Practice: For the circuit below, find the current through each of the 3 branches.

Subjects

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Intro to Current | 7 mins | 0 completed | Learn |

Resistors and Ohm's Law | 14 mins | 0 completed | Learn |

Power in Circuits | 11 mins | 0 completed | Learn |

Microscopic View of Current | 8 mins | 0 completed | Learn |

Combining Resistors in Series & Parallel | 37 mins | 0 completed | Learn |

Kirchhoff's Junction Rule | 4 mins | 0 completed | Learn |

Solving Resistor Circuits | 32 mins | 0 completed | Learn |

Kirchhoff's Loop Rule | 86 mins | 0 completed | Learn |

Additional Practice |
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!! Resistor-Capacitor Circuits |

Concept #1: Intro to Kirchhoff's Loop Rule

**Transcript**

Hey guys. So, now we're going to start talking about Kirchhoff's loop rule which is going to be massively important when you start solving more complicated circuits. Now, we're going to build this up a little bit at a time over a few videos and I'm going to take a little bit slow because I want you to have a strong foundation so that when you get to problem solving it can be a breeze, this can be a little hairy so we're going to build it up a little bit in the time, let's check it out. Alright, so far we've dealt only with resistor, with circuits that has a single source like a single battery, okay? But now, we're going to start to get into circuits with multiple batteries. So, we're going to need new tools and what we mean by new tools is new rules and new equations, okay? So, the new rule we're going to use is Kirchhoff's loop rule and it states that the sum of all voltages in a loop is simply 0, okay? It's conceptually very simple but it's pretty tricky to write with equations. So, one way to summarize this entire thing here is just to say that the sum of all voltages in a loop is 0, this rule is also called Kirchhoff's or a Kirchhoff's voltage law, okay? Voltage law. So, loop rule because you're going to be going in a loop adding voltages or voltage law because you're adding voltages. So, whatever you want to call it, I want point out that this rule actually works for any circuit. So, we could have used this to solve some surface earlier but it's going to be especially useful when you have multiple sources, it's too hard for the easy stuff but it's necessary for the hard stuff, got it? So, let's see. So, what we're going to do is for each loop in a circuit, we're going to write one loop equation. So, for example, if you look here, these wires and all these elements form one loop. So, we're going to get one loop equation. So, what we're going to focus on now, is how to write this loop equation, okay? What the equation will do is that we either add or subtract voltages of the batteries and the resistors, okay? We have a voltage of the battery here. Now, the voltage of the resistor is simply i times R, i times are either current and R is the resistance and this comes from Ohm's law V equals iR, okay? So, that's the voltage of the resistor is going to be i R. So, for example, I have a current i here. So, it's going everywhere here, okay? You know, what we're going to do is we're going to write the sum of all voltages which by the way is going to equal 0, I'm going to write all the voltages here, I'm going to list all the voltages, okay? So, I'm going to pick a point here, let's say, we're going to start here and we're going to make a loop all the way around this circuit here and I can go in any direction, I can go this way or I can go the other way, I'm going to go this way because it's the way that the current going so we're just going to go with the current and so the first element that you run through here is R1. So, what you're going to do is you're going to write i, R1, okay? And then when you get to V2 you're going to just write plus v2 and then you're going to keep going here, imagine you're moving along, we're going to cross this R2s. So, you're going to write plus the current which is i everywhere i, R2 and then you're going to get over to here and you're going to write plus v1 now. Notice here, that I said that the equation is going to add or subtract the voltages and here I just added all of them which is actually wrong, these voltages are going to be either adding or subtracting and I don't know yet, we don't know yet which one, okay? And, that's what we're going to talk about next but I just want to give you a basic idea first before we start that you're going to be adding these things and then setting them equal to 0. Now, are they going to be adding or subtracting is what we're going to talk about now, okay? They're going to be added or subtracted depending on two things, on the direction of the current and on the direction of the loop, okay? So, direction the current in this case was, was clockwise and the direction of the loop was also clockwise, cool? Just to show you what I mean by direction current interaction loop, so the first thing we're going to do and we have to do this before we start writing the equation is we're going to use a direction of current to put positive and negative signs on the ends of each resistor, okay? So, let me show you what I mean. So, here for a resistor, the positive sign the positive m is where the current, where the current enters the resistor. So, look at this diagram down here, the current is going this way it enters a resistor here. So, I'm going to put a little positive on this side of the resistor and therefore the negative goes on the other side. The current keeps going and keeps flowing and the current enters this resistor right here. So, we put a positive here and make this negative bigger and a negative here, okay? So, that's that first step, got it, for batteries the positive end is always going to be the positive terminal so that actually does not depend on the direction of the current so the positive terminal is a longer one here. So, I'm just going to put a positive on this side so the other side's are negative and I'm going to put a positive on this side and these other sides are negative, okay? So, the first thing we do is we put these signs everywhere.

Now, we're going to do is we're going to choose the direction of the loop. Remember, I mentioned, when we start going around the circuit up there that I could have gone clockwise or counterclockwise, we're going to choose the direction and I'm going to choose to loop this way, I'm going to write the loop here and all that means is the sequence in which you're going to add things, right? So, we're going to start here and we're going to go this way so the first element I'm going to cross is R1 then when I cross V2 and so and so forth and it says here, when crossing elements in this direction, this is the direction of the loop, you're going to add the voltage, if you crossed that element going from negative to positive, okay? Once you get this it's going to be super simple but I want to move slowly here. So, let's just see here, if you're, if you're going from negative to positive. So, if you follow the current here, you're going from positive to negative, that's the oposite here, right? So, if you add when you're going from negative to positive this means you will subtract, subtract if you're going from positive to negative, okay? So, this R1 here, the voltage of that R1 will be subtracted. So, I'm going to write this, sum of all voltages is negative i, R1, okay? Now, let's go to the next element, I'm going to keep going here and I'm going from positive to negative again. So, when I cross this battery, I'm going from positive to negative. So, I'm going to subtract that voltage there, okay? V2. Now, let's keep going. Now, I'm going to go again from positive to negative. So, because I'm going into the negative I'm going to subtract the voltage of R2 and remember the voltage of every resistor is always i, R so this is going to be i, R2 and finally I get to this last element here and I'm going to cross it and when I cross it, right? Imagine your sort of charge is going through, I'm going from negative to positive finally go into a positive. So, this voltage will be a positive voltage when I listed here, V1 and this entire thing has to be set to 0 okay, if you want to make this a little neater I can move some things around. Notice that everyone's negative except V1. So, I can move everything to the other side and it's going to look like this V1 equals V2 plus i1, r, I'm sorry I meant i, R1 plus i, R2. So, that's the loop equation for this loop, okay? So, what I want to do now, is I want to write a loop equation for the same circuit above which I draw down here, but now we want to go in the opposite direction of the loop, let's go in the opposite direction the loop. So, I'm going to put my little starting point here but now we're going to go this way and what this means is that you're first going to encounter this guy and jump through it, okay? Now. Remember, the first thing you do is you put these signs everywhere, these pluses and minuses, the battery is the easiest one to do because the big one is always plus, this is always minus, always plus, always minus, the resistor, the plus goes where the, the plus goes where the current enters the resistor, this is one of the most important things to remember, okay? For the current enters a resistor. So, if current hasn't change direction so it enters here so positive to negative and the current keeps going till here and it enters here, positive to negative. Now, we're ready to start going around the loop, sum of all voltages in this loop, we're going to start here. So, the first element you jump across is the V1 and it's going from positive to negative. So, it's going to be negative V1 and then you keep going you jump R2 from negative to positive. So, it's going to be positive of voltage of the R2 voltage of a resistor is i, R. So, i R2 and then we're going to keep going here and we're going to jump from negative to positive. So, it's going to be positive voltage of that battery which is V2 and lastly we're going to get here and jump from negative to positive and because I'm going into a positive it's going to be positive the voltage of I, of R1 which is i, R1, don't forget at the end to set this entire thing to 0, okay? And you end up with this. Now, if I want to clean this up a little bit I can move, the only guy it's negative here is V1. So, I can move you one to the other side and I get that V1 equals V2, actually I'm gonna write V1 on the left, okay? I'm gonna write V1 of the left, V1 equals, imagine that all these guys go to the other side and you get rid of all the negatives. So, you get something like this after a little bit of moving around and we can order to put it here is because I wanted to show you, this the final equation, that this equation is exactly identical to this equation and the point here that you absolutely have to remember is that direction of the loop does not matter, okay? The direction of the loop does not matter and that it's going to give you the same equation whether you go clockwise or counterclockwise. So, just pick one okay, cool? So, this is a quick introduction of how you write these loop equations, we're going to do a little bit more to build up the concept so that you are a beast at this, let's keep going.

Example #1: Direction of Current in Loop Equations

**Transcript**

Hey guys. So, in this video we're going to keep talking about writing loop equations except that I'm going to add one complexity which is the fact that you won't always know the direction of the current, let's check it out. Alright, so there are more complex circuits that we're about to start seeing, you're typically not going to know the direction of the current. So, what you're going to have to do is you're going to have to assume the direction of the current, okay? This is the word are you typically hear but what this effectively means is that you're guessing them in other words, if you don't know, if you have no way of knowing you just pick one and then you solve the problem that way, okay? And don't worry about picking the wrong one it's going to be okay. I'll talk about that at the end, okay? So, before I talked about having there being these two steps to writing these equations but there's actually a new step here that you have to take before you can even get steps one and two and it is that you have to assume, assume the direction of all currents unless it's given to you, right? So, if you don't know you assume, if you're given a direction then you use that direction, okay? So, then you're going to label positive and negative signs and you're going to cross elements in the chosen direction of the loop.

First, let's talk about number one real quick and we've done this before but just as a refresher for every one of these elements you want to put pluses and minuses to the left and right or on the opposite ends of the elements. So, plus or minus here, plus or minus here and we're going to do this based on this rule here. So, the battery is going to be positive on its longer terminal. So, you put a little positive on the longer terminal and remember. So, I'm going to posit here, which means you put a negative on the other side. So, positive here, negative on the other side, okay? Same thing here by the way, positive here, negative on this side positive here, negative on this side, okay? Resistor is going to be positive where the current enters the resistor, where the current enters a resistor. So, this depends on the direction the current, if you don't know the direction of the current then you're going to use the assumed direction of the current, okay? Alright, cool. So, in this question here, before we get into number two, we're being asked to write loop equations. So, we're going to be following these steps over here to do that, okay? And here, it's asking to write a loop equation, if the current is clockwise. So, here current is clockwise means it goes this way so the current is entering this resistor here. So, I'm going to put a positive and then this one must be a negative the current keeps going and it enters this resistor here. So, this is the positive end and this is the negative end and then here we want the current to be counterclockwise which means it is going this way and it's entering this resistor here. So, that's positive, the other ones negative and then it's entering this resistor here. So, this is positive and this is negative, okay? So, step one is to label the positive and negative signs, we got that done and step two is to cross elements in the chosen direction of loop, in other words you're going to go around the loop and then you're going to sort of jump over each one of these elements and as you do that you're going to add the voltage, if you're crossing from negative to positive and you're going to subtract the voltage if you're crossing from positive to negative, okay? Remember that. So, now let's just do that, let's pick, let's pick a direction of the loop here, we're going to loop this way for both of them and we're going to start right here, okay? Right here, and right here, that's my starting point. So, if I'm going in this direction. Remember, the direction of the loop it's just a sequence in which you're going to sort of walk around this circuit it's not the direction of the car necessarily, okay? So, it's just a sequence that you're going to add these things. So, I'm going to write the sum of all voltages in this loop is a bunch of stuff equals 0, okay? And, what we have to do is write all these different things, they're going to be four numbers here or four elements here adding or subtracting, so the first one we get to is this true resistor then I'm going from positive to negative, I'm going from positive to negative. So, I'm going to subtract that voltage.

Remember, the voltage of a resistor comes from Ohm's law V equals i, R. So, here I'm going to write, we're going to the negative. So, I'm going to write negative i, R and the resistance here is 2 ohms. So, I can do this, I cannot plug anything for i because we're looking for the current, okay? We're looking for the current through the whole thing. Alright, now we're going to keep walking here, I'm going to jump from a positive to a negative therefore that voltage is also going to be subtracted negative for V, I'm going to keep going and we're going to jump from a positive to a negative, this voltage is also going to be subtracted negative, this is a resistor. So, it's going to be i, R, i is i and R is 1 and then finally I'm going to get over here and jump over this guy from it's going to go from negative to positive because I land on a positive, this is a positive 10 volts, okay? Everything is equal to 0 and this is going to allow us to solve. So, I have 4 volts and 10 volts or 10 minus 4 volts, if I combine these two, this is going to be 6 volts and then I have negative 2, actually let's move these guys to the other side. So, if you move negative 1i is going to be positive 1i, and then this is 2i it's going to become positive 2i so this is obviously just 3i and at this point, I'm going to be able to solve for i, I is just 6 over 3 which is 2 amps and we're done, that's it for Part A.

Now, we're going to do the same thing here, for Part B, write the sum of all voltages in this loop equals a bunch of stuff equals to 0 and then it's going to be, it's going to allow us to solve for i, what I've loved for you to do is pause the video right now, try to emulate the steps that I had in Part A see, if you can get something for Part B, I'm going to keep rolling here, but hopefully you gave this a shot, it's important that you're following and then you're able to do this yourself. So, we're going to start here and we chose to go in the same direction. So, I'm actually going against current now. I'm going to sort of walk around the circuit against the direction of card which is totally okay, right? So, I'm going to go here and the first thing I gotta jump over is this and I'm jumping from a negative to a positive so this is going to be positive, the voltage of that resistor and the voltage of that resistors i, R. So, it's I times 2, this is a new problem by the way so I can't just use this to over these two as current, right? We've got to find the current in this one, I'm going to keep walking here and jump from positive into a negative terminal. So, this would be negative 4 volts and we're going to keep walking here and jump from a negative to a positive and because end in the positive thi is going to be positive voltage of the resistor, voltage of resistors r or is i, R. So, it's going to i times R which is 1 and finally we get over here and we're going to jump from a negative to a positive and because I'm landing on a positive this is positive 10, okay? This may even be already a little repetitive for some of you hopefully, which means you are getting, that'd be awesome. So, here I can add up these i's, I have one i and two i's, it's a lot of i's. So, this will be 3i and then 10 minus 4 is 6 plus 6 volts equals 0. So, if I'm solving for i, I'm going to move with 6 volts the other way 3i equals 6 volts therefore i, I'm sorry, negative 6 volts, almost messed up and that means that i is negative 2 amps, okay? Negative 2 amps Now, notice that this here was a 2 amps positive and this here was a 2 amps negative do you think that's a coincidence? It isn't, it's not a coincidence at all, in fact, you should have expected that because these are the exactly the same circuits that you would get the same current, this negative here, what's up with that guy? All that negative, that means is that you've chosen or the direction that was assumed for you, you didn't choose, it was chosen for you, was wrong. So, if you have chosen this direction it just means that you chose the wrong direction but the cool part is the answer is still right, you just now know that it was actually the current was actually physically moving in the opposite direction, okay? So, the answer to what is the magnitude of the current, Well, the current is 2 amps, that's the magnitude of the current but the direction of the currents we now know is opposite to what we were working with, where we're going with the current being counterclockwise and we know the direction of the current is actually clockwise, okay? Here, I got a positive which means the directional is working with was actually a correct direction, cool? So, you picked, you were either given the direction or you assumed/guess one and if you, you know, if you pick the wrong one, it's okay? You'll just end up with a negative current at the end, the number will still be right and you would just know, oh I guess the wrong one, that's, cool? Flip it you don't have to resolve the problem again, you would just know that the direction of the current is opposite to the assumed one, okay? I said that same thing a few times, hopefully that sticks, let's keep going,

Example #2: Solving Circuits with Multiple Sources

**Transcript**

Hey guys. So in this video we're going to put together everything we know about kirchoff's rules to solve a full problem, let's check it out. Alright, so we're going to combine Kirchhoff's or Kirchhoff's Junction rule which is the simpler one, it says that the current in equals current out, we're going to also use the loop rule which is the more complicated one, where we have to write the loop equations and we're going to use them together to solve more complicated circuits with multiple resistors. Here are the three steps, we're going to follow. Now, we've used lots of these parts individually. Now, we're just putting everything together, cool? First step, we're going to label directions, we're going to first label junctions. Remember, Junction is a split on the wire. So, this point here is a junction, I'm going to call that a and this point down here is a junction, I'm going to call that. B, okay? So, label the junction, easy, we're going to label loops and remember that loops are arbitrary, this is just a sequence in which you're going to walk through the circuit and you can pick the direction. So, here's a loop and we're just going to go this way, we're just going to go this way, I'm going to call this loop1 and we're going to go this way, we'll call this loop2, okay? So, this step is done this step is done. Now, we want to label the direction of current and remember current will be the assumed direction you can try to figure out which one might make most sense or you can just randomly guess them. So, here notice that this battery has a positive here which means current typically would flow out of here if this was a single battery circuit but you can't really know that for sure but either way I'm going to just use that to sort of dictate that I'm going to assume that the currents is going to go this way, I'm gonna call this current one. Now, notice that there are three branches here, right? This is a branch, this is a branch and this is a branch. So, because there are three branches I'm going to have three different currents, okay? Let me clean this up a little bit. So, I'm going to call that i1, there's another battery here and this is a positive terminal here. So, you could think that, okay, the current is probably going this way. So, I'm going to draw it right here, i2, this bit of branch has no batteries just a resistor. Remember, the junction rule says that current in equals current out. So, if I have two currents going into the A the third current has to be coming out of the A so that we're at least consistent, we're guessing all these directions but at least, let's do it in a somewhat consistent way. Alright, so i3 we're going to assume that it's going that way. So, we got our directions of current assumed, once we know that we can label positive and negative on voltage sources and resistors so the batteries will be positive on the long side, positive and negative, positive and negative but the resistor will be where the current will enter, okay? So, let's look at every resistor, this resistor right here, the current is entering from this side so this is the positive, this is the negative and then this resistor right here, if you sort of backtrack i2 you can see that i2 goes like this and it goes like this right? So, i2 entering right here. So, this is the positive of the resistor and this is the negative of the resistor, you will be very careful setting this up because if you set this up wrong you're going to get the wrong answer, cool? So, don't screw this part up. Alright, so we're done here, we put our little pluses and minuses everywhere. Now, we're going to write some equations.

First, we're going to write junction equations, we're going to write one for each Junction, I have two junctions, two junctions therefore I'm going to have two equations, I'm going to do this first because they're easier and then later, we're going to write loop equations for each loop, I have two loops therefore I'm going to have two equations, okay? But, let's focus on the junction first. So, for junction A I'm simply going to write that current in equals current out. Now, look at A right here, i1 goes in and i2 goes in. So, in as i1 plus i2 and then out is i3 coming out, okay? So, that's it for Junction B same thing, I'm going to write in equals out. So, look at D here, i3 is going on, let make this green, i3 is going here, because it just keeps going through, i3, we had already drawn i2 is coming out of B and if you backtrack I want you to sort of go backwards here, right? i1 is obviously going to look like this and like this so that it's just flowing. Now, if you look at B, i3 is going in and the other two guys are going out. Soit's going to be i3 equals out which is i1 plus i2 and we're done at that equation, you might notice that these two equations are actually identical, they're both saying i3 equals the other two, it's the same which means you effectively don't really have two equations you just have one equation. So, I'm going to cross that out so we don't make confused, it doesn't mean it's wrong it just means that it's excessive, we don't need both of them, okay? Cool, so we got one equation out. Now, we're going to write these two loop equations. So, let's start with loop1 over here, loop1 and then we're going to write is that the sum of all voltages equals 0, okay? Equals a bunch of stuff which then equals 0 and this is what we have to fill in. Now, loop1 is this guy right here, okay? I'm going to pick a starting point like in the little corner here, actually I'll start over here so we hit that battery first. So, we're going to start here and go in that red direction which looks like this, okay? So, you're walking over here and then you get to this battery, you have to cross that, you're crossing from negative to positive. So, you're landing on the positive so this is going to be positive 9 volts and I'm just going to write a 9 because everything here is a voltage so everything os in volts so positive 9 and then you keep walking, keep walking, no one here and then being loop is just this little square here. So, we're actually going to turn here and then we're going to jump over, cross over this guy, we're going to go from positive to a negative so this is going to be a negative voltage. Now, voltage of a resistor is given by Ohm's law, voltage equals i, R okay? So, this is going to be the current through this resistor which is i3 times the resistance which is 15, okay? And that's it, there's nothing else to keep going around the loop, you're done, okay? So, just get back to the original point. So, now we're done, I'm going to move i3 that way so I get 9 equals i3 times 15 for i3 equals 9 divided by 15 which is 0.6, I'm actually just save time I'm going to save space, I'm going to put it over here, 9 over 15 is 0.6 amps, cool? So, we got that first one down and this is i3 and, in writing this equation we already solved for one of the currents and if you look at the problem, the problem is asking for the current through each one of the three branches, one main down. So, let's get the second one.

We're going to write the loop2 here and hopefully we can get a current out of this as well, sum of all voltages equals a bunch of stuff which equals to 0, let's look at loop2, we're looping this way, I'm going to start over here and we're going to go like that, okay? I don't want you to confuse that with the current so this is the loop direction. So the first element that we're going to hit up while we're walking here is we're going to jump over this resistor and we're going from a negative to a, from a negative to a positive, so this is going to be positive i, R, the voltage of resistors are R, the current here is i2 and the resistance, actually I can just plug in a resistance, it's i2 times 10. Now, notice that we're walking, we're looping opposite to current, that's fine, okay? Forget about that just go, just look at the positive and negative. So, we're jumping into a positive. So, it's positive and then here we get this guy right away and we're jumping from a positive to a negative, we're landing on a negative so this is negative or minus 5 volts, okay? Because, we're going from positive to a negative and we keep going, we keep going, we keep going and then we complete the loop, we will go all the way back to this point right here. So, we have to, we're going to go this way because that's the loop we chose and we're going to go from negative to positive, we're going to go from negative to positive, so this is going to be positive the voltages of the resistor, voltage of resistors i, R, the i here is i3 and the R is 15, by the way, we already know i3 it is 0.6, okay? So, we're going to be able to plug that in there. So, 10 i2 minus 5 plus 0.6 times 16 equals 0, this is simply 9. So, I'm going to write 10 i2 equals, I'm going to put the i and then the 9 to the other side negative 9, I'm going to move the 5 to the other side, it becomes a positive so this is going to be negative 4, negative 4, okay? So, this is going to be a negative 4. So, i2 is going to be negative 4 divided, running out of space there, negative 4 divided by 10 which is negative 0.4 amps of course. Alright, so that's that, I got the second current, I got that i2 is negative 4 amps. Now, what does this negative mean, okay? I got a positive here, which is, cool? But, what does that negative mean? Well, what that means is that the direction of i2 that we assumed is actually wrong, okay? So, assumed, let's write this here, very important, assumed incorrect direction of i2. So, i2 is actually the other way and I'm going to write this over here, actual i2 is in this direction right here, okay? Now, that doesn't mean anything, that doesn't mean you stop doing the problem or you do it again or you flip the, you flip the arrow and start over. Now, you just know that the direction was wrong but the answer is still correct, the magnitude of the current is 0.4, in fact, we're going to keep using the wrong number. Alright, as we keep going here but deep inside you know that the current is the other way. Alright, cool. So, we want to find, we have i2, we have i3, we want to find i1 but we ran out of loop equations, we already made two loops. So, what we're going to do now, is we can actually use this equation right here, we can use this equation right here to find i1 because we already have the other two. So, i1 plus i2 equals i3. So, i1 is simply i3 minus i2, i3 is 0.6 amps minus i2, i2 is negative 0.4. Notice that you plug as a negative even though you know that that means that it's in the other way just keep rolling with it, okay? Negative minus, so I have two negatives here si this is going to be a positive 0.6 plus 0.4 is 1 so the current through i1 is 1amp, okay? Now, we're basically done I mean, we are done, what I want to do is real quick just show you something with these currents. So, i1 right here. Now, we know was 1 amp and I know that i2 the actual direction of i2 is this way and I know.

Practice: For the circuit below, find the current through each of the 3 branches.

Concept #2: Combining Voltage Sources in Series

**Transcript**

Hey guys. So, this short video we're going to talk about how we can merge voltages in a circuit to make that circuit simpler, let's check it out. Alright, so you can combine voltage sources if the sources are connected in series, this only works, if they're connected in series you can do this to simplify a circuit. So, for example, I have a 10 volt battery here, in a 5 volt battery here, they are in series because there's a direct path between all these 4 guys here, the Y doesn't split so they're all in series. So, I can combine these 2 so instead of having two batteries I can just have one and generally when you do you just add up the voltage is something like v1 plus v2. Now, if the voltage sources are in opposite directions their voltages are actually not going to add they're going to subtract, okay? And we're going to do these two quick examples and I think this is going to make a ton of sense, totally obvious. So, look at this battery here, the positive terminal is this way so it's pushing current this way, right? The current is leaving, the charge is leaving the battery through this side, this battery is here. So, the the positive sign is here, that's for positive and positive. So, this guy is also pushing this side. So, you might imagine, think of this as force as almost, right? This thing is pushing clockwise and then this thing is also pushing clockwise, they're helping each other they're both pushing this interaction. So, I can redraw this and just say that these two act as a single battery of voltage 15, it doesn't matter that there's a resistor in between them, got it? similarly by the way if you can remember we can also merge these two resistors, they're in series. So, they're also just going to add up. So, an equivalent circuit here, would be one with a 15 volt battery, I'm actually just going to draw in this direction here, that 10 instead, I'm going to put a 15 volt here and the resistances will add 2 plus 3 is 5, it would be this 5 ohms, okay? And by this example here is asking us to find the magnitude and the direction of the current because the both currents are going to same direction, the direction of the current will simply be, let's write here, direction is going to be clockwise.

Now, what about the magnitude of that current? The current goes this way, well, to find a magnitude of a current in a circle like this you can just write V equals i, R. So, I is going to be V over R, the voltage is 15, the resistance is 5 so it's just 3 amps very simple. Now, here it's a little bit different because of this guy over here is pushing current or at least it's trying to push current this way, something that'd be really cool is if you could pause this video and try to do in your own, I think a lot of you are going to get this right because, it's very straightforward but I'm going to keep going here, So, this guy's pushing this way, this guy's pushing this way they're clashing, well, who do you think is pushing current harder, right? So, the 10 volt is putting more of a pressure or providing more influence for the current to move. So, that one is going to win which means that despite the effect that they're pushing against each other the currents will overall move in this direction, okay? So, the 5 is actually fighting against the 10 volts and making it weaker therefore, we can just subtract the 2, we're going to say 10 minus 5 is just 5 which means the effective, the equivalent of the voltage here is going to be 5 in this direction, 5 volts this way, again, going up against a 5 ohm resistor, the direction is still clockwise but the current will be different, current is going to be V equals i, R, we're going to solve for i and this is going to be V over R, the voltage is a 5, the R is a 5 so the answer is 1 ampere, cool? That's it for this one, super simple, let's keep going.

Example #3: Find Two Voltages (3 sources)

**Transcript**

Hey guys. So, let's check out this example. So, here we have a circuit with three batteries and three resistors and we want to find two of the voltages, one voltage is given to us and then we're asked to find V1, V2, we're also given two of the three currents there are three branches one branch, two branch three branches. So, they're going to three currents and two of them are already given to us. Now, remember the general steps to solve these problems are that first, we're going to label a lot of stuff and then two, we're going to write as many equations as we need to find everything, we're looking for, okay? So, we're labeling? We're labeling junctions we're labeling loops, we're labeling currents and we're also putting at the end pluses and minuses on the resistors and the batteries which is the setup to be able to write these equations. Alright, so junctions are easy it's just points where the wires split. So, let's call this A and B, you might notice that this 4A right here, if I keep extending it it's going to go this way 4A and then this is going to go this way 4A, right? And right away because of loop because of Junction rule you might already be able to see here that if I have a 5 coming in and a 4 coming in there has to be a 9 coming out this way because current in equals current out. So, 9 in means there's going to be 9 out and that's this current. So, now I actually already know all three currents which is good. Alright, so let's label our, we got the junctions out of the way, we actually already calculated one of the currents which is going to be helpful, what about loops? So, the current on this loop is going this way 5 amps 5 amps, right? I got all the currents labeled. So, I like to loop. Remember, the direction of the loop is arbitrary but I like to write loops that follow the current just because everything's going the same way, it's just easier in the brain, right? So, we're going to do a loop this way which if you notice it will follow the 5 amps all the way around and then it's actually to follow the 9 amps as well, not necessary it's just a little nicer but don't sweat it too much you can just make a loop in any direction, okay? Same thing here, I'm going to start this loop right here, at point B and I'm also going to spin this way which will neatly go along the direction of 4 and the 9 doesn't have to, I just like doing that okay, cool? So, those are the loops and the currents are actuality all labeled. So, we got this, this and this and now we're going to put pluses and minuses everywhere. the plus on a battery will always be on the long stick right here. So, plus, plus, minus, minus and the plus on the resistor will be where the current answer so the current is entering from this side. So, that's a plus so the other is a minus, this is a plus because the 9 amps name is right here, this is a minus and this is a plus because of the 4 amps entering that way and this is a minus.

So, we're done setting this up, really important thing is that you get this way so you can write the equation properly. Now, let's write equations. Remember, there are Junction equations and there are loop equation so the junction equations will help you figure out the currents and the loop equations will help you figure out anything but typically voltages or currents, right? They'll help you figure out either one of those, we already know all the currents so we don't have to write a junction equation at all, in fact, when we said that 5 and 4 gave you a 9, we're essentially writing that in our heads, okay? So, let's just write loop equations. So, there are two loops so we can write two equations. So, for loop one I can write that the sum of all voltages equals a bunch of stuff which equals 0 and what we got to do is fill in this bunch of stuff, loop one, if you go around loop one this way there are 4 elements that are going to hit up, right? Two resistors and two batteries. So, should expect they're going to be 4 terms in this thing here. So the first thing we'll do is we're going to go through the 18 volts. So, we're walking over here, let's make this a different color, I'm going to make it green, we're walking here and then you're going to jump from a positive to a negative. So, remember that means we are going to add 18 volts, over here you're jumping for a negative, from a positive to a negative, sorry, negative to a positive so that's positive 18 and then positive to negative so that's going to be negative in the voltage of the resistor. Remember, the voltage of a resistor is given by Ohm's law V equals i, R. So, I'm going to write, I'm going to make two little spaces here, this is for my i, this is for my r, the i here is 5 and the r is 8, okay? I'm going to make a little bit more space here, just in case. Now, we're going to keep walking down this path here, turn over here, we're going to jump from a negative to a positive and this is going to be positive, I don't know that voltage is what I'm looking for positive V1 and, when you keep walking jump from positive to negative. So, negative, this is a resistor, negative i, r. The current is 9 and the resistance is 4. So, you get this, cool? So, let's clean this up, 18 minus 40 plus V1 minus 36 equals 0, if you notice, good news there's only one, is one variable here, only one unknown. So, you can solve for this and if you move some stuff around, I got it here, you're going to get you're going to get 58, the first voltage is going to be 58 volts, cool? We already got V1. Now, we just got to get V2 and we're also done with the equation for the first loop. So, now it's going to write an equation for the second loop and hopefully the V2 will come out of there so the sum of all voltages of the second loop it's going to be a bunch of stuff equals 0, second loop is also going to run into two batteries and two resistors. So, we expect four things to show up in this equation and let's keep going. So, we're going to jump here, wrong color, we're running around with green I forgot to put a plus and a minus here, we know the plus is the big one, we're going to jump from positive, from negative to positive. So, it's going to be positive V2 finally, that's the variable I'm looking for right there on you're going to keep going here, jump from positive to negative. So, it's going to be negative, this is a resistor. So, I'm going to put negative i, r, i is 4, r is 6 and then you're going to keep walking here, you're going to get here, you're going to jump from here to here from negative to positive. So, it's going to be positive V1, positive V1 by the way v1 is 58. So, we could plug that in as well and I'm going to keep walking and then going to cross over, I'm going to get negative i, r where i is the 9 and r is the 4, okay? So, if we write this out you get V2 minus 24 plus 58 minus 36 and if you move everything around carefully you get the V2 is just 2 volts, okay? So, that's it for this one, we got V1, we got V2 along the way, we also got all the currents. So, we actually know everything we need to know about this resistor, about this circuit rather. Alright, that's it for this one, let's keep going.

Practice: For the circuit below, calculate

(a) the voltage V shown, and(b) the current through the 6-Ohm resistor.

Practice: For the circuit below, calculate the voltage across the 100-Ohm resistor.

Concept #3: How to Check Your Work (Kirchhoff's Rules)

**Transcript**

Hey guys. So, now that we've seen how to solve these complex circuits using kirchoff's rules I wanna show you a simple map that you can use to double check your work and the idea here is that solving these circuits is a long process there's a lot of steps to it. So, there's a lot of room for small mistakes and it'd be great to be able to quickly double check. So, let's talk about that in this video. Alright, so once you know everything, you know all the voltages all the currents and all the resistances you can double check your work with a simple rule and you actually already know this rule which is that all branches must have the same magnitude of voltage, same magnitude of voltage, okay? So, you might remember, we talked about this many times, if you have something like this and this voltage of the source in 9 volts then the voltage of this resistor is 9 volts as well and that's because they're sort of opposite to each other on the two branches, right? You can think of these as, this is one branch and this is another branch so the voltages have to be the same on opposite sides. Now, the direction must always be the same as well and direction is actually not the right word, the right word here is polarity but it really just means direction. So, what does that mean? So, if you have a 9-volt battery this way and then 9 volt battery this way it's the same voltage but this one's positive on the left and this one is positive on the right so they have different directions, the directions have to match and the total magnitude has to match. So, let's do this real quick, we're going to here check if all the numbers match up. Now, just to be clear you're not going to get a test question that says here's a complete circuit is this right or not, right? I haven't seen those in tests but I'm getting you this just to sort of build up the skill that might be useful if you have enough time at the end of the test, right? So, or if you're doing your homework. Alright, so we're going to check that all the voltages and all the bridges add up and the branch, remember, is all of this but really you just have to worry about the top part because there's only, you only have certain elements up here, there's not the sides, same thing here. So, does that branch have the same voltage here as this branch and does it have the same voltage as this branch, cool? So, let's look at all the voltages, this guy is 18, what about this guy? This is a resistor, the voltage of a resistor, remember, comes from Ohm's law V equals i, R. So, I can just multiply the i with the R, same thing here, I can multiply the i against with the R. So, I times R they're multiplied, this is going to give you the voltage R times i, okay? So, let's multiply these numbers, this is a 40, this here is a 36 volts, volts and this here is it 24 volts, okay? So, now we're going to do is we're going to put the polarities and I'm going to actually, I'm going to write it over here. So, this 18 has a positive and a negative, right? Remember, the positive and negative on the resistor depends the direction of the current and if the current is coming this way, this is a positive and negative. So, if you go here to the side, I'm going to write that you have 18 v positive and negative and then here you have 40v positive and negative and you can think of it as 18 volts this way and 40 volts this way, what do you think is that net voltage or the equivalent voltage of those two? If you have, if they're going in oposite directions they are going to subtract, the 40, the 40 wins over the 18. So, it's 40 minus 18, 40 minus 18 which is 22. So, this guy is the winner which means this entire branch has the equivalent voltage of a 22 with the positive pointing to the right, okay? Now, all the other branches have to have the same thing, let's look here, this is positive and negative. So, I got a 58 volts positive, negative and this is a 36, the currents look at these two currents here, look at these two currents the third one, they're both going into here. So, the other one has to be this way which means, this is a positive and this is a negative. So, here I have positive 36 volts negative the 58 this way is going to overpower the 36 and then you do 58 minus 36, that's 22 as well, awesome. So, I have a 22 to the right, cool? So, so far it's matching up and then here at the end I have a 24, 24 here and a 2 here, the 24 wins. So, it's 24 minus 2, they're also going in oposite directions, it's 24 minus 2. So, you end up with 22 here and here, okay? I drew it this way by yet but I can also have drawn 22 this way, 22 this way, 22 this way, maybe that's a little bit easier to sort of make sense.

All these things have a polarities are 22 to the right, okay? Or the positive is on the right side over here of the whole branch, that's it, that's all you have you. So, let's look at this one and it's might be a good idea for you pause the video and do exactly right as before and double-check it, this might seem a little long because the first time I'm explaining but once you get the hang of it you can do it really quickly, cool? So, I'm going to keep rolling here and I'm going to go as fast as I can to show you that this can be actually pretty fast. So, if you're doing the calculator 6 times this to get the voltage, it's 4 volts and if you multiply these two numbers to get the voltage, by the way this is a period not a comma, hopefully you caught that, this is going to be 14 volts, okay? This current is leaving, this current is leaving. So, this current must be entering which means this is positive, negative, positive, negative, positive, negative, positive, negative, positive, negative by the way, if you solve this yourself you already would have all these little negatives and positives everywhere, cool? So, what do we have here, I have a 4 to the right and then I have a 12 to the right and to the right means it's going negative to positive, okay? So, this is obviously a 16 to the right, here I have 4 times 4, that's 16 and it's to the right, cool? And then here I have 14 to the right and then I have 32, well, was I saying to the right, I don't know, I don't remember, but this is to the left, I might have said to the right, If I did I'm sorry. So, obviously the 30 wins, 30 minus 14 is 16. So, 16 to the left, okay? That's it all you gotta do is the whole to just make sure that it works by the way, if you had a resistor here you could have just moved it over here. So, that you're comparing sort of rows or just columns, right? So, it's a little bit easier to visualize but a resistor here, for example, is part of this entire branch. So, you could have just moved it over here, to make it easier to sort of see you're comparing to the rows of voltage, cool? That's it for this one, hopefully makes sense, hopefully helps, let's keep going.

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Concept #1: Intro to Kirchhoff's Loop Rule

Example #1: Direction of Current in Loop Equations

Example #2: Solving Circuits with Multiple Sources

Practice #1: Find All Currents Using Kircchoff's Rules

Concept #2: Combining Voltage Sources in Series

Example #3: Find Two Voltages (3 sources)

Practice #2: Find One Voltage and One Current (2 sources)

Practice #3: Find Voltage Across Resistor (2 sources)

Concept #3: How to Check Your Work (Kirchhoff's Rules)

Present one junction and two loop rule equations that you could solve simultaneously to obtain values for the three currents, I1, I2 and I3, in the dc circuit shown below. Do not solve the equations!

Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is 2.0 A, in the direction shown.
What is I1, the current through the 4.0 Ω resistor?

Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is 2.0 A, in the direction shown.
What is the resistance of the resistor R3?

In the following figure, the configuration of resistors is known as a Wheatstone Bridge, which is used to measure the resistance of an unknown resistor (let's say R1 in our case). This is accomplished by varying the resistance of one of the outside resistors (let's say R5 in our case) until the resistance through the center of the bridge is zero. For our bridge, the battery produces 20 V and the fixed resistances are R2 = 10 Ω, R3 = 20 Ω, and R4 = 30 Ω. If the current through the center of the bridges is zero, and the current produced by the battery is 1 A, when R5 reaches a value of 15 Ω, what is the unknown resistance?

In the following circuit, the resistances are R 1 = 5 Ω, R2 = 10 Ω, R3 = 10 Ω, R4 = 25 Ω, and R5 = 15 Ω. If the current through R3 is 1 A to the right, what is the voltage of the battery?

In the following circuit,V= 25 V, R 1 = 5 Ω, R2 = 15 Ω, R3 = 25 Ω, R4 = 10 Ω, and R5 = 20 Ω. Find each current in the circuit.

The circuit shown in the sketch consists of a battery with emf ε and internal resistance r connected to two resistors as shown. R1 = 3.0 Ω, R2 = 6.0 Ω and r = 0.50 Ω. The terminal voltage Vab of the battery is 14.0 V. At what rate is electrical energy dissipated in R2?

Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is I1 = 3.0 A. What is the current through 4.0 Ω resistor?

The circuit shown in the sketch consists of a battery with emf ε and internal resistance r connected to two resistors as shown. R1 = 3.0 Ω, R2 = 6.0 Ω and r = 0.50 Ω. The terminal voltage Vab of the battery is 14.0 V. What is the emf of the battery?

Consider the circuit shown in the sketch. What is the current through the 3.0 Ω resistor?

Consider the circuit shown in the sketch. What is the current through the 10.0 V battery?

A 10 Ω lightblub is connected to a variable voltage source. The way to vary the voltage of the source is not by altering the source itself, but rather by altering the internal resistance of the source, thus changing the terminal voltage, Vab. If the internal resistance of the source were zero, the termal voltage would be 120 V, and at some value r, the power output by the lightbulb is 100 W.
(a) What is the terminal voltage, Vab?
(b) If the terminal voltage were doubled, what would be the power output by the lightbulb?
(c) What is the internal resistance, r, of the lightbulb when the power output is 100 W?

In Figure 3, consider the circuit sketched. The two batteries have negligible internal resistance and emf's ε1 = 28.0 V and ε2 = 42.0 V. The three resistors have resistance R1 = 2.00 Ω, R2 = 5.00 Ω, R3 = 1.00 Ω. Calculate the potential difference Va - Vb between points a and b.

What current will a 3.0 V battery cause to flow if it is short circuited? The internal resistance of the battery is 14.0 Ω.
A) 42 A
B) 130 A
C) 0.21 A
D) 4.7 A

Your car will not start. You open the hood and measure the voltage on the battery as 14 volts. You then have a friend try to start the car while you continue to monitor the voltage on the battery. When your friend turns the ignition key, the battery voltage drops to 7.5 volts. If the car is drawing 150 A from the battery, what is the battery's internal resistance?
a. 0.01 Ω
b. 0.14 Ω
c. 0.04 Ω
d. 0.09 Ω
e. 0.05 Ω

The circuit shown in the figure has resistors R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω, ε1 = 10 V, ε2 = 20 V. What is the current through resistor R1?
(1) 1 A
(2) 2 A
(3) 1.5 A
(4) 0.5 A
(5) 0.67 A

The two light bulbs in the circuit on the right are identical. When the switch is closed,
A) they both go out
B) the intensities of both A and B decrease
C) the intensities of both A and B increase
D) nothing changes
E) some combination of (A)-(C) happens

The following equation is true for the right loop of the circuit.
a) E2 - I3R3 - I2R2 - E1 = 0
b) -E2 - I3R3 - I2R2 - E1 = 0
c) E2 - I3R3 - I2R2 + E1 = 0
d) E2 + I3R3 + I2R2 - E1 = 0
e) None of the above

The following equation is true for the right loop if the circuit.
E2 – I3R3 – I2R2 – E1 = 0
–E2 – I3R3 – I2R2 – E1 = 0
E2 – I3R3 – I2R2 + E1 = 0
E2 + I3R3 + I2R2 – E1 = 0
None of the above

For the circuit below, determine all the currents. Assume ε1 = 10 V, ε2 = 15 V, R 1 = R 3 = 40 Ω, R 2 = 60 Ω, R 4 = 100 Ω.

In the circuit shown in the figure (Figure 1) find(a) the current in resistor R(b) the resistance R(c) The unknown emf ℰ(d) If the circuit is broken at point x, what is the current in resistor R?

Use Kirchhoff's rules to find the current in each of the three resistors in the circuit shown. Also calculate the voltage drop across all three resistors.

The batteries shown in the circuit in (Figure 1) have negligibly small internal resistances. Assume that & 8.00 negligibly small internal resistances. Assume that-8.00 V and R = 21.9 Ω.1.Find the current through the 30.0 Ω resistor.Express your answer in amperes to three significant figures2.Find the current through the 21.9 Ω resistor.Express your answer in amperes to three significant figures3.Find the current through the 8.00 V battery.Express your answer in amperes to three significant figures

What is the current through the 10 Ω resistor in the figure? Express your answer to two significant figures and include the appropriate units. Is the current from left to right or right to left?a) left to right b) right to left

For the following circuit, you ultimately want to measure the power dissipation of the 2.0 Ω resistor:a) First, redraw the diagram on the page you are turning in.b) Define the directions of the currents arriving at the junction point a. Physically draw the directions of these currents on the diagram.c) Write an equation for the Kirchhoff Junction Rule for the current directions arriving at the junction point a as chosen in part a).d) Choose two Kirchhoff loops that include the 2.0 Ω resistor. Physically draw these two loops on the diagram. e) Write out a simple equation for each circuit element in your two Kirchhoff loops (note: you should have two equations). Clearly denote for each term whether it is a rise (+) in voltage or a drop (-) in voltage for your Kirchhoff loops. f) Combine your equations from part c) and part e) to solve for the current through the 2.0 Ω resistor.g) Use your result from part f) to solve for the power dissipation of the 2.0 Ω resistor.

Although Kirchhoff's junction law is needed only when there are one or more junctions in a circuit. Kirchhoff's loop law is used for analyzing any type of circuit, as explained in the following tactics box.(a)What is the absolute value of the potential difference |ΔV| across the unknown element in the circuit shown in the figure. Express your answer in volts.(b) Is the unknown element a battery or a resistor?A. the unknown element is a battery.B. The unknown element is a resistor.C. Whether it is a battery or a resistor cannot be determined.

Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large.(a) A voltmeter with resistance Rv is connected across the terminals of a battery of emf "EMF" and internal resistance r. Find the potential difference Vmeter measured by the voltmeter.(b) If emf = 7.50 v and r = 0.45 Ω, find the minimum value of the voltmeter resistance Rv for which the voltmeter reading is within 1.0% of the emf of the battery.

a. What is the magnitude of the current in the 18 Ω resistor in the figure? b. What is the direction of the current—to the left or to the right?

The circuit diagram below shows two emf sources and a bulb connected in parallel. Also connected in the circuit is a resistor with resistance R = 0.2Ω. The resistance of the bulb is Rb = 0.5 Ω , and each of the sources has internal resistance: r1 = 0.025 Ω and r2 = 0.02 Ω.If ε1 = 13.0 V and ε2 = 5.0 V , Part A. Calculate the current I2 flowing in emf source ε2Part BCalculate the current I1 flowing in emf source ε1.Express your answer in amperes to three significant figures.Part CA useful strategy to evaluate your answer is to consider a loop other than the ones you used to solve the problem; if the sum of potential drops around this loop isn’t zero, you made an error somewhere in your calculations. Begin by calculating the current I1 flowing through the emf source ε1.Express your answer in amperes to three significant figures.

An idealized ammeter is connected to a battery as shown in the figure (Figure 1).Part A. Find the reading of the ammeter.Part B. Find the current through the 4.00-Ω resistor.Part C. Find the terminal voltage of the battery.

Consider the circuit shown below. Suppose four resistors in this circuit have the values of R1 = 13 Ω, R2 = 7.5 Ω, R3 = 7.7 Ω and R4 = 10 Ω, and that the emf of the battery is 18 V Find the current through each resistor using Kirchoff's rules

For the circuit shown in the figure below, we wish to find the currents I1, I2, and I3.(a) Use Kirchhoff's rules to complete the equation for the upper loop. (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.)(b) Use Kirchhoff's rules to complete the equation for the lower loop. (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.)(c) Use Kirchhoff's rules to obtain an equation for the junction on the left side. (Use any variable or symbol stated above as necessary. All currents are given in amperes.)(d) Solve the junction equation for I3.(e) Using the equation found in part (d), eliminate I3 from the equation found in part (b). (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.)(f) Solve the equations found in part (a) and part (e) simultaneously for the two unknowns for I1 and I2, respectively.(g) Substitute the answers found in part (f) into the junction equation found in part (d), solving for I3.

Part A. Find the potential of point a with respect to point b in the figure. Express your answer using two significant figures.Part B. If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 10.0 V battery. Express your answer using two significant figures. Express your answer using two significant figures.

Calculate the currents in the below resistors of the figure.Part CCalculate the current in 64 ohm resistor.Express your answer to two significant figures and include the appropriate units.Part DCalculate the current in 110 ohm resistor.Express your answer to two significant figures and include the appropriate units.Part ECalculate the current in 82 ohm resistor.Express your answer to two significant figures and include the appropriate units.

A) For the circuit shown in the figure, determine the current in the 7.0-Ω resistor.B) For the circuit shown in the figure, determine the current in the 8.0-Ω resistor.C) For the circuit shown in the figure, determine the current in the 4.0-Ω resistor.Please show detailed work.

A. In the circuit shown in the figure, find the magnitude of current in the upper branch.B. Find the magnitude of current in the middle branch.C. Find the magnitude of current in the lower branch.D. What is the potential difference Vab of point a relative to point b?

In the following figure, what is the value of the potential at points a and b? (Figure 1) Express your answer using two significant figures.

Calculate the currents in the below resistors of the figure.Part ACalculate the current in 25 ohmresistor.Express your answer to two significant figures and include the appropriate unitsPart BCalculate the current in 120 ohm resistor.Express your answer to two significant figures and include the appropriate units.

a) What is the potential difference Vad in thecircuit?b) What is the terminal voltage of the 4 V battery?c) A battery with emf 10.3 V and internal resistance .5ohm isinserted in the circuit d with its negative terminal connected tothe negative terminal of the 8 V battery. WHat is the difference ofpotential Vbc between the terminals of the 4 V batterynow?

In Fig. 27-57, R1 = 2.96 R, the ammeter resistance is zero, and the battery is ideal. What multiple of ε/R gives the current in the ammeter?

Part A. What is the magnitude of the current in the 30-Ω resistor in the figure? Answer in AmperesPart B. What is the direction of the current?

What is the magnitude of the current in the 30 Ω resistor in the figure (Figure 1)?

The ammeter in the figure reads 3.0 A.(a) Find I2. Express your answer using two significant figures, in Amps.(b) Find ε. Express your answer using two significant figures, in Volts.

What are the magnitude and direction of the current in the 20 Ω resistor in (Figure 1)?

(a) Find the emf ε1 in the circuit of the figure (Figure 1) .(b) Find the emf ε2 in the circuit of the figure.(c) Find the potential difference of point b relative to point a

Find the unknown emfs ε1 and ε2. The current through the 3.00-Ω resistor is 8.00 A.

Find the resistance R. (Note that three currents are given.)

Two batteries and three resistors are connected as shown in the figure.How much current flows through 6.0 - battery when the switch is open?I = ____A

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