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 talk about current and resistance.
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Our objectives include understanding the definition of electric current.
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Relating magnitude and direction of the current to the rate of flow of electric charge.
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Relating current flow with drift velocity and the density of charge carriers in a conductor.
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Relating current and voltage for a resistor.
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Writing a relationship between electric field strength and current density in a conductor.
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Describing how the resistance of a resistor depends upon its length and cross sectional area, as well as the material it is made out of.
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Finding the resistance of a resistor of uniform cross section from its dimensions and the resistivity of that material.
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Let us dive right in.
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An electric current is the flow rate of electric charge, units are in C/s
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which we also know as amperes which are given the symbol A and oftentimes you will hear that referred to as amps.
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Positive current flow is the direction of the flow of positive charges.
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It can be a little bit confusing realizing that in most of the circuits we are going to talk about is actually electrons which are moving.
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The direction of the charge carrier flow in most circuits, electrons is opposite what we call the direction of positive current flow.
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Formally, current I is the amount of charge passing through a point at a given time.
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Or DQ DT the time rate of change of charge.
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Let us talk a little bit about drift velocity.
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In a conductor, electrons are in constant thermal motion.
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Net electron flow, however is 0, because that motion is in all directions.
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It is random so they all cancel out.
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When an electric field is applied, a small net flow in a direction opposite the electric field is observed.
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The average velocity of these electrons due to the electric field is known as the electron drift velocity VD.
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This is typically much smaller than that is the speed of the constant random thermal motion.
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To give you an idea, let us take a look at the derivation of current flow.
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Consider a uniform conductor of cross sectional area A and apply some electric field E.
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Let us try to draw that in here, an electric field.
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We will define the N as the volume density of charge carriers in this material.
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Electrons in the conductor move randomly with thermal velocities.
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We talked about that on the last slide.
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Roughly 1,000,000 m/s, they are moving pretty quick but it is on random directions.
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When we apply this electric field however, there is some small net movement of electrons opposite the direction of the electric field.
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And that speed might be on the order say ½ cm/s compared to the thermal motion of 1,000,000 m/s.
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If we define N as that volume density of charge carriers, that means the electrons contained in some volume, let us highlight it here in yellow.
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Let us say that that is their drift velocity, VD × some time interval Δ T × that cross sectional area A.
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The electrons in that volume are going to pass surface A in time T.
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From that, the total charge it is passing, A is equal to the product of the volume passing surface A.
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The carrier density and the charge on each carrier, which we are going to call e.
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We have got this amount and we have to deal with the VD TA.
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If they want the charge that passes A in that period of time, that is going to be that carrier density ×
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the charge e that goes with each of those charged carriers.
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Typically, an elementary charge × that volume VD TA.
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Since current is charge per unit time, we can say that current flow then is going to be N ×
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our elementary charge × that drift velocity × that cross sectional area.
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Or I oftentimes write this E as Q as well, so you may see in this form NQ VD A, the current flow derivation.
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We can also look at this from the perspective of current density.
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Current density through a surface is the current per area and it is a vector quantity usually given the symbol J.
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J is the current density would be that carrier density × the amount of charge per carrier × the drift velocity VD,
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which implies then as well that current flow I, is going to be the integral / that cross sectional surface of J ⋅ DA.
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You can relate current flow to current density.
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As we talked about this, we are also going to bring resistors in the play.
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Resistance is the ratio of the potential drop across an object, the current flowing through the object.
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Objects which have a fixed resistance that is not a function of the current potential drop are known as Ohmic materials.
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And they are said to follow Ohm’s law, an empirical law.
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Now R = V/ I, therefore, Ohm’s law V = IR, just a rearrangement of that.
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The potential drop across a resistor is equal to the current flowing through it × its resistance.
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This is a constant slope, a constant resistance regardless of the current or potential drop, we say that the material is Ohmic.
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If we did not have a straight line, we would call that material a non Ohmic material.
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What happens when you have a wire?
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The resistance of the wire depends on the geometry of the wire.
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As well as the property of the material the wire is made out of, known as its resistivity given the symbol ρ.
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The units of ρ are ohm’s ω × m.
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Resistivity relates to the ability of a material to resist the flow of electrons.
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If we have a resistor of some length L and cross sectional area A, we can find its resistance R
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is the resistivity × its length ÷ that cross sectional area.
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You can almost think of it to look kind of like water in a pipe.
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If you have a thicker diameter pipe, you have less resistance to water flow.
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Thicker diameter wire, less resistance.
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A is in the denominator.
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The longer it is, the harder it is to push things out, the higher the resistance.
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Same thing with the water pipe, the longer the pipe the more resistance to water flow.
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Very similar and a nice analogy for helping to understand these qualitatively.
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Let us see if we can refine Ohm’s law just a little bit.
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We start with V = IR but we also just said that R = ρ L / A, our cylindrical resistor.
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Then we have V = I ρ L ÷ A.
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But we also know, because we have a uniform material that the electric field is going to be the potential drop ÷ the length.
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That our electric field then is going to be ρ × I / A.
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But this I/ A current per area is the current density.
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This implies then because current density = current flow ÷ area, we can write then that our electric field
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is equal to our resistivity × our current density vector.
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Going further, we can even talk about the conversion of electrical energy to thermal energy.
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The work done or energy used is charge × potential difference which implies that the time rate of change of that,
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the rate of change of the work done with respect to time is going to be the derivative with respect to time of QV.
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We also know that power is the time rate of change of work done, DW DT.
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We could write then that power is equal to, potential should be a constant V DQ DT.
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DQ DT we said that was current flow.
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Therefore, we can write that power is current × voltage.
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Or using Ohm’s law, V = IR, replacing V with IR power = I² R.
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Or using Ohm’s law rearranged again.
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If I = V/ R, let us replace I² with V²/ R².
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We determine that power is also D² / R.
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We have a couple different derivations for the rate of change at which energy is expanded which we call power.
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Let us take a look at an example having to do with a silver wire.
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The silver wire with 1/2 mm radius cross section is connected to the terminals of a 1V battery.
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If the wire is 0.1 m long, determine the resistance of the wire.
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It gives us some information, the resistivity of silver, its molar mass, and its mass density.
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Resistance is ρ L / A, where our resistivity here is 1.59 × 10⁻⁸ ohm meters, our length is 0.1 m and our cross sectional area
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is going to be π × our radius² which is 0.0005 m² for resistance of 0.00202 ohms.
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Let us see if we can extend this example a little bit further.
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A silver wire with 1/2 mm radius cross section connected to the same battery, same length of wire, determine the current flowing through the wire.
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Current is potential ÷ resistance.
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We have a 1V potential and we just found our resistance 0.00202 ohms, gives us a current flow of around 494 amps.
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Let us take this even further in a slightly more detailed calculation.
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The same wire but now we are asked to find the drift velocity of the free electrons and the wire assuming 1 free electron per atom.
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In order to find the drift velocity, we first need to know the charge carrier density and
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we will determine this by dividing Avogadro’s number by the volume of a mol of silver.
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Then, we can find the drift velocity from our formula for current.
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Let us start there with our charge carrier density is Avogadro’s number ÷ our volume which implies then,
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since we know the resistivity of silver or molar density of silver is going to be our molar mass ÷ volume.
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Or volume then is going to be our molar mass ÷ ρ.
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Therefore, N is going to be equal to Avogadro’s number ρ silver / its molar mass.
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We can look at the current flow I, we know as N × the charge for carrier V drift velocity A.
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We want drift velocity so VD drift velocity will be I ÷ N EA.
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But we just found N up here, so we will plug that in to determine that VD = IM/ NA ρ silver EA.
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Or solving numerically that is going to be, we have got our 494 amps for our current.
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We have our molar mass 0.1079 kg/ mol, making sure to put this into our standard units.
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By the way, 10.5 g/ cc that is going to be 10,500 kg/ m³.
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We have got to divide all this by Avogadro’s number 6.02 × 10 ⁺23 × our mass density for silver which we said 10,500 kg/ m³ ×
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our charge per carrier, that is our elementary charge 1.6 × 10 ⁻19 C × our area which is π R².
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Π × 0.0005².
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Put that all very carefully in your calculator, I come up with a drift velocity of about 0.067 m/ s.
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A little bit more to do on that one.
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Alright let us go on step further here with a silver wire.
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Determine the average time required for electrons to pass from the negative terminal of the battery to the positive terminal.
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We found the drift velocity and velocity is distance ÷ time.
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Then time is going to be distance ÷ drift velocity.
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And if we have to cross 0.1 m and our velocity is 0.067 m/ s,
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that means it is going to take right about 1.5, 1.49 s for those electrons to travel that distance.
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Let us take a look at one last example problem.
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The 12 gauge aluminum wire with a cross sectional area of 3.31 × 10⁻⁶ m² carries a 4 amp current.
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The density of aluminum is 2.7 g/ cc.
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Find the drift velocity of the electrons and the wire assuming each aluminum atom supplies 1 conduction electron.
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Starting off with what we know, our area is 3.31 × 10⁻⁶ m².
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Our current is 4 amps, our ρ is going to be 2.7 g/ cc our density.
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Or if we convert that into kg/ m³ that is going to be 2700 kg/ m³ standard units.
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And our molar mass is 27 g/ mol for aluminum which is 0.027 kg/ mol.
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We can go back to what we did in example 3 to give us at least part way there in this problem.
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We know that is current is N EV DA.
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We went through and we also found then that the drift velocity is current × that molar mass /
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Avogadro’s number × our density of aluminum × our charge per charge carrier × A.
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And if I substitute in my values, we end up with our current 4 amps, our molar mass was 0.027 kg/ mol.
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Avogadro’s number 6.02 × 10 ⁺23.
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We had our density 2700 kg/ m³, our charge per carrier 1.6 × 10 ⁻19 C our elementary charge.
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And our cross sectional area it gives it to us here 3.31 × 10⁻⁶ m².
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I come up with a drift velocity of right around 1.25 × 10⁻⁴ m/ s.
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Hopefully that gets you a good start with current resistance.
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Thank you so much for watching www.educator.com.
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We will see a very soon, make it a great day everyone.