Physics 2112 Unit 9: Electric Current Today’s Concept: Electric Current Electricity & Magnetism Lecture 9, Slide 1.

Physics 2112Unit 9: Electric Current

Today’s Concept:Electric Current

Electricity & Magnetism Lecture 9, Slide 1

A Big Idea Review

Electric PotentialPotential energy per

unit charge

b

a

baba ldE

q

UV

Electric PotentialScalar Function that can be used to determine E VE

Gauss’ LawFlux through closed

surface is always proportional to charge

enclosed

0

encQAdE Gauss’ Law

Can be used to determine E field

SpheresCylinders

Infinite Planes


CapacitanceRelates charge and potential for two

conductor system V

QC

Electric FieldForce per unit charge

Electric FieldProperty of Space

Created by ChargesSuperposition q

FE

Coulomb’s LawForce law between

point charges q21,2r

1,2

F

q12,122,1

212,1 r̂

r

qkqF

ConductorsCharges free to move

SpheresCylinders

Infinite Planes

Gauss’Law

Field Lines & Equipotentials

ABCD

ABCD

ABCD

ABCD

Fiel

d Lin

es

Equipotentials

Work Done By E Field

b

a

b

a

ba ldEqldFW

b

ababa ldEqWU

Change in Potential Energy

Capacitor Networks

Series: (1/C23) = (1/C2) +

(1/C3)Parallel

C123 = C1 + C23

Applications of Big Ideas

What Determines How They Move?

They move until E = 0 !

E = 0 in conductor determines charge

densities on surfaces


Today’s Plan:

1) Review of resistance & preflights

2) Work out a circuit problem in detail

Key Concepts:

1) How resistance depends on A, L, s, r

2) How to combine resistors in series and parallel

3) Understanding resistors in circuits


I A

V

L

s

V = ELI = JA

Observables:

R = L sA

Ohm’s Law: J = s E

Conductivity – high for good conductors.

I/A = sV/L I = V/(L/sA)

I = V/RR = ResistanceR = 1/s


R = L sA

I is like flow rate of waterV is like pressureR is how hard it is for water to flow in a pipe

Civil Engineering Analogy

To make R big, make L long or A small

To make R small, make L short or A big


Battery is like a pump

Same current through both resistors

Compare voltages across resistors

A

LR

141

212 4 VVAA 1212 22 VVLL

1 CheckPoint: Two Resistors 2


A

LIRV


CheckPoint: Current DensityThe SAME amount of current I passes through three different resistors. R2 has twice the cross-sectional area and the same length as R1, and R3 is three times as long as R1 but has the same cross-sectional area as R1. In which case is the CURRENT DENSITY through the resistor the smallest?

Circuit

Unit 9, Slide 9

VB

R1

R2

a

b c

d

E is conservative force go completely around circuitVf = Vo

V

a b c d a

+VB-IR1

-IR2

Voltage

Current

Resistance

Series Parallel

Different for each resistor.Vtotal = V1 + V2

IncreasesReq = R1 + R2

Same for each resistorItotal = I1 = I2

Same for each resistor.Vtotal = V1 = V2

Decreases1/Req = 1/R1 + 1/R2

Wiring Each resistor on the same wire.

Each resistor on a different wire.

Different for each resistorItotal = I1 + I2

R1 R2

R1

R2

Resistor Summary


CheckPoint: Resistor Network 1


Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R1= R2 = R3 = R.

Compare the current through R2 with the current through R3:

A. I2 > I3

B. I2 = I3

C. I2 < I3




Compare the current through R1 with the current through R2:

A. I1/I2=1/2B. I1/I2=1/3C. I1 = I2

D. I1/I2=2E. I1/I2=3




Compare the voltage across R2 with the voltage across R3:

A. V2 > V3

B. V2 = V3 = VC. V2 = V3 < VD. V2 < V3

R1 = R2 = R3 = R

CheckPoint 2Compare the current through R1

with the current through R2

I1 I2

CheckPoint 3Compare the voltage across R2

with the voltage across R3

V2 V3

CheckPoint 4Compare the voltage across R1

with the voltage across R2

V1 V2





Compare the voltage across R1 with the voltage across R2.

A. V1 = V2 = VB. V1 = 1/2 V2 = VC. V1 = 2V2 = VD. V1 =1/2 V2 =1/5 VE. V1 =1/2 V2 = 1/2 V

Example 9.1

In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, and R4 = 4W.

What is V2, the voltage across R2?Conceptual Analysis: Ohm’s Law: when current I flows through resistance R, the potential drop V is given by:

V = IR. Resistances are combined in series and parallel combinations

Rseries = Ra + Rb

(1/Rparallel) = (1/Ra) + (1/Rb)

Strategic Analysis: Combine resistances to form equivalent resistances Evaluate voltages or currents from Ohm’s Law Expand circuit back using knowledge of voltages and currents

V

R1 R2

R4

R3


Example 9.1

V

R1 R2

R4

R3


V

R1

R234

VR1234

V

R1 R2

R24R3

Quick Follow-Ups


What is I3 ?

A) I3 = 2 A B) I3 = 3 A C) I3 = 4 A

V

R1

R234

a

b

V

R1 R2

R4

R3 =

V3 = V234 = 12V

What is I1 ?

We know I1 = I1234 = 6 A

V = 18V

R1 = 1W R2 = 2W R3 = 3W R4 = 4WR24 = 6W R234 = 2WV234= 12VV2 = 4VI1234 = 6 AmpsI3 = V3/R3 = 12V/3W = 4A

I1 = I2 + I3 Make Sense?

I1 I2

I3

Power In Resistors

Unit 9, Slide 19

dq

dW

Electro-Motive Force (EMF),

Energy provided to make charges move, units of V

Not a force!!

dt

dq

dt

dWPower

=VI (for a battery)

= I2R (for resistor)

Example 9.2

In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, and R4 = 4W.

How much electrical energy does the battery put into the circuit every second in the previous problem?

V

R1 R2

R4

R3

How much electrical energy does each resistor turn into thermal energy every second?

Parallel and Series (with color)

Unit 9, Slide 21

V

R1 R2

R4R3

If every electron that goes through one element must go through another, those two are in series.

If two sides of two elements can be connected by different colored lines, those two are in parallel. .

If two points are connected by a line not containing any circuit elements those point are at the same potential.

Physics 2112 Unit 9: Electric Current Today’s Concept: Electric Current Electricity & Magnetism Lecture 9, Slide 1.

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