WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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This lesson is on current and resistance.
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Our objectives are going to be to define and calculate electric current, to explain the factors and calculate the resistance of the wire and to determine the equivalent resistance for resistors in series and in parallel.
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Let us start by talking about electric current.
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An electric circuit is a path through which current flows and current is the amount of charge passing a point per unit time.
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The units of current are coulombs per seconds also known as amperes or amps (A).
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Average current is the charge passing a given point in a set amount of time.
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Conventional current is defined as a direction of positive current flow.
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This is a little non-intuitive. Typically, it is the electrons that are actually carrying the charge that are causing the currents.
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Usually it is negative charges that are moving.
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Conventional current flow says that the current is going the opposite direction of the electrons.
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Most of the time you will not even have to worry about it, but if you are calculating currents and somebody says, "Well, which way are the electrons flowing?"
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It is going to be the opposite of the direction of positive current flow.
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That dates back to Ben Franklin and some of the initial work done on current electricity.
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Current through a resistor - a charge of 30 coulombs passes through a 24 ohm resistor in 6 s.
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Find the current through the resistor.
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Well, current is just Δ(Q)/Δ(T) and our charge that passes is going to be 30 coulombs that happens in 6 s for 5 amperes of current.
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How about current due to elementary charges (e)?
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A charge flowing at a rate of 2.5 times 10^16 elementary charges per second is equivalent to what current?
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Same formula, I = Δ(Q)/Δ(T), but now we have to convert this into coulombs first.
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That is going to be 2.5 × 10^16e/s and if I want elementary charges to go away, 1 elementary charge is equal to 1.6 × 10^-19 coulombs.
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Units of elementary charges will cancel out and I will be left with a current of about 0.004 amperes or amps.
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Let us take a look at charges in a light bulb.
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The current through a light bulb is 2 amperes. How many coulombs of electric charge pass through the light bulb in one minute?
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Well, if I = ΔQ/ΔT, then that implies that ΔQ = I(ΔT) or 2 A × 1 minute -- our standard unit of time is seconds -- 1 minute = 60 s.
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So that is going to be 120 coulombs of charge.
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Let us look at a slightly more complex example.
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A 1.5 volt triple A-cell supplies 750 milliamperes or milliamps (mA) of current through a flash light bulb for 5 min. while a 1.5 volt C-cell supplies 750 mA of current through the same flash light bulb for 20 min.
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Compared to the total charge transferred by the triple A-cell through the bulb, what is the total charge transferred by the C-cell through the bulb?
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Let us figure out the charge in each of these cases.
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Our first case for the triple A-cell -- ΔQ = I(ΔT).
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Our current is 750 mA or 0.75 A and our time is 5 minutes -- 5 × 60 min./s, which is going to be 300 s or 225 coulombs.
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Now for our C-cell -- ΔQ = I(ΔT) -- We have the same current of 750 mA or 0.75 A, but now our time, it does it for 20 minutes, so 20 min. × 60 s/min. is going to be 1200 s or 900 coulombs.
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Compared to the total charge transfer by the triple A-cell, what was the charge transferred by the C-cell -- well 900 is four times bigger than 225, so our correct answer must be D -- 4 times as great.
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Let us talk a little bit about conductivity and resistivity.
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Electrical charges move easily in some materials -- we call those materials conductors and less freely in others and we call those materials, insulators.
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A material's ability to conduct an electrical charge is known as its conductivity.
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And conductivity is given the symbol, Greek letter, σ.
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Conductivity of a material depends on the density of free charges available to move and the mobility of those free charges.
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A material's ability to resist the movement of electrical charge, the inverse of conductivity, is known as its resistivity (ρ).
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The units of resistivity are ohm-metres, so resistivity is 1 over conductivity or conductivity is 1 over resistivity and most of the time we are going to be talking about resistivity.
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It is a little bit more popular as a measurement compared to conductivity.
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Let us talk about resistance versus resistivity versus resistors.
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Resistivity is a material property. Its units are ohm-metres.
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You could look up the resistivity for a specific type of material, whether it is gold, silver, copper, wood, glass -- you name it, you could look up resistivity, a material property.
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Resistance is a functional property of an element in a circuit and that tells you how that item is going to behave.
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The units of resistance are ohms (ω) and it is a function of the material that the resistor is made out of and the geometry of the device.
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If we look at a fairly basis resistor and let us take something like a wire -- I will make a cylinder out of it, so there is our material and we will attach conducting leads to both ends.
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We have a cross-sectional area (A) and we have some length of our material (L) and it is made out of some material that has a resistivity (ρ).
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The resistance of this resistor is equal to the material property resistivity times the length of the resistor divided by the cross-sectional area.
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You can almost think about it in terms of something like a water pipe -- the longer a water pipe is, the more resistant to water current flow it is going to have and the skinnier the pipe is, as it gets smaller and smaller cross-sectional area, you are going to have more and more resistance.
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So as (A) gets smaller, you have greater resistance, so if you want a very good conductor, a very low resistance, you want a very wide, fat pipe that is very short.
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If you want a lot of resistance, you want a long, skinny narrow pipe or a long, skinny narrow wire.
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Now circuit elements designed to impede the flow of current are known as resistors and resistors have some amount of resistance.
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It is a circuit element. The resistance of a resistor tells you how well it impedes the flow of charges.
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The resistance of the resistor depends upon the resistors geometry again, as we just said and its resistivity, that materials property.
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Just a picture of some resistors -- over here we have some ceramic resistors and the color codes (bars) here tell you what the resistance is -- you have to have a decoder table that you could look up.
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Over here on the right, a light bulb is typically modeled as a resistor too.
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What you have is a wire and then you have the resisting element up here and then you go back down and the circuit schematic for resistor is going to be that.
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So calculating resistance -- let us say we have a 3.5 meter length of wire with a cross-sectional area of 3.14 × 10^-6 square meters and at 20 degrees (C) it has a resistance of .0625 ohms.
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Find the resistivity of the wire in the material from which it is made.
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Well, the first thing we need to point out is why is it telling us the resistivity is at 20 degrees (C).
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Well, resistivity actually is somewhat of a function of temperature and as temperature goes up, oftentimes resistivity will go up a little bit.
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As temperature goes down, resistivity goes down in general, so oftentimes when you look up a resistivity it will give you the resistivity at a specific temperature, so that you are looking at a common-based line.
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If we want to look and to find out what the resistivity is so we can find its material, let us start out with R = ρL/A and we are going to solve for the resistivity (ρ).
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So ρ is going to be equal to RA/L, where our resistance is .0625 ohms × the cross-sectional area, 3.14 × 10^-6 square meters divided by the length (3.5 m).
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That should give us a resistivity of about 5.61 × 10^-8 ohm-metres.
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Based on that and looking at the table and I am assuming that the resistivity we have here -- our answer should correspond to something on the table and it looks like tungsten must be our answer.
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So tungsten is the material from which this resistor is made.
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The electrical resistance of a metallic conductor is inversely proportional to its -- well we have to pick one.
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Well let us start out by looking at our relationships. If R = ρL/A -- well as temperature goes up, we just said resistivity is going to go up a little bit.
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That means that resistance is going to go up and that is a directional relationship, not A.
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As length goes up, resistance goes up -- direct, not inverse -- not B.
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Cross-sectional areas -- as cross-sectional area gets bigger, resistance goes down and that is an inverse proportionality, so the correct answer must be C. Cross-sectional area.
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Now when we talk about configurations of resistors -- when you place them in a circuit, resistors can be organized in what are known as both serial and parallel arrangements.
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For analysis purposes, it is oftentimes useful to determine some equivalent resistance, which could be used to replace a system of resistors with just one resistor that has the equivalent resistance value of whatever you had in that multiple resistor arrangement.
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If we talk about resistors in series -- series circuits have one single current path.
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For example, if we have current flowing to the right through these two resistors, we have one current path that is a series circuit.
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The equivalent resistance for resistors wired in series is found by just adding up the individual resistances.
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So for this example, the equivalent resistance -- our equivalent is going to be R1 + R2 and if we had more resistors we would keep adding them and in this case that is going to be 200 ohms + 300 ohms for an equivalent resistance of 500 ohms.
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You just add them up -- that simple.
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Now for the circuit down here, we have 1, 2, 3, 4, 5 resistors, so our equivalent resistance is going to be R1 + R2 + R3 + R4 + R5...
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...which is going to be 200 ohms (R1) + 300 ohms (R2) + 100 ohms (R3) + 100 ohms (R4) + 100 ohms (R5) for a total of 800 ohms in this configuration.
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Very easy to find the equivalent resistance of resistors in series.
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Let us take a look at resistors in parallel, where parallel circuits are configurations where currents can take more than one path.
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In this case, the equivalent resistance for resistors in parallel is 1 over the equivalent resistance is 1/R1 + 1/R2 + 1/R3 for however many resistors you happen to have.
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Now it is important to note that the equivalent resistance of resistors in parallel is always going to be less than the smallest resistance in that configuration.
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So here we have two resistors and if you look -- if we have current flowing this way, part of the current can go that way, part can go that way and it will all re-combine and come back through here.
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But you have multiple current paths. It is a parallel configuration.
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We already know the equivalent resistance must be less than the smallest resistor here and if they are both 200 then we should expect an equivalent resistance that is less than 200 ohms.
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How do we solve it?
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One over our equivalent is going to be 1/R1 + 1/R2, therefore, 1 over our equivalent is 1/200 + 1/200 or you could say 1 over our equivalent -- 1/200 + 1/200 = 0.01.
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So if 1 over our equivalent is 0.01 then that means our equivalent is going to be 1/0.01, which is 100 ohms, which is less than 200.
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So the equivalent resistance is 100 ohms.
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You could take these two 200 ohm resistors and replace them by one 100 ohm resistor and have the same functional resistance.
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Now that is a lot of work, but there is a trick.
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If there are only two resistors in parallel, here is a short-cut formula.
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Our equivalent only works if you have two resistors as R1 × R2/R1 + R2, so in this case it would be 200 × 200/200 + 200, which is going to give you again, 100 ohms.
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This equation always works, however many resistors you have in parallel.
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This one is only good when you have two resistors.
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Let us take a look and practice some equivalent resistance analysis.
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Find the equivalent resistance of each of the following circuits.
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Well, we will start over here with Number 1 -- We have two resistors and as I look we are going to go from the positive side of our battery and the current will flow this way and it has two different paths before it re-combines and comes back into the battery.
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That must be a parallel circuit configuration.
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So our equivalent -- we only have two resistors, so I will use that formula for the two that I showed you a minute ago and we know the answer should be less than two -- is going to be R1 × R2/R1 + R2.
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That will be 2 × 2/(2 + 2), 4/(2 + 2), so 4/4 = 1 ohm and the equivalent resistance of that is 1 ohm.
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Let us take a look and do Number 2 now down here.
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Now we have just one current path through the resistors as we go back to the batteries, which is a series configuration.
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So for (2) our equivalent equals R1 + R2, which is going to be 2 + 2 or 4 ohms.
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Looking at Number 3 here, we have -- Well our current can go that way, it can go that way, it can go that way through the resistor or that way and then it is all going to keep coming back and re-combining.
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With multiple current paths there will be a parallel resistor configuration.
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We have four resistors in parallel, so for (3) -- 1 over our equivalent is going to be 1/2 + 1/2 + 1/2 + 1/2 or 1 over our equivalent is going to be equal to 1/2 + 1/2 = 1 and 1/2 + 1/2 = 1 for a total of 2.
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One over our equivalent is equal to 2, therefore our equivalent must be equal to 1/2 and our answer is going to be 1/2 of an ohm which is less than the smallest resistor in that configuration.
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Now for Number 4 down here -- We have a series circuit.
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The currents are going to flow from the positive side of our battery through those resistors and back in one current path, therefore our equivalent is going to be equal to...
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We have four resistors all of 2 ohms each, so that will be 2 + 2 + 2 + 2 = 8 ohms -- pretty straightforward.
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Let us take a look at the relationship between length and resistance.
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Which graph here best represents the relationship between resistance and length of a copper wire of uniform cross-sectional area at a constant temperature?
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Well, if I want to relate resistance and length, let us write a formula that shows that to me mathematically -- R = ρL/A.
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I am specifically worried about the resistance and the length.
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As length goes up, resistance goes up, so that gets rid of Number 1 and 2 and that looks like a direct linear relationship, therefore the correct answer here must be (3).
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Let us take a look at some resistors here.
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Which of the following resistors made of the same material has the highest resistance?
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Well, we will go back to our formula R = ρL/A and if they all have the same material -- they all have the same resistivity -- if we want the biggest resistance, we want something that has the biggest Ls and we want a very little cross-sectional area to give us a very big resistance.
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Which one of these has the biggest length and the smallest cross-sectional area?
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Obviously that one right there, which you could think of as a water pipe.
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Of all of these -- if they all carried water -- the one that would carry the least water -- have the most resistance to water flow is the long, skinny one.
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Let us take a look at one more here -- Going to compare some wires again.
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A length of copper wire and a 1 m long silver wire have the same cross-sectional area and resistance at 20 degrees (C).
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Calculate the length of the copper wire.
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Well we will start off by writing our relationship -- R = ρL/A and if that is all for copper and they have the same resistance that must equal ρL/A for silver.
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Now they both have the same -- Well, let us expand this first -- Let us write down this in a slightly different form.
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If that is all for copper, that is the resistivity of copper times the length of the copper over the cross-sectional area of the copper must equal the resistivity of silver times the length of the silver divided by the cross-sectional area of the silver.
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But we know that they have the same cross-sectional area, so we could multiply both sides by -- the cross-sectional areas will cancel out.
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And if we are looking for the length of the copper wire, let us now rearrange this to get the length of the copper all by itself.
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That means that the length of our copper wire is going to be equal to the resistivity of silver times the length of the silver wire divided by the resistivity of copper.
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Now I can substitute in my values the resistivity of silver that I get from my table and that is going to be 1.59 × 10^-8 ohm-metres and the length of my silver wire is 1 m and the resistivity of copper, from my table is 1.72 × 10^-8 ohm-metres.
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A little bit of division and calculator work and I come up with a length of about 0.924 m.
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Very good! All right, hopefully this gets you a good start on current and resistance.
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Thank you for watching Educator.com and make it a great day!