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Department of Physics and Applied Physics PHYS.1440 Lecture 11 Danylov Lecture 11 Chapter 28 Resistors Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsII Physics II

L11 Ch27 Resistors - University of Massachusetts Lowell

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Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Lecture 11

Chapter 28

Resistors

Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsII

Physics II

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Today we are going to discuss:

Chapter 28:

Section 28.1-6 Resistors 28.7 (Example 28.29)

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Circuit Elements

Slide 31-22

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Resistorsin series/parallel

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

+ +

Resistors in ParallelConsider three resistors connected in parallel.

I

Rea

l cir

cuit

Equ

ival

ent c

ircu

it

ΔV

Resistors in parallel have the same potential difference, ΔV

I + +

;

We have replaced 3 resistors with an “equivalent” resistor.

+ +

Conservation of current

Req is inserted without changing the operation of the circuit, so I and ΔV are same as in the real circuit

Equivalent resistance of resistors in parallel.

=

I1

I2

I3

Ohm’s law;

ΔV

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

+ +

Resistors in SeriesConsider three resistors connected in series.

Rea

l cir

cuit

Equ

ival

ent c

ircu

it

ΔV

+ +

Ohm’s law ∆

Req is inserted without changing the operation of the circuit, so I and ΔV are same as in the real circuit

Equivalent resistance of resistors in series.

ΔV

ΔV1 ΔV2 ΔV3

∆∆

+ +

Are headlight wired: A) in parallel?

B) in series?

ConcepTest Headlights

ConcepTest Resistors I

The battery current I is

A) 3 A

B) 2 A

C) 1 A

D) 2/3 A

E) ½ A

=2/3 A

+

ConcepTest Resistors II

The battery current I is

A) 3 A

B) 2 A

C) 1 A

D) 2/3 A

E) ½ A

=3 A

+

=4

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Exa

mpl

e:

Ana

lyzi

ng a

com

plex

cir

cuit

a)Find the equivalent resistance.b)Find the current through and the potential difference across each of the resistors in the circuit.

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Real batteries

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Give me a break! I do it as fast as I

can!

Real Batteries. Internal resistanceTo drive a current in a circuit we need a “charge pump”, a device that by doing work on the charge carriers maintains a potential difference.Let’s look at a gravitational analog of a battery:

A person does work to maintain a steady flow of balls through “the circuit”.However, this guy cannot move balls instantaneously. It takes time. So there is a natural hindrance to a completely free flow. To describe this hindrance we can introduce the internal resistance, r.It is inside a battery and it cannot be separated from the battery.

Pot. difference of a battery without an internal resistance is called an electromotive force.(EMF, ε)

∆Terminal voltage

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Why is electric energy useful?

It can be easily transformed into other forms of energy.

Electric energy

Mechanical energy Thermal

energy

LightE/M waves

How to find the power transformed by these electrical devices

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Electric PowerConsider any electrical device:

ΔVSome charge Q

Δt (for Q to go through the device)

As some charge Q moves through the potential difference ΔV, the potential energy of the charge changes by: ∆ ∆Let Δt be the time required the charge to move through the potential difference ΔV. Then, the power P (the rate energy is transformed, Physics I) is:

∙ ∆∆ ∆ ∙ ∆ ∙ ∆

power transformed by an electrical device

The power transformed by a resistor can be written like this: ∙ ∆ ∆

∙ ∆ /

∆∆

A) 2 AB) 3 AC) 4 AD) 5 AE) 25 A

Most loudspeakers are designed to have a resistance of 4 Ω. If it is connected to an amplifier with a rating of 100 W, what is the current to the loudspeaker?

ConcepTest Electric Power

∙ ∆ ∆ ∙

Department of Physics and Applied PhysicsPHYS.1440 Lecture 11 Danylov

Thank youSee you next time