For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

### Current & Resistance

- Current is the flow rate of electric charge. Conventional current flows in the direction that a positive charge moves. If negative electrons are the charge carriers, conventional current flows opposite the direction of electron flow.
- In a conductor, electrons are in constant thermal motion. The net electron flow, however, is zero because the motion is random. When an electric field is applied, a small net flow in a direction opposite the electric field is observed. The average velocity of these electrons due to the electric field is known as electron drift velocity.
- Resistance is the ratio of the potential drop across an object to the current flowing through the object.
- Objects which have a fixed resistance are known as ohmic materials and follow Ohm’s Law (R=V/I).
- The current density through a surface is the current per area, and is a vector quantity (J).
- The resistance of a wire depends on the geometry of the wire as well as the resistivity of the wire, a material property relating to the ability of the material to resist the flow of electrons.

### Current & Resistance

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Electric Current
- Drift Velocity
- Derivation of Current Flow
- Current Density
- Resistance
- Ratio of the Potential Drop Across an Object to the Current Flowing Through the Object
- Ohmic Materials Follow Ohm's Law
- Resistance of a Wire
- Depends on Resistivity
- Resistivity Relates to the Ability of a Material to Resist the Flow of Electrons
- Refining Ohm's Law
- Conversion of Electric Energy to Thermal Energy
- Example 1
- Example 2
- Example 3
- Example 4
- Example 5

- Intro 0:00
- Objectives 0:08
- Electric Current 0:44
- Flow Rate of Electric Charge
- Amperes
- Positive Current Flow
- Current Formula
- Drift Velocity 1:35
- Constant Thermal Motion
- Net Electron Flow
- When Electric Field is Applied
- Electron Drift Velocity
- Derivation of Current Flow 2:12
- Apply Electric Field E
- Define N as the Volume Density of Charge Carriers
- Current Density 4:33
- Current Per Area
- Formula
- Resistance 5:14
- Ratio of the Potential Drop Across an Object to the Current Flowing Through the Object
- Ohmic Materials Follow Ohm's Law
- Resistance of a Wire 6:05
- Depends on Resistivity
- Resistivity Relates to the Ability of a Material to Resist the Flow of Electrons
- Refining Ohm's Law 7:22
- Conversion of Electric Energy to Thermal Energy 8:23
- Example 1 9:54
- Example 2 10:54
- Example 3 11:26
- Example 4 14:41
- Example 5 15:24

### AP Physics C: Electricity and Magnetism Online Course

I. Electricity | ||
---|---|---|

Electric Charge & Coulomb's Law | 30:48 | |

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

Electric Potential & Electric Potential Energy | 1:14:03 | |

Electric Potential Due to Continuous Charge Distributions | 1:01:28 | |

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

RC Circuits: Transient Analysis | 1:01:07 | |

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

Forces on Current-Carrying Wires | 17:52 | |

Magnetic Fields Due to Current-Carrying Wires | 24:43 | |

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

Faraday's Law & Lenz's Law | 1:04:33 | |

IV. Inductance, RL Circuits, and LC Circuits | ||

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

1998 AP Practice Exam: Multiple Choice Questions | 32:33 | |

1998 AP Practice Exam: Free Response Questions | 29:55 |

### Transcription: 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/s*0046

*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 R*0390

*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 field*0482

*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 area*0612

*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 and*0702

*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

0 answers

Post by Parth Shorey on December 3, 2015

Thank you for getting back on Q&A and I appreciate the help that I am getting regardless of what date it is.

I still don't understand why "N" would be a volume density of charge carriers? And Why would you use Q=Ne?

3 answers

Last reply by: Professor Dan Fullerton

Tue Jun 9, 2015 6:38 AM

Post by Sagar Rathee on June 3, 2015

sir,

we calculated acceleration of electron in a conductor, but according to classical electrodynamics, any charged particle while accelerating looses energy through EM radiations, So does these electrons also emit radiation, if not then please tell me WHY?

1 answer

Last reply by: Professor Dan Fullerton

Tue Apr 7, 2015 2:07 PM

Post by Thadeus McNamara on April 7, 2015

and can you also explain why resistivity = molar mass / volume

1 answer

Last reply by: Professor Dan Fullerton

Tue Apr 7, 2015 2:07 PM

Post by Thadeus McNamara on April 7, 2015

at around 12:30, how did you get N = avogadros number/ V ?

i thought N was # of atoms / V ? So why must the # be exactly avogadros number? can't it be any number, depending on the material is?

1 answer

Last reply by: Professor Dan Fullerton

Mon Apr 6, 2015 12:25 PM

Post by Thadeus McNamara on April 6, 2015

why is there chemistry in these problems? do we need to know how to solve questions with moles and avagadros number on the ap