AP Physics C: Electricity & Magnetism Current & Resistance

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about current and resistance.0003

Our objectives include understanding the definition of electric current.0008

Relating magnitude and direction of the current to the rate of flow of electric charge.0013

Relating current flow with drift velocity and the density of charge carriers in a conductor.0017

Relating current and voltage for a resistor.0022

Writing a relationship between electric field strength and current density in a conductor.0025

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.0029

Finding the resistance of a resistor of uniform cross section from its dimensions and the resistivity of that material.0036

Let us dive right in.0044

An electric current is the flow rate of electric charge, units are in C/s0046

which we also know as amperes which are given the symbol A and oftentimes you will hear that referred to as amps.0051

Positive current flow is the direction of the flow of positive charges.0059

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.0063

The direction of the charge carrier flow in most circuits, electrons is opposite what we call the direction of positive current flow.0070

Formally, current I is the amount of charge passing through a point at a given time.0079

Or DQ DT the time rate of change of charge.0086

Let us talk a little bit about drift velocity.0096

In a conductor, electrons are in constant thermal motion.0099

Net electron flow, however is 0, because that motion is in all directions.0103

It is random so they all cancel out.0107

When an electric field is applied, a small net flow in a direction opposite the electric field is observed.0109

The average velocity of these electrons due to the electric field is known as the electron drift velocity VD.0115

This is typically much smaller than that is the speed of the constant random thermal motion.0121

To give you an idea, let us take a look at the derivation of current flow.0129

Consider a uniform conductor of cross sectional area A and apply some electric field E.0135

Let us try to draw that in here, an electric field.0142

We will define the N as the volume density of charge carriers in this material.0148

Electrons in the conductor move randomly with thermal velocities.0153

We talked about that on the last slide.0157

Roughly 1,000,000 m/s, they are moving pretty quick but it is on random directions.0159

When we apply this electric field however, there is some small net movement of electrons opposite the direction of the electric field.0164

And that speed might be on the order say ½ cm/s compared to the thermal motion of 1,000,000 m/s.0172

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.0180

Let us say that that is their drift velocity, VD × some time interval Δ T × that cross sectional area A.0191

The electrons in that volume are going to pass surface A in time T.0200

From that, the total charge it is passing, A is equal to the product of the volume passing surface A.0207

The carrier density and the charge on each carrier, which we are going to call e.0214

We have got this amount and we have to deal with the VD TA.0218

If they want the charge that passes A in that period of time, that is going to be that carrier density ×0224

the charge e that goes with each of those charged carriers.0231

Typically, an elementary charge × that volume VD TA.0236

Since current is charge per unit time, we can say that current flow then is going to be N ×0243

our elementary charge × that drift velocity × that cross sectional area.0252

Or I oftentimes write this E as Q as well, so you may see in this form NQ VD A, the current flow derivation.0260

We can also look at this from the perspective of current density.0273

Current density through a surface is the current per area and it is a vector quantity usually given the symbol J.0276

J is the current density would be that carrier density × the amount of charge per carrier × the drift velocity VD,0284

which implies then as well that current flow I, is going to be the integral / that cross sectional surface of J ⋅ DA.0297

You can relate current flow to current density.0310

As we talked about this, we are also going to bring resistors in the play.0315

Resistance is the ratio of the potential drop across an object, the current flowing through the object.0318

Objects which have a fixed resistance that is not a function of the current potential drop are known as Ohmic materials.0324

And they are said to follow Ohm’s law, an empirical law.0330

Now R = V/ I, therefore, Ohm’s law V = IR, just a rearrangement of that.0335

The potential drop across a resistor is equal to the current flowing through it × its resistance.0346

This is a constant slope, a constant resistance regardless of the current or potential drop, we say that the material is Ohmic.0351

If we did not have a straight line, we would call that material a non Ohmic material.0360

What happens when you have a wire?0366

The resistance of the wire depends on the geometry of the wire.0369

As well as the property of the material the wire is made out of, known as its resistivity given the symbol ρ.0371

The units of ρ are ohm’s ω × m.0378

Resistivity relates to the ability of a material to resist the flow of electrons.0386

If we have a resistor of some length L and cross sectional area A, we can find its resistance R0390

is the resistivity × its length ÷ that cross sectional area.0399

You can almost think of it to look kind of like water in a pipe.0405

If you have a thicker diameter pipe, you have less resistance to water flow.0411

Thicker diameter wire, less resistance.0419

A is in the denominator.0424

The longer it is, the harder it is to push things out, the higher the resistance.0426

Same thing with the water pipe, the longer the pipe the more resistance to water flow.0431

Very similar and a nice analogy for helping to understand these qualitatively.0436

Let us see if we can refine Ohm’s law just a little bit.0442

We start with V = IR but we also just said that R = ρ L / A, our cylindrical resistor.0444

Then we have V = I ρ L ÷ A.0456

But we also know, because we have a uniform material that the electric field is going to be the potential drop ÷ the length.0462

That our electric field then is going to be ρ × I / A.0470

But this I/ A current per area is the current density.0478

This implies then because current density = current flow ÷ area, we can write then that our electric field0482

is equal to our resistivity × our current density vector.0492

Going further, we can even talk about the conversion of electrical energy to thermal energy.0504

The work done or energy used is charge × potential difference which implies that the time rate of change of that,0510

the rate of change of the work done with respect to time is going to be the derivative with respect to time of QV.0519

We also know that power is the time rate of change of work done, DW DT.0527

We could write then that power is equal to, potential should be a constant V DQ DT.0535

DQ DT we said that was current flow.0544

Therefore, we can write that power is current × voltage.0551

Or using Ohm’s law, V = IR, replacing V with IR power = I² R.0557

Or using Ohm’s law rearranged again.0567

If I = V/ R, let us replace I² with V²/ R².0570

We determine that power is also D² / R.0575

We have a couple different derivations for the rate of change at which energy is expanded which we call power.0579

Let us take a look at an example having to do with a silver wire.0591

The silver wire with 1/2 mm radius cross section is connected to the terminals of a 1V battery.0596

If the wire is 0.1 m long, determine the resistance of the wire.0602

It gives us some information, the resistivity of silver, its molar mass, and its mass density.0606

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 area0612

is going to be π × our radius² which is 0.0005 m² for resistance of 0.00202 ohms.0632

Let us see if we can extend this example a little bit further.0650

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.0656

Current is potential ÷ resistance.0665

We have a 1V potential and we just found our resistance 0.00202 ohms, gives us a current flow of around 494 amps.0670

Let us take this even further in a slightly more detailed calculation.0685

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.0692

In order to find the drift velocity, we first need to know the charge carrier density and0702

we will determine this by dividing Avogadro’s number by the volume of a mol of silver.0708

Then, we can find the drift velocity from our formula for current.0712

Let us start there with our charge carrier density is Avogadro’s number ÷ our volume which implies then,0715

since we know the resistivity of silver or molar density of silver is going to be our molar mass ÷ volume.0731

Or volume then is going to be our molar mass ÷ ρ.0742

Therefore, N is going to be equal to Avogadro’s number ρ silver / its molar mass.0749

We can look at the current flow I, we know as N × the charge for carrier V drift velocity A.0759

We want drift velocity so VD drift velocity will be I ÷ N EA.0768

But we just found N up here, so we will plug that in to determine that VD = IM/ NA ρ silver EA.0777

Or solving numerically that is going to be, we have got our 494 amps for our current.0797

We have our molar mass 0.1079 kg/ mol, making sure to put this into our standard units.0803

By the way, 10.5 g/ cc that is going to be 10,500 kg/ m³.0815

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 ×0828

our charge per carrier, that is our elementary charge 1.6 × 10 ⁻19 C × our area which is π R².0848

Π × 0.0005².0856

Put that all very carefully in your calculator, I come up with a drift velocity of about 0.067 m/ s.0861

A little bit more to do on that one.0874

Alright let us go on step further here with a silver wire.0878

Determine the average time required for electrons to pass from the negative terminal of the battery to the positive terminal.0884

We found the drift velocity and velocity is distance ÷ time.0892

Then time is going to be distance ÷ drift velocity.0897

And if we have to cross 0.1 m and our velocity is 0.067 m/ s,0903

that means it is going to take right about 1.5, 1.49 s for those electrons to travel that distance.0912

Let us take a look at one last example problem.0923

The 12 gauge aluminum wire with a cross sectional area of 3.31 × 10⁻⁶ m² carries a 4 amp current.0927

The density of aluminum is 2.7 g/ cc.0935

Find the drift velocity of the electrons and the wire assuming each aluminum atom supplies 1 conduction electron.0939

Starting off with what we know, our area is 3.31 × 10⁻⁶ m².0946

Our current is 4 amps, our ρ is going to be 2.7 g/ cc our density.0955

Or if we convert that into kg/ m³ that is going to be 2700 kg/ m³ standard units.0967

And our molar mass is 27 g/ mol for aluminum which is 0.027 kg/ mol.0976

We can go back to what we did in example 3 to give us at least part way there in this problem.0988

We know that is current is N EV DA.0993

We went through and we also found then that the drift velocity is current × that molar mass /0998

Avogadro’s number × our density of aluminum × our charge per charge carrier × A.1007

And if I substitute in my values, we end up with our current 4 amps, our molar mass was 0.027 kg/ mol.1017

Avogadro’s number 6.02 × 10 ⁺23.1031

We had our density 2700 kg/ m³, our charge per carrier 1.6 × 10 ⁻19 C our elementary charge.1036

And our cross sectional area it gives it to us here 3.31 × 10⁻⁶ m².1049

I come up with a drift velocity of right around 1.25 × 10⁻⁴ m/ s.1056

Hopefully that gets you a good start with current resistance.1072

Thank you so much for watching www.educator.com.1074

We will see a very soon, make it a great day everyone.1076