit is 18. and then know that is equal. So this is in volts and then would be in microum. So equals CV. So can then multiply those together in order to get my total voltage in the circuit. mean sorry my total charge in the circuit. So 9 * 18 give me my total charge in the circuit. And then think about the total charge in the circuit. But it doesn't really map to any individual capacitor because my last step was parallel step meaning that had my voltage and then have C12 in parallel with C3 and that's the most kind of simplified version of the circuit before the equivalent capacitance. So my total charge in the circuit is going to not map directly onto any of these. What does map is my total voltage. So the voltage of the battery is the voltage across C12 which is the voltage across C3. So can take this 18 and it's going to map onto C12 and C3. So that voltage maps down here. and so now have some things that can calculate for my charge. and can see over here that then could calculate the total charge. And that might come in handy later when I'm ranking these charges. here. So that's my charge on three and on that combination one two. and so every time calculate something in an intermediate branch, say the charge on this, then actually want to say okay that then can deconstruct this and notice that the charge on Q1 is going to equal the charge on Q2 and that equals the charge on that total branch because it is series branch. So can actually now take this value and map that onto the charge for one and the charge for two. So both of these have 3 * 18 as their total charge. And that's now going to allow me to do my last step which is to calculate the voltage for each of these. So from this equation Q= CV, we see that is equal to over and so can just see that then what get for the box for V2 is that V2 will equal my charge on capacitor 2 over my capacitance of 2. So get 3 * 18, that's the charge, all over the capacitance, which was 4. So that is where we get the answer of the voltage across V2 being equal to 13.5 vol. Now for these last two, we're simply ranking the charges on each capacitor and we're ranking the voltage drops across each capacitor as well. so we can actually do this pretty easily from the table. where we can see that if we're going to rank the charges then we actually have that the one is going to equal two because they're in series with each other and that so we can see that the charge on three is actually more and so we would end up here with this choice And the reason that that can make sense in this scenario is because note that this is branch of C12 that we found the equivalent capacitance to be three microfarads and then this one is six microfarads. So when we look at that branch then we know that the more capacitance literally what capacitance is is how much charge it can hold for some voltage and these both have the 18 volts across them and so that C12 gets 18 volts it's 3 microfarads and the C3 gets 18 volt volts and it's 6 microfarad farad. So the higher capacitance has higher capacity to store charge and therefore is going to have that higher charge in this ranking for the voltages. We can also look at our table and we can see you know that when we do this last step, right? that we're going to be dividing the voltages between V1 and V2 and we're going to see that V2 gets the higher C2 gets the higher voltage drop. and so if we kind of look across C2 and C3, we get here that this is you know our V2 and then this is our V1. What you see from the table is that V2 is greater than V1. And in fact V1 + V2 equals 18 vol because that's again the voltage across that whole branch. V3 also equals my 18 volts because it gets the entire amount of the battery. So right away we know that the most one is going to be C3, right? The voltage across the capacitance the third capacitor. and then we know that that's going to be greater than V2 which is greater than V1. Now the reason that V2 is larger than V1 is because the C1 and C2 hold the same charge but they have different capacitances. In order to hold that charge on weaker capacitor, it has to get more energy. It has to get more voltage to do that. and so the smaller capacitor gets more of that energy. And so we get that ranking here for the voltage V3 being that of the battery greater than V2 greater than V1. All those come directly from the table but conceptually they should also make sense. Okay, moving on to the next problem with electric potential. We have non-uniform linear charge density given by this lambda. And we want to have the electric potential at the origin to only this long bar. So I'm going to go ahead and I'm going to do what we like to do, which is to just draw in some dq, which is equal to lambda dx. have some distance here, which equals some variable And then can get my total potential by adding up by integrating all of my tiny dq's dq over And so in this scenario, I'm going to add up lambda dx over And I'm going to add up from the start of the bar, which is at zero, to the end of the bar, which is at And so when write this out, simply get that from the bar is equal to the integral 0 to of lambda is to the 4th. But I'm going to cancel out one of my x's with the on the bottom. So cubed dx. And so that ends up being x4 over 4 evaluated from 0 to And so we see here that we get choice All right. Now we're adding in point charge. So this point charge plus is distance over2 away. It doesn't matter that it's lower. Remember we are out of vector world now. everything is just scalar. It's just distance away that matters. and so when we want to have that fixed in place, the total potential at the origin, we're not adding vectors is simply going to be equal to bar plus the of the point charge. And again, there's no vectors involved in this. We're in energy, right? But is energy per charge. So just like with energy and mechanics, we didn't have any vectors. So total is simply going to be my bar, which is just phrased as it is, plus the of point charge, which is and then plus over where is the distance. It's it's positive value because it's literally just distance away. So KQ over and so from that choice we see that we get choice So that's now my total potential at this origin. so at this point we have another point charge plus is now passing at the origin that has this potential total at that location of the origin. So in terms of that total what is the electric potential energy associated with the charge? so very simply this is just the definition of what potential is right potential is energy per charge. and so we have this equation here that is So if write this as = over and so we get is equal to So the total energy electric potential energy associated with that charge plus is simply the charge times the total potential. So that's how we get potential and we turn it into potential energy. And then lastly, we have conservation of energy question where the same point charge has that has mass right? And it's at the origin. so the energy we're going to have our initial kinetic energy. So we could say initial plus initial equals kf final plus final. and initial, it has an initial velocity, but we don't care the direction because we want the speed for kinetic energy. So it's 12 v^ squar that jhat direction is not important it's just the speed that's important plus initial was this qv total but in this problem we're just writing that as kn what we have found in part number 20 and then equals kfal that's where it has the speed that we want to solve for 12 vf final squared where that's what we want and then plus 0 because goes to zero as goes to infinity. So now we just solve this for the correct answer and we do in fact get All right. And then lastly, we have circuit that involves multiple resistors. and we're going to want to again find the equivalent resistance, find the voltage across R4, and rank the currents. So, we might as well just set up our table and and start to analyze this circuit as well. so have that already started over here. and if we look at this circuit, we see that we have R1 is in series with and then here have R2 in series with R3 which is in parallel with R4. so again here have noticing my last step and in this case my last step is series step. So I'm going to make note of that when start doing my table. So want to find these equivalent resistances. So R23 is R2 in series with R3. R2 is 2. R3 is 6. So that's 2 + 6 is equal to 8. and then we can put that value into our table here. So that's our 8 ohms. And then have R234 which is going to be given by 1 / R2 34 = 1 / R2 3 + 1 / R4. So that's going to equal in this case 1 over8 + 1 over 4. So 1 over 24. We can look at that as 3 over 24 + 1 over 24. So we end up with this equaling 4 over 24 = 1 / 6. So 2 3 4 = 6 ohms. And so can put that into my table. And then lastly, EQ is going to be equal to R1 + R2 34 because those are in series. and so R1 is 9. so 9 ohms + 6 ohms and end up with 15 ohms. So that gives me my first answer here that my equivalent resistance is 15 ohms. All right. So now what want to do in the next question is voltage V4 across R4. So I'm going to just make note here that this is the voltage that the question is asking for. And then I'm also noticing that want to rank my currents through each resistor. So I'm going to start through the path where again start with my total circuit. do know that the total circuit has voltage of given in the problem of 15 volts. So that's in volts here. And so can then calculate my total current. And I'll calculate this in amps. and that's simply going to be equal to 1 amp. And here I'm just using that V= IIR. So is equal to / So that's my total circuit. So my total circuit current is going to be 1 amp coming into R1. and so right away we can see okay my last step if simplify this circuit is R1 in series with R2 34 and so then that's going to have the current total headed into it and so it's now in this scenario the total current that gets mapped onto I1 R1 and also R2 34. So, it gets mapped all the way down onto R234. At this point, can calculate value here cuz can get the voltage across 234 from 6 * 1. And so, that's just 6. And again, that's because this equ= So that gives me the voltage drop across 2 3 4 which if we look at what that would be on the larger diagram we see that that voltage is the voltage drop across R23 and it is the voltage drop across R4. So we go over here and we get six volts. And so we can right away see that that's the answer for the next question. And then so now for ranking the currents, we have to work our way further through this table. so in order to do that, I'm going to calculate each of these currents. And again, can get the current from V= IIR. So I= / So this current here will be 6 over 24. So that's essentially 1/4. and I'll just write it as as such. so that can be written as 14th. And then this would be 6 over 8. So that would be 34s for the one through the current through the 23 branch. So the current through the 23 branch. We notice then that we're going to be able to take that value and map it onto the current through two. So that's 3/4s and the current through 3. So 34 and 3/4s because it's they're in series with each other. The current through that branch has to be the current through R2, which has to be the current through R3. So now we're ready to do our ranking. and we can actually see straight away that we have the highest current is going to be through R1. So the current through R1 is going to be the largest. And then that's larger than the current through I2 which equals the current through I3. And then that's larger than the current through I4. And if we think about this, that's going to be because R4 is very large value. It's 24 ohms. And R23, this whole thing was 8 ohms. And so we're actually going to have more current go through the lower resistance path. And so you can also reason through this ranking conceptually by knowing that the current through the lower resistance path the 8 ohms is going to be more than the current through the high resistance path through R4. so that gives us our choice here I1 greater than I2= I3 greater than I4. And then finally the last question say that we cut the wire at point What happens to the voltage drop across R1? Okay, so if we look at this one here, we can either work it out kind of the long way and just do whole another table, but what I'm going to do is talk through it conceptually. so in this case, what we're looking for is the voltage cross drop across V1. and then notice here, I'm going to have this and I'm going to call it V2 34. That's my voltage drop here. So notice that Vub1 plus V2 3 4 is equal to my total voltage of the battery. This is what stays the same, right? when cut the that that wire at point it's the total voltage that stays the same. and so when look at this scenario, would have here V1 which is just R1, right? and R1 was like 9 ohms. and then my equivalent resistance of this other path is now going to be so that's 9 ohms and then this is 6 ohms for R2 34. So what that means is that in terms of the voltage right V1 and then V2 3 4 that we're going to be actually getting more than V2 34 in that scenario. We being R1, right? Because that's the one we care about. So it's larger resistance. The higher resistance gets more of the voltage drop in series. And conceptually that's because there's going to be more collisions. There's going to be more energy loss relative to smaller resistance in that same path for the same amount of current flow. So now cut the wire. So this is my initial state. And then cut the wire. And now I'm going to have simply two and three go away. So now simply have R1 and then R4. And so now V1 and V4. If we look at these values, R1 is still the same value that it's always been 9 ohms. R4 is the giant 24 ohms. So now V4 is going to get is going to get more. So now V4 gets more than V1. So V1 goes down because now V4 is the giant one in this series. It's going to have to use up more of the energy for all of its collisions. and so what happens is that V1 has to go down. It used to get substantially more than half. Now it's getting substantially less than half of that voltage of the battery because they have to share that energy and the higher resistance gets more. So that would be down here. it decreases but it is still more than zero. It doesn't get no energy. it just gets less energy. Okay. hope this helps and yes, good luck on the rest of the semester.