AP Physics 1 & 2 Current & Resistance

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This lesson is on current and resistance.0002

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.0005

Let us start by talking about electric current.0018

An electric circuit is a path through which current flows and current is the amount of charge passing a point per unit time.0021

The units of current are coulombs per seconds also known as amperes or amps (A).0028

Average current is the charge passing a given point in a set amount of time.0035

Conventional current is defined as a direction of positive current flow.0043

This is a little non-intuitive. Typically, it is the electrons that are actually carrying the charge that are causing the currents.0048

Usually it is negative charges that are moving.0055

Conventional current flow says that the current is going the opposite direction of the electrons.0057

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?"0062

It is going to be the opposite of the direction of positive current flow.0069

That dates back to Ben Franklin and some of the initial work done on current electricity.0072

Current through a resistor - a charge of 30 coulombs passes through a 24 ohm resistor in 6 s.0080

Find the current through the resistor.0086

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.0090

How about current due to elementary charges (e)?0108

A charge flowing at a rate of 2.5 times 10¹⁶ elementary charges per second is equivalent to what current?0110

Same formula, I = δ(Q)/δ(T), but now we have to convert this into coulombs first.0117

That is going to be 2.5 × 10¹⁶e/s and if I want elementary charges to go away, 1 elementary charge is equal to 1.6 × 10^-19 coulombs.0124

Units of elementary charges will cancel out and I will be left with a current of about 0.004 amperes or amps.0140

Let us take a look at charges in a light bulb.0154

The current through a light bulb is 2 amperes. How many coulombs of electric charge pass through the light bulb in one minute?0157

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.0163

So that is going to be 120 coulombs of charge.0183

Let us look at a slightly more complex example.0193

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.0196

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?0213

Let us figure out the charge in each of these cases.0220

Our first case for the triple A-cell -- δQ = I(δT).0223

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.0230

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.0245

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.0267

Let us talk a little bit about conductivity and resistivity.0281

Electrical charges move easily in some materials -- we call those materials conductors and less freely in others and we call those materials, insulators.0285

A material's ability to conduct an electrical charge is known as its conductivity.0295

And conductivity is given the symbol, Greek letter, σ.0298

Conductivity of a material depends on the density of free charges available to move and the mobility of those free charges.0303

A material's ability to resist the movement of electrical charge, the inverse of conductivity, is known as its resistivity (rho).0311

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.0319

It is a little bit more popular as a measurement compared to conductivity.0331

Let us talk about resistance versus resistivity versus resistors.0336

Resistivity is a material property. Its units are ohm-metres.0340

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.0344

Resistance is a functional property of an element in a circuit and that tells you how that item is going to behave.0357

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.0363

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.0373

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 (rho).0386

The resistance of this resistor is equal to the material property resistivity times the length of the resistor divided by the cross-sectional area.0397

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.0407

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.0424

If you want a lot of resistance, you want a long, skinny narrow pipe or a long, skinny narrow wire.0435

Now circuit elements designed to impede the flow of current are known as resistors and resistors have some amount of resistance.0443

It is a circuit element. The resistance of a resistor tells you how well it impedes the flow of charges.0450

The resistance of the resistor depends upon the resistors geometry again, as we just said and its resistivity, that materials property.0456

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.0466

Over here on the right, a light bulb is typically modeled as a resistor too.0480

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.0485

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.0497

Find the resistivity of the wire in the material from which it is made.0511

Well, the first thing we need to point out is why is it telling us the resistivity is at 20 degrees (C).0516

Well, resistivity actually is somewhat of a function of temperature and as temperature goes up, oftentimes resistivity will go up a little bit.0522

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.0531

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 (rho).0545

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).0556

That should give us a resistivity of about 5.61 × 10^-8 ohm-metres.0579

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.0589

So tungsten is the material from which this resistor is made.0601

The electrical resistance of a metallic conductor is inversely proportional to its -- well we have to pick one.0609

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.0615

That means that resistance is going to go up and that is a directional relationship, not A.0626

As length goes up, resistance goes up -- direct, not inverse -- not B.0631

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.0637

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.0651

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.0659

If we talk about resistors in series -- series circuits have one single current path.0675

For example, if we have current flowing to the right through these two resistors, we have one current path that is a series circuit.0681

The equivalent resistance for resistors wired in series is found by just adding up the individual resistances.0688

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.0695

You just add them up -- that simple.0713

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...0717

...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.0731

Very easy to find the equivalent resistance of resistors in series.0750

Let us take a look at resistors in parallel, where parallel circuits are configurations where currents can take more than one path.0755

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.0765

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.0779

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.0787

But you have multiple current paths. It is a parallel configuration.0797

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.0801

How do we solve it?0812

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 = 0.01.0814

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 200.0835

So the equivalent resistance is 100 ohms.0846

You could take these two 200 ohm resistors and replace them by one 100 ohm resistor and have the same functional resistance.0849

Now that is a lot of work, but there is a trick.0860

If there are only two resistors in parallel, here is a short-cut formula.0862

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.0866

This equation always works, however many resistors you have in parallel.0889

This one is only good when you have two resistors.0895

Let us take a look and practice some equivalent resistance analysis.0901

Find the equivalent resistance of each of the following circuits.0905

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.0908

That must be a parallel circuit configuration.0924

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 + R2.0927

That will be 2 × 2/(2 + 2), 4/(2 + 2), so 4/4 = 1 ohm and the equivalent resistance of that is 1 ohm.0941

Let us take a look and do Number 2 now down here.0957

Now we have just one current path through the resistors as we go back to the batteries, which is a series configuration.0961

So for (2) our equivalent equals R1 + R2, which is going to be 2 + 2 or 4 ohms.0968

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.0970

With multiple current paths there will be a parallel resistor configuration.0993

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 2.1002

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.1020

Now for Number 4 down here -- We have a series circuit.1034

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...1039

We have four resistors all of 2 ohms each, so that will be 2 + 2 + 2 + 2 = 8 ohms -- pretty straightforward.1049

Let us take a look at the relationship between length and resistance.1061

Which graph here best represents the relationship between resistance and length of a copper wire of uniform cross-sectional area at a constant temperature?1067

Well, if I want to relate resistance and length, let us write a formula that shows that to me mathematically -- R = ρL/A.1075

I am specifically worried about the resistance and the length.1081

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).1088

Let us take a look at some resistors here.1102

Which of the following resistors made of the same material has the highest resistance?1104

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.1109

Which one of these has the biggest length and the smallest cross-sectional area?1131

Obviously that one right there, which you could think of as a water pipe.1136

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.1140

Let us take a look at one more here -- Going to compare some wires again.1151

A length of copper wire and a 1 m long silver wire have the same cross-sectional area and resistance at 20 degrees (C).1157

Calculate the length of the copper wire.1163

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.1166

Now they both have the same -- Well, let us expand this first -- Let us write down this in a slightly different form.1183

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.1193

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.1206

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.1216

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.1222

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.1236

A little bit of division and calculator work and I come up with a length of about 0.924 m.1257

Very good! All right, hopefully this gets you a good start on current and resistance.1266

Thank you for watching Educator.com and make it a great day!1271