WEBVTT physics/ap-physics-c-electricity-magnetism/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson, we are going to start talking about series circuits.
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Our objectives include understanding the behavior of both series
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and parallel combinations of resistors in order to find the current voltage resistance and power dissipated.
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We will continue to work with parallel circuits into our next lesson.
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Applying Ohm’s law in Kirchhoff's rules to DC circuits.
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Determine unknown circuit characteristics.
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Understanding the properties of voltmeters and ammeters.
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Understanding the properties of both ideal and real batteries and we that will carry into our next lesson as well.
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Let us start by revisiting Ohm’s law.
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Ohm’s law relates resistance potential difference and current flow.
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V = IR or you may see it written as R = V/ I or I = V/ R.
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A material that obeys this is known as in Ohmic material.
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It is an empirical law but you can do a pseudo derivation that comes close to approximating what and where it really comes from.
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Although it is not a pure law of physics, it is extremely useful in analyzing circuits.
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Let us start with a couple of simple examples to get us going.
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The current on a wire is 24 amps when connected to a 1.5V battery, find the resistance of the wire.
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R = V/ I which is 1.5 V/ 24 amps or 0.06250 ohms or we could write that as 62.5 milliohms.
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Nice, straightforward, quick example.
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Or slightly differently, in a simple electric circuit at 24 ohm resistor is connected across a 6V battery, what is the current in the circuit?
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This time we are given resistance and potential and we asked to find current flow.
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I = V/ R which is 6V/ 24 ohms or 0.25 amps, which we can also write as 250 ma.
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A constant potential difference is applied across the variable resistor held at constant temperature,
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which graph best represents the relationship between the resistance of the variable resistor and the current through it?
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V = IR or I = V/ R, if we wanted to put that in a form that is similar to Y = NX the equation of the line.
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It does not really work that way.
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If we look at our Y, it is equal to a constant in this case.
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A constant potential difference divided by resistant.
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We are looking for an inverse relationship.
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The correct answer here must be number 1.
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And example 4, the graph shows the relationship between the potential difference V across a resistor and the current I, through that resistor.
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Through which interval does the resistor obey ohms law?
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Where it obeys Ohm’s law, where it is known as an ohmic material is where we have this linear relationship.
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Here from B to C, because that is where our IV graph is linear.
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Current flowing through a circuit causes a transfer of energy into different types.
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The rate at which electrical energy is transforming the other types is the electrical power dissipated.
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For example of a light bulb, electrical energy is converted to light and heat.
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When a television electrical energy becomes light, sound, and heat.
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All examples of energy conversions.
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We can look at this analytically, we take a look at a toaster.
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A 110V toaster oven draws a current of 6 amps at its highest setting converting electrical energy into thermal energy.
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They can use some nice tasty toast.
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What is the toaster’s maximum power rating?
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Power is current × potential which will be 6 amps × 110V or 660W.
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And we have also talked about power = I² R or power = V²/ R.
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An electric iron operating 120V draws 10 amps of current, how much heat energy is delivered by the iron in 30 seconds?
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Let us start off by finding the power that is going to be current × voltage or 10 amps × 120 V, which is 1200 W.
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But we do not want the power, we want to know how much energy.
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Energy is power × time which will be 1200 W, × that 30 seconds which is 36,000 J or 36 kJ.
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The potential drop of 50V is measured across a 250 ohm resistor, what is the power developed in the resistor?
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We will use V² / R because we are given potential and resistance, that will be 50V²/ 250 ohms which is 10 W.
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Let us try and put this all together to start making some electrical circuits.
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An electrical circuit is a closed loop path through which current can flow.
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If you do not have a closed loop path, you are not going to get any current flow.
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It has to be at closed loop.
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It can be made up of most any materials but typically what we are talking about are things like wires, batteries, resistors, switches.
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We will talk about capacitors in a little bit, inductors towards the end of the E and M course.
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Conventional current flows from high potential to low potential, the direction a positive charge would move.
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Keep in mind, in most circuits it is actually electrons that are moving which are going in the opposite direction from low to high potential.
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Therefore, electron current is in the opposite direction of what we call positive conventional current.
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As we talk about circuits, we are going to talk about symbols.
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We are going to represent them with schematics on paper,
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because trying to draw 3 dimensional circuit on paper and making it look really good is very difficult.
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We have come up with these symbols to help us diagram these.
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You can see much more easily how they operate.
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Lines represent wires and these symbols over here are fairly standard symbols for different elements in the circuit.
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Things like the voltaic cells, batteries, cells, voltage sources, often× use the same symbols.
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A switch whether it is open or closed, a voltmeter and ammeter, a resistor look like Charlie brown shirt.
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A variable resistor has an arrow that goes to that Charlie brown squiggly line thing.
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A lamp is just a resistor that also gives off light and heat, as do all resistors but it is designed to do that.
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Sometimes, you will see a lamp portrayed as a resistor.
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Other times, you will see the lamp symbol like this in the circuit schematic.
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In order for current to flow, the first thing you need is a source of potential difference.
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Those voltaic cells, batteries, power supplies, any of those are required in order to have a circuit with current flow.
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If you do not have a potential difference, you are not going to have any current flow.
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Current only flows in complete conducting paths.
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Here in the circuit to the left, we have a source of potential difference.
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We have a switch which is open at the moment and we have a lamp.
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No current will flow because we do not have a closed path.
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Over on the right here, once we have closed that path, close that switch,
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current is going to flow from high potential, the positive side through our lamp.
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And that is the direction of conventional current flow.
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Note that the longer side is the positive side.
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The way I remember that, is if you draw this, it takes more ink to make the longer side.
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It takes more ink to make a + , the short side looks like a minus.
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Let us take a look at voltmeters, use to measure the potential difference between two points in a circuit.
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Voltmeters are always connected in parallel with the element to be measured.
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The two lids on the voltmeter go to either side of the object you want to measure the potential difference across.
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What you do not want to do is have a voltmeter in the circuit, where all of the current goes through the voltmeter.
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It is very bad for the voltmeter and your circuit.
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If the voltmeter is connected correctly, you can always remove it from the circuit without breaking the circuit.
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And voltmeters have a very high resistance and the reason they do that, is that they have a negligible effect on the circuit.
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Although they take the tiny amount of energy in order to make them operate, you really do not want it affecting the circuit as a whole.
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The way you do that is by using them in parallel and they have a very high resistance.
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You know it is connected correctly because in this diagram, we can actually go
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and we could disconnect it and our circuit would still function.
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Our voltmeter would not do anything but it would still work as a whole.
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If we wanted to put that voltmeter back in, now it is connected in parallel on either side of the element we want to measure the voltage across.
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Looking at the ammeters, they measure the current flowing through an element of the circuit.
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Unlike voltmeters, ammeters are designed to go right in line with the circuits.
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Any current going to the element also goes to the ammeters so we can measure the entire current going through it.
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Ammeters have very low resistance so that they have a negligible effect on the circuit.
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That they do not cause a large potential difference, a large voltage drop.
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Therefore, affect the rest of the circuit.
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They are designed to have a negligible effect.
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In order to put an ammeter in correctly, if you break the circuit, the circuit is not going to function correctly.
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If you put an ammeter in a circuit in parallel, because it has such a low resistance, chances you are going to break it.
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You are going to smoke it and burn it up.
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Be very careful when inserting ammeters.
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Ammeters always go in series with what you are trying to measure.
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Let us take an example where we are looking at ammeter and voltmeter placement in a circuit.
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In the circuit diagram here, possible locations of an ammeter and voltmeter are indicated by 123 and 4.
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Where should we place an ammeter to measure the total current and the voltmeter to measure the total voltage?
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As I look at this, it seems to me if we want the total current flow, we would have to put our ammeter here at 1.
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Here is our source of potential difference, we are going to have current flowing this way and around that way.
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The only place where we get all of the current is if we place the ammeter here at 1 and that is in series with the rest of the circuit.
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That eliminates choice B and C.
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And the voltmeter, we are left with choice 4 or 2.
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If we put the voltmeter at 4, that would be bad, that is in series.
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Voltmeters go in parallel so it cannot be that when you are in a circuit.
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Instead, if we put it here at 4, our voltmeter right there is going to measure
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the potential difference across our voltage source which should be just about perfect.
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I'm going to pick choice A.
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Which circuit diagram below correctly shows the connection of ammeter A and
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voltmeter V to measure the current through and potential difference across resistor R.
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Let us take a look at each of these here.
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In number 1, we have an ammeter connected in parallel with the resistor.
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That is not going to be good at all, that is going to get nice and smoked and crispy.
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We do not want one.
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As we look at 2, our ammeters now in series but our voltmeters in series is not parallel, that is not going to work.
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As we look at 3 here, we have our voltmeter in parallel with R but our ammeter is connected
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on the same portion of the circuit as the voltmeter.
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It is not going to measure the current going through R, it is going to measure the current going through
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the voltmeter which we are trying to make as small as possible.
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Although that is probably not going to hurt anything, it is not going to give you the information you want.
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Number 4, here now we have our ammeter connected so that all the current has to flow through our ammeter.
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That is what we want, to measure the total current and the current that is going to resistor R.
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Our voltmeter is connected in parallel with each lead connected to either side of R to measure the potential difference across it.
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EA 4 must be our answer.
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Looking at voltmeters again, a student uses a voltmeter to measure the potential difference across a resistor.
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To obtain a correct reading, the student must connect the voltmeters go in parallel with the element you want to measure.
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Which statement about ammeters and voltmeters is correct?
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The internal resistance of both meters should be low.
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That is not good, if you did that with the voltmeter, you will pull a lot of current and change the way your circuit works.
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It cannot be A.
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Both meters should have a negligible effect on the circuit being measured.
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Yes, that is the whole point.
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We want to get information from our circuit, about how it is operating without affecting the way it operates.
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B is going to be our correct answer there.
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A little more on series circuits.
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Series circuits have only a single current path.
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You do not have multiple current paths.
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The removal of any single circuit element causes an open circuit.
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Think of it like Christmas or holiday lights.
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Have you ever seen where 1 ball goes out and they all go out?
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That would be a great example of the series circuit.
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A diagram of one, if we started with a source of potential difference and we will put a couple of resistors here.
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There is one, there is another, and let us say one more resistor here.
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We have one current path in which our current can flow.
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Remove any single element and all of the current stops, that is a series circuit.
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What we can do is we can find the equivalent resistance for resistors in series, if we have a bunch of resistors all lined up,
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we could replace them with a single resistor where the equivalent resistance value is just the sum of all your individual resistors.
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Now in a series circuit, anywhere in the circuit the current is the same.
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Whatever current enters a point must leave a point, by the law of conservation of charge.
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The total current anywhere in the series circuit is going to be the same.
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The current through resistor 1, let us make it smaller here.
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R1, R2, R3, R4, total current the current through I1 must be equal to the current through R2 must be equal to I3,
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the current through R3 and so on, for however many resistors you have in that series circuit.
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The potential drop is going to be equal to the drop across R1 + R2 + R3 + R4.
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The potential drops across each of these elements all add up to give you your total.
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That will be V1 + V2 + V3, for however many resistors you happen to have.
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I did this a moment ago but we have a couple laws or rules that will help with analyzing circuits.
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Kirchhoff’s laws are tools utilized in analyzing circuits and there are 2 of them.
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Kirchhoff’s current law, which you will often× hear me abbreviate as KCL.
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It is also known as the junction rule, states that the summit enters any point of the circuit,
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the current that enters any point in the circuit also has to leave that point, by the law of conservation of charge.
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Kind of like if you pour water into a cup, you keep pouring water in.
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Whatever you pour in eventually is going to come out.
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Same thing with current, whatever you put into a point has to come out.
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Restatement of the law of conservation of charge.
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Kirchhoff's voltage law which I call KBL, states that the sum of all the potential drop in any closed loop of the circuit has to equal 0.
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It is a restatement of the law of conservation of energy.
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Basically, what goes up must come down.
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As you go around any closed loop in a circuit, you have to come back to the exact same energy or the same potential.
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And that is oftentimes called the loop rule.
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And you will see how we apply these here in just a moment.
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Taking a look at voltage across a resistor, you have a 3 ohm resistor and a 6 ohm resistor connected in series.
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Let us draw those, there is a 3 ohm resistor, there is a 6 ohm resistor.
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If the current through the 3 ohm resistor is 4 amps, so we have 4 amps going this way.
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What is the potential difference across the 6 ohm resistor?
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Kirchhoff's current law KCL says is that whatever comes into a point in the circuit has to leave that.
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Which means we must have 4 amps going through the 6 ohm resistor.
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The potential drop then across the 6 ohm resistor, using ohms law is the current × the resistance or 4 amps × 6 ohms which gives us a potential drop.
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Potential difference of 24 V across the 6 ohm resistor.
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Taking another look at Kirchhoff's current law.
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We have a diagram that shows current in the segment of an electric circuit.
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What is the reading of ammeter A?
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To do this, I'm going to take a look at everything that is entering this point and everything that is leaving this point.
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It looks like we have 3 amps and 2 amp coming in, so we have 5 amps coming in.
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We also have coming out 4 + 1 + 2 or 7 amps out.
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Whatever comes in, must come out so we are missing 2 amps in.
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That means that this must be 2 amps entering the circuit.
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The ammeter reading 2 amps.
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Let us take a look a little bit of basic series circuit analysis.
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As we do these, I’m going to start off for simpler circuits using what I call A VIRP table.
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What that is, is just the way of tracking different characteristics of different elements in the circuit.
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I'm going to label the column for potential current flow resistance and power and then do that for the different elements in my circuit.
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Where here we have resistor 1, resistor 2, resistor 3, and a make a line for total.
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We will just make ourselves a nice little table.
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R1, R2, R3, R4, and the nice thing about these VIRP tables is as you fill them in,
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any time you know 2 things in a row you can always solve for the other items in a row using ohm’s law or a power equation.
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Let us start off by filling in what we know.
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Right away, I know since I have a 12V battery that my total and there is only one of them that might total voltage must be 12.
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Let me make that line there to make it a little meter.
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I know my resistance is R1 is 2000 ohms, 2 kl ohms.
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R2 is 2000 and R3 is 2000.
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Now we also know for a series circuit that the total resistance, the equivalent resistance is the sum of the individual resistances.
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My total resistance here must be 6000 ohms.
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Once I know that, I can start filling in the other items because that 6000 and that is 12,
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I can find the current here using ohm’s law I = V/ R, 12 /6000 gives me a current of 0.002 amps.
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That means that I must have 0.002 amps flowing through that portion of the circuit.
00:20:32.100 --> 00:20:38.000
By Kirchhoff's current law, that means I have to have the same current through R1, through R2, and R3.
00:20:38.000 --> 00:20:44.200
These all must be 0.002 amps as well.
00:20:44.200 --> 00:20:55.100
When I know the current and the resistance, I can then use ohm's law to solve for the potential drop V = IR, 2000 × 0.002 is just going to be 4V.
00:20:55.100 --> 00:20:57.100
Same there and same there.
00:20:57.100 --> 00:21:08.500
I'm dropping 4V here, 4V here, 4V here, that makes sense because I'm lifting up the source of potential difference is raising our potential 12V.
00:21:08.500 --> 00:21:13.700
We drop 4, we drop 4, we drop 4, we come back to 0, we raise 12 again.
00:21:13.700 --> 00:21:18.500
If you are possible to draw, you make a closed loop in the circuit, you come back to where you start.
00:21:18.500 --> 00:21:22.400
The sum of all the potential drops in a closed loop must be 0.
00:21:22.400 --> 00:21:25.000
To find the power, we have our choice of equations.
00:21:25.000 --> 00:21:30.600
We could use power = V × I, power = I² R, power = V²/ R.
00:21:30.600 --> 00:21:40.600
For any of these, we will get the same result which will be for R1 0.008, 0.008, 0.008.
00:21:40.600 --> 00:21:44.800
And of course, we are expanding 8 mw.
00:21:44.800 --> 00:21:51.700
In each of these, the total will have to add up those powers to be 0.024 or 24 mw
00:21:51.700 --> 00:21:57.300
which by the way we can also get using V × I, I² R, or V²/ R.
00:21:57.300 --> 00:22:05.700
There is our VIRP table which tells us just about anything we could ever want to know about this simple series circuit.
00:22:05.700 --> 00:22:08.800
Let us take a look at some more sample problems.
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A 2 ohm resistor and a 4 ohm resistor are connected in series with a 12V battery.
00:22:13.800 --> 00:22:19.100
If the current through the 2 ohm resistor is 2 amps, what is the current through the 4 ohm resistor?
00:22:19.100 --> 00:22:24.100
That is a trick question because right away if it is a series circuit, we know the current is the same everywhere.
00:22:24.100 --> 00:22:30.300
It is just going to be 2 amps.
00:22:30.300 --> 00:22:35.900
Here, we have got a circuit connected below with two 4 ohm resistors connected to a 16V battery.
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Fill in the VIRP table for the circuit and determine the rate at which electrical energy is expanded in the circuit.
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That is what we call power.
00:22:43.700 --> 00:22:52.700
Let us call that R1, R2, and make our VIRP table.
00:22:52.700 --> 00:23:05.500
Our circuit elements, we have R1, R2, and total.
00:23:05.500 --> 00:23:07.500
Let us start to fill in what we know.
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Our total potential is 16V, we know R1 is 4 ohms, we know R2 is 4 ohms, and because it is a series circuit our total resistance is the sum of those or 8 ohms.
00:23:19.800 --> 00:23:26.200
I = V/ R, 16/8 tells us that is 2 amps.
00:23:26.200 --> 00:23:33.000
That current must be 2 amps and anywhere in the series circuit you must have the same current so those must be 2 amps as well.
00:23:33.000 --> 00:23:41.100
We can now find our potential difference V = I × R, which is 8 and 8, and those add up as we expect.
00:23:41.100 --> 00:23:44.100
And for the powers, we have our choice of formulas again.
00:23:44.100 --> 00:23:48.900
V × I, will be simply 16 and 16 is 32.
00:23:48.900 --> 00:24:03.700
The rate at which electrical energy is expanded is the total power, power = 32 W answers the question that we were asked.
00:24:03.700 --> 00:24:12.400
Another problem, we have got a 50 ohm resistor, some unknown resistor R, 120V source and the ammeter connected as shown.
00:24:12.400 --> 00:24:19.400
The ammeter reads 0.5 amps so we know the current right there is 0.5 amps.
00:24:19.400 --> 00:24:25.800
Find the equivalent resistance of the circuit, the resistance of resistor R, and the power dissipated by the 50 ohm resistor.
00:24:25.800 --> 00:24:29.700
The way I find out all of these things is I make the VIRP table and then just look it up.
00:24:29.700 --> 00:24:35.800
Once we have a VIRP table made, everything else we should be able to look up right on there.
00:24:35.800 --> 00:24:46.700
We will write our VIRP, I'm going to call this R1 and this R2.
00:24:46.700 --> 00:25:03.800
R1, R2, total, we do not have to make a line for the ammeter because it is supposed to have a negligible effect on the circuit.
00:25:03.800 --> 00:25:06.000
We will start filling in what we know.
00:25:06.000 --> 00:25:16.900
We have a total of 120 V, we know the current is 0.5 amps, and that is the same anywhere in the circuit because it is series.
00:25:16.900 --> 00:25:23.400
We know R1 is 50 ohms and I think that will get us started.
00:25:23.400 --> 00:25:29.800
Our voltage through R1 is I × R, half of 50, that is just going to be 25V.
00:25:29.800 --> 00:25:37.900
Since we know that potentials add up in a series circuit, we know that this must be 95V.
00:25:37.900 --> 00:25:51.400
The resistance R = V/I, 95/ ½ means that this is going to be 190 ohms and resistance is add in a series circuits or total resistance must be 240 ohms.
00:25:51.400 --> 00:25:56.100
We can find our power as well with any of the formulas that you are most happy with.
00:25:56.100 --> 00:26:05.500
V × I, 25 × ½ will be 12.5, 95 × 1/2 will be 47 ½, and 120 × 0.5 will be 60 W.
00:26:05.500 --> 00:26:10.600
Of course, the total power is the sum of the individual powers from those elements.
00:26:10.600 --> 00:26:12.500
What did it ask us?
00:26:12.500 --> 00:26:20.600
Find the equivalent resistance in the circuit, right from our table that is going to be 240 ohms.
00:26:20.600 --> 00:26:25.400
Find the resistance of resistor R, that is what we call R2.
00:26:25.400 --> 00:26:29.300
R2 we said was 190 ohms and our power dissipated by the 50 ohm resistor.
00:26:29.300 --> 00:26:44.100
The power dissipated by R1 is 12.5 W, right from my VIRP table.
00:26:44.100 --> 00:26:48.600
Let us take a look at 1 more simple series circuit here.
00:26:48.600 --> 00:26:54.500
In the circuit represented in the diagram, what is the reading of the voltmeter V.
00:26:54.500 --> 00:26:56.500
First thing, let us label some of these.
00:26:56.500 --> 00:27:08.000
Let us call this R1 and this R2, make our VIRP table.
00:27:08.000 --> 00:27:23.400
We have resistor 1, resistor 2, and total.
00:27:23.400 --> 00:27:25.900
Start filling in what we know.
00:27:25.900 --> 00:27:29.600
We have a 60V source so there is our total.
00:27:29.600 --> 00:27:35.200
We know that R1 is 20 ohms, we know that R2 is 10 ohms.
00:27:35.200 --> 00:27:43.300
Once we have done that, we know that the total resistance in our circuit is the sum of the individual resistances because it is a series circuit.
00:27:43.300 --> 00:27:50.000
30 ohms I = V/ R, this must be 2 amps.
00:27:50.000 --> 00:27:56.700
That means that anywhere in our circuit, because it is a series circuit we have the same current 2 amps and 2 amps.
00:27:56.700 --> 00:28:00.800
We can find our voltage using Ohm’s law, I × R so that is 40.
00:28:00.800 --> 00:28:08.900
The potential drop across R1 this must be 20, we drop 20V across R2 and those add up to the total, that make sense.
00:28:08.900 --> 00:28:16.200
And for the powers, take your choice of equations V × I, 80, 40, and 120.
00:28:16.200 --> 00:28:21.200
And of course, our 2 individual powers expanded add up to a total power expanded.
00:28:21.200 --> 00:28:22.500
What was the question asks?
00:28:22.500 --> 00:28:25.300
What is the reading of voltmeter V?
00:28:25.300 --> 00:28:31.100
If we look up here at V, V is measuring the potential difference across R1.
00:28:31.100 --> 00:28:39.800
We come down to my VIRP table and I find that the potential difference across R1 is 40V.
00:28:39.800 --> 00:28:47.000
Hopefully that gets you some good fleeting, a start on series circuits and I know this is a calculus based physics course.
00:28:47.000 --> 00:28:51.500
And by all means, these are simple problems that we are starting with.
00:28:51.500 --> 00:28:56.300
We are just laying the groundwork for as we get to more complex circuits and more complex topics.
00:28:56.300 --> 00:28:58.600
We got to have these basics down first.
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We will start getting to those into the next lesson and further on as well.
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Thank you for your patience, thank you for watching www.educator.com.
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We will see you again real soon, make it a great day.