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April 19, 2023April 19, 2023 11CURRENT ELECTRICITYCURRENT ELECTRICITY

Demonstration 4Demonstration 4Independent and Non-independent BranchesIndependent and Non-independent Branches

What changes will you see in A, B C

when RHS (w&x or y&z) is disconnected?

C

Circuit 2

Series & parallel

B y

z

A

Circuit 1

Parallel

w

x


Demonstration 4 circuit 1Demonstration 4 circuit 1

OBSERVATION• A is unchanged when branch w&x is disconnectedCONCLUSION• Parallel branches across battery are INDEPENDENT

A

n

o A

AA

ww

xx AA



OBS: brightness B decreases and C increases

B

y

zC

y

zC

B

y&z branch y&z branch disconnecteddisconnected

Loops are NOT INDEPENDENT Loops are NOT INDEPENDENT if they share a resistor (eg Bulb B)if they share a resistor (eg Bulb B)


CONCL: B and C are not independent of y & z.


Brightness B > C > y = z Battery current is shared

by C and yzC has less resistance than yz,

so C more current

B

y

zC

y

zC

B

y&z branch y&z branch disconnecteddisconnected

B and C B and C now in now in seriesseries

B and C have different B and C have different current, so not in series current, so not in series


When yz disconnected B (common to both loops) becomes

dimmer – now yz has no current C becomes brighter – battery current

not shared Now current B = C (series)

TAKE NOTE: Series and parallelTAKE NOTE: Series and parallelA and B in seriesB and F in parallelB and E in seriesA, E and F in series B in parallel to CD

1. False2. False3. False4. True5. True

April 19, 2023April 19, 2023 CURRENT ELECTRICITYCURRENT ELECTRICITY 66

C

B

D

A

F

E

Current splits Current splits at nodeat node

Geometry is not electrical

parallelCurrent joins Current joins

at nodeat node

Demonstration 5: Kirchoff’s 1Demonstration 5: Kirchoff’s 1stst Law Law

• Predict the relationship between currents at A,B,C,D,E,F

E

F

A

B

C

D

i


Demonstration 5 Compare the currentsDemonstration 5 Compare the currentsResultsResults

PositionPosition Current(A)Current(A)

AA 0.240.24

BB 0.230.23

CC 0.160.16

DD 0.150.15

EE 0.080.08

FF 0.240.24

A

E

B

C

D

F

NodeNode

NodeNode A = B = F (battery A = B = F (battery current)current)

C = D (series current)C = D (series current)B = C + EB = C + E


ConclusionsConclusionsCurrent can split or combine at a node Current can split or combine at a node (eg B or D) (eg B or D) Current Current mustmust split or combine at a node split or combine at a node

Kirchoff’s First Law (KI)Kirchoff’s First Law (KI)

The total current going into a node equals the total current going out of a node

AA

Circuit 1Circuit 1

BB xx

yy

CC

Circuit 2Circuit 2

nn

oo

nodenode

nodenodeApril 19, 2023April 19, 2023 99CURRENT ELECTRICITYCURRENT ELECTRICITY

Current relationshipsCurrent relationships

• iA = iB = iF = battery current

• iC = iD (series)

• iB = iC + iE (at node x)

• iD + iE = iF (at node Y)

Kirchoff’s first law (K I) The total current going into a node equals the total current going out of a node

E

F

A

B

C

D

i Node XNode X

Node YNode Y


Exercise 1 page 8Exercise 1 page 8

IA = 8 A (battery current)

At node q: Iin = Iout (K I law)

6 A + IB = 8 A

IB = 2 A

IC = 2 A (series with B)

ID = 6 A (given)


Current Splitting: Exercise 1Current Splitting: Exercise 1Current is inversely proportional to resistanceCurrent is inversely proportional to resistance

i2

6.0 A

2 Ω

3 Ω

In these examples you are given one In these examples you are given one current and find another by inverse current and find another by inverse

proportionproportionApril 19, 2023April 19, 2023 1212CURRENT ELECTRICITYCURRENT ELECTRICITY

For 3 or more resistors, For 3 or more resistors, compare them in pairscompare them in pairs


Exercise 2: Splitting up the TOTAL currentExercise 2: Splitting up the TOTAL current

i5

3.0 A

r

ri6

For equal resistors

Current splits equally

i5 = i6 = ½ × 3.0 A

= 1.5 A


CHECKiTOTAL = 9.0 + 6.0 = 15

A

For unequal resistors, current splits inverselyFor unequal resistors, current splits inversely

I4 : I6 = 6 : 4 = 6 : 4 (10

parts)

i7

15 A

4

6 i8

II1 1 :: I I22 = R = R2 2 :: RR11


Exercise 3 (a)Exercise 3 (a)Identical bulbsIdentical bulbsID = 0.200 A (parallel to C)

IA = IB = ½ × 0.200 A = 0.100A

(A, B in series; 2 R gets ½ I )

At node s: K I law

0.200 A + 0.200 A = 0.400 A

At node r: K I law

0.400 A + 0.100 A = 0.500 A

Battery current = 0.500 A


0.400 A

0.500 A

200 mA

0.200 A

0.100 A

0.100 A

0.200 A

iA = i battery (series) = 120 mAShare 120 mA in 2 branchesiB : iCD = 2R : R = 2 : 1 (3 parts inv prop) iB = = 80 mA

Exercise 3 (b) Exercise 3 (b) Identical Identical bulbsbulbs


A

120 mA

C

D

i

B

CHECK : CHECK : IItotal total = 80 + 40 = = 80 + 40 = 120 mA 120 mA

0.40 A

G

F

J

K

p

0.20 A

r

q

s

E3R

R

R RH

m n

Exercise 3 (c)Exercise 3 (c)iK = iJ (parallel, same bulbs) = 0.40 AiF = iR = × 0.20 A = 0.60 A

iG = iH (series) = i2R = × 0.20 A = 0.30 AApril 19, 2023April 19, 2023 1818CURRENT ELECTRICITYCURRENT ELECTRICITY

Using Kirchoff’s 1st lawAt n: Inq = 0.20 A + 0.60 A + 0.30 A = 1.10 AAt s: Isq = 0.40 A+ 0.40 A = 0.80 AAt q: Ibattery = 1.10 A + 0.80 A = 1.90 A

0.40 A

0.80 A

1.10 A

0.60 A

0.30 A

Summary of current: General Summary of current: General


• Rate of flow of charge in a closed circuit

• Measured in amperes (A). 1 A = 1 C s-1

• Inversely proportional to resistance of circuit

• Battery current is not fixed; it depends on the components of the circuit and how they are connected.

• The total current going into a node is equal to the total current coming out of that node (Kirchoff’s first law).

t

QI

RcircuitIbattery

1


Series Circuits Parallel Circuits

Has no nodes Has at least two nodes

More R will increase Rtotal

And decrease Ibattery

More R will decrease R||

And increase Ibattery

Ibattery = I1 = I2 = I3 … I battery = I1 + I2 + I3 + … Kirchoff’s Law

Ibattery inversely proportional to Rtotal I splits inversely proportional to R

Equal resistors: I splits equallyOR

RA and RB are NOT IN SERIES

RB is parallel to RC & RD

Combined Series & ParalleCombined Series & ParallellIbattery = IA = IE (series)

IC = ID (series)

At x: IA = Ibranch1 + Ibranch2

(Kirchoff I)

At y: Ibranch1 + Ibranch2 = IE

(Kirchoff I)

For parts given

I1 : I2 = R2 : R1 or

For Whole Given


IA

IE

I2 =

IC =

ID

RA

RB

I1 = IB

x

y

RC

RD

Now you needPart 2: Voltage

To get back to the computer press ESC key (top left of

keyboard)April 19, 2023April 19, 2023 CURRENT ELECTRICITYCURRENT ELECTRICITY 2222

Well done.Well done.

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