Lecture 07

Current & CircuitsOctober 11, 2005

Dis week …

Monday – finish resistance and current and begin electric circuit.

There is a new WA on board. Friday – Quiz on Monday-

Wednesday’s material NEXT FRIDAY – Examination #2

Studying is a good idea!

J and E

EJ

and 0A

1

let

EL

VJ

L

V

AL

A

V

RA

V

A

iJ

JAi

Consider a wire

d

d

d

neJ

Anet

Necurrent

tAnN

Micro-View “Resistivity”

m

eEa

m

eEa

d

a

depends on the material and is the mean time between collisions

ease of motion – mobilityresistance to motion - scattering

m

ne

EEEm

nenevJ d

2

2 1

Power

V i

+

-

ELECTRONs

Battery supplies energy to the resistor which, in turn, dissipates it in the form of heat.

Work done on charge Q = Q x V

RiiRiiVVt

Q

t

QVP

PPOWERtimeWork

2

/

REMEMBER: P=iV and P=i2R

The figure below gives the electrical potential V(x) along a copper wire carrying a uniform current, from a point at higher potential (x=0m) to a point at a lower potential (x=3m). The wire has a radius of 2.45 mm. What is the current in the wire?

copper

12 volts 0 volts

What does the graph tell us??

*The length of the wire is 3 meters.*The potential difference across the

wire is 12 volts.*The wire is uniform.

Let’s get rid of the mm radius and convert it to area in square meters:A=r2 = 3.14159 x 2.452 x 10-6 m2

orA=1.9 x 10-5 m 2

Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters

We have all we need….

ma 49.41067.2

1012

R

Vi

:Law sOhm' From

67.2 109.1

0.3m-ohm 1069.1

3

6

5

8

ohms

volts

mx

mx

A

LR

Series CombinationsR1 R2

i i

V1 V2V

iiRseriesR

general

RRR

iRiRiRVVV

and

iRV

iRV

)(

:21

2121

22

11

Parallel Combination??

R1, I1

R2, I2

V

i iRR

general

RRR

so

R

V

R

V

R

Viii

iRV

11

111

..

21

2121

What’s This???

In Fig. 28-39, find the equivalent resistance between points (a) F and H and [2.5]  (b) F and G. [3.13]  

Moving on …..

Fun and Frolic With Electric Circuits

Power Source in a Circuit

V

The ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher.

A REAL Power Sourceis NOT an ideal battery

V

E or Emf is an idealized device that does an amount of work E to move a unit charge from one side to another.

By the way …. this is called a circuit!

Internal Resistance

A Physical Battery

Internal Resistance Rr

Emfi

Back to Potential

Represents a charge in space

Change in potential as one circuitsthis complete circuit is ZERO!

Consider a “circuit”.

This trip around the circuit is the same as a path through space.

THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!!

To remember

In a real circuit, we can neglect the resistance of the wires compared to the resistors. We can therefore consider a wire in a circuit to

be an equipotential – the change in potential over its length is slight compared to that in a resistor

A resistor allows current to flow from a high potential to a lower potential.

The energy needed to do this is supplied by the battery.

VqW

NEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.

LOOP EQUATION The sum of the voltage drops (or rises)

as one completely travels through a circuit loop is zero.

Sometimes known as Kirchoff’s loop equation.

NODE EQUATION The sum of the currents entering (or

leaving) a node in a circuit is ZERO

TWO resistors againi

R1 R2

V1 V2

V

jj

21

21

RR

Resistors SERIESfor General

RRR

or

iRiRiRV

A single “real” resistor can be modeledas follows:

R

a b

V

position

ADD ENOUGH RESISTORS, MAKING THEM SMALLERAND YOU MODEL A CONTINUOUS VOLTAGE DROP.

Take a trip around this circuit.

Consider voltage DROPS:

-E +ir +iR = 0or

E=ir + iRrise

Circuit Reduction

i=E/Req

Multiple Batteries

Reduction

Computes i

Another Reduction Example

PARALLEL

1212

1

600

50

30

1

20

11

RR

NOTICE ASSUMED DIRECTION OF TRAVEL

Voltage Drops:-E1 –i1R1 + i2R2 + E2 +i1R1 = 0

From “a”

-i3R1 + E2 – E2 –i2R2 =0 NODEI3 +i2 = i1

In the figure, all the resistors have a resistance of 4.0 and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R?

The Unthinkable ….

RC Circuit

Initially, no current through the circuit

Close switch at (a) and current begins to flow until the capacitor is fully charged.

If capacitor is charged and switch is switched to (b) discharge will follow.

Close the Switch

I need to use E for E

Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)

Really Close the Switch

I need to use E for E

R

E

RC

q

dt

dq

or

EC

q

dt

dqR

C

qiRE

dt

dqi since

0

Equation Loop

Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)

This is a differential equation.

To solve we need what is called a particular solution as well as a general solution.

We often do this by creative “guessing” and then matching the guess to reality.

You may or may not have studied this topic … but you WILL!

RC

REaeCE

R

E

RC

q

dt

dq

CEq

R

E

RC

q

dt

dq

Keqq

at

p

atp

1

RCE

Ea

E/R0CEa

0for t

/)e-CE(1)(

)e-CE(1q and

-CEK

KCE0

solution from and 0q 0,When t

and 0dq/dt charged,fully is device When the

:solution particularat Look

Solution General

at-

at-

Time Constant

RC

Result q=CE(1-e-t/RC)

q=CE(1-e-t/RC) and i=(CE/RC) e-t/RC

RCteR

Ei /

Discharging a Capacitor

qinitial=CE BIG SURPRISE! (Q=CV)i

iR+q/C=0

RCt

RCt

eRC

q

dt

dqi

eqq

solutionC

q

dt

dqR

/0

/0

0

In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase of thermal energy Eth in the resistor as a function of time t.

(a)What is the electric potential across the battery? (60)(b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39)(c) Did you put your name on your paper? (1)

Looking at the graph, we see that theresistor dissipates 0.5 mJ in one second.

Therefore, the POWER =i2R=0.5 mW

ma 88.41088.4

1038.2Ω 21

mW 0.5

3

252

ampi

ampR

Pi

mVamp 10221104.88iRV

or iR reisitor theacross drop Voltage3-

If the resistance is doubled what is the power dissipated by the circuit?

mJRiP

ma

R

248.0

43.242

10102

R

Vi

mV 102V 42

2

3