electricity magnetism lecture 8: kirchhoff’s rules

Electricity & MagnetismLecture 8: Kirchhoff’s Rules

Today’sConcept:

Kirchhoff’sRules

Electricity&Magne<smLecture10,Slide1

News‣ DeadlineforUnit23Ac<vityguideandWriKenHomeworkispushedtoMonday,Feb5duetoMidtermsinotherclasses.

‣ SomeotherFlipItDeadlineshavebeenpostponed‣ Youcans<lldothemearlyifyouwant

‣ YoushoulddefinitelyfinishtheKirchhoffAc<vitybeforemovingontoUnit24.

‣ OurmidtermisscheduledforFriday,Feb2‣ WillcoverupthroughFriday’sFlipItandAc<vityGuide.‣ Ac<vityGuidesanswersheetscanbeused.‣ AleKersizedformulasheetisallowed.‣ D100AllTables:RoomSUR5140

If the batteries are ideal andVA = 1.5 VA)VAB = 0.0 VB)VAB = 0.5 VC)VAB = 1.5VD)VAB = 3.0 VE)something elseF)

V

V

VV

V

Figure 23-2: Voltmeters connected to measure the potential difference across (a) a single battery, (b) a single battery and two batteries connected in series, and (c) a single battery and two batteries connected in parallel.

Activity 23-2: Combinations of Batteries(a) Predict the voltage for each combination of batteries in Fig 23-2. Write

you prediction beside the meter symbols.(b) Measure the voltages you predicted and write them below the predicted

values on the figure.

Using a MultimeterA digital multimeter (DMM) is a device that can be used to measure either current, voltage or resistance depending on how it is set up. We have already used one to measure voltage. The following activity will give you some practice in using it as an ohmmeter. You will need:! ! • A digital multimeter! ! • A D-cell alkaline battery w/ holder ! ! • A SPST switch! ! • 4 alligator clip wires! ! • 1 resistor, 10 Ω

VΩCOMMAA

Ω

V A

MA

Figure 23-6: Diagram of a typical digital multimeter that can be used to measure resistances, currents, and voltages

Page 23-12 Workshop Physics II Activity Guide SFU

© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by S. Johnson, N. Alberding, 2014.

V

V

VV

V






VΩCOMMAA

Ω

V A

MA




VAVAVAB

V

V

VV

V






VΩCOMMAA

Ω

V A

MA




VAV

V

VV

V






VΩCOMMAA

Ω

V A

MA




VAVAB

If the batteries are ideal andVA = 1.5 VA)VAB = 0.0 VB)VAB = 0.5 VC)VAB = 1.5VD)VAB = 3.0 VE)something elseF)

Comments"Can we do problems using the junction and loop rule with number values for the battery and resisters?"

(%i5) solve([%o1,%o2,%o4],[I1,I2,I3]);

(%o5)

[[I1 =E1 R3 + (

E1 E2)R2

R2 (R3 + R1) + R1 R3

, I2 =E1 R3 + E2 R1

R2 (R3 + R1) + R1 R3

, I3 =(E1 E2)

R2 E2 R1R2 (

R3 + R1) + R1 R3]]

This symbolic expression of the answer is very useful. We’ll need to substitute somevalues in order to get the values of the currents that we’ll be measuring. In order toassign a numerical value to a symbol use the colon (:).

First enter the voltages. We use volts as default unit.

(%i6) E1:4.5;

(%o6) 4.5

(%i7) E2:1.5;

(%o7) 1.5

Now specify the resistors in ohms.

(%i8) R1:68;

(%o8) 68

(%i9) R2:100;

(%o9) 100

(%i10) R3:39;

(%o10) 39

In order to evaluate the symbolic expressions for the currents type the label of theequations and append ”numer” to force a numerical evaluation.

(%i11) %o5,numer;

4

Comments"Please explain Kirchhoff in human language."How can you have a voltage drop across a battery and a voltage gain across a resistor”

"What way does current flow from A to B?". Also, if charges flow through a resistance, then why does I(before) = I(after)? Does the resistance not slow down the current (charge speed)?"The Blue Wire"The whole concept of the joined 2 parallel curcuits

water, pipes, pumps, tanks ...

will talk about these

Currentthroughissame.

VoltagedropacrossisIRi

Resistorsinseries:

Voltagedropacrossissame.

CurrentthroughisV/Ri

Resistorsinparallel:

SolvedCircuits

V

R1 R2

R4

R3V

R1234I1234=

Last Time


THEANSWER:Kirchhoff’sRules

I1234

New Circuit


Kirchhoff’s Voltage Rule

Kirchhoff'sVoltageRulestatesthatthesumofthevoltagechangescausedbyanyelements(likewires,baKeries,andresistors)aroundacircuitmustbezero.

WHY?Thepoten<aldifferencebetweenapointanditselfiszero!


Kirchhoff'sCurrentRulestatesthatthesumofallcurrentsenteringanygivenpointinacircuitmustequalthesumofallcurrentsleavingthesamepoint.

WHY?ElectricChargeisConserved

Kirchhoff’s Current Rule


Kirchhoff’s Laws

1)LabelallcurrentsChooseanydirec<on

2)Label+/−forallelements Currentgoes+⇒−(forresistors)

3)Chooseloopanddirec<onMuststartonwire,notelement.

4)Writedownvoltagedrops Firstsignyouhitissigntouse.

R4

I1

I3I2 I4

+

+

+ +

+

−

−

−

−

−

+

+

+

−

−

−

R1

E1

R2

R3E2

E3

R5

A

B

5)Writedownnodeequa<onIin = Iout

I5

We’lldocalcula<onfirsttodayIt’sactuallytheeasiestthingtodo!


CheckPoint: Gains and Drops


Inthefollowingcircuit,considertheloopabc.Thedirec<onofthecurrentthrougheachresistorisindicatedbyblackarrows.

IfwearetowriteKirchoff'svoltageequa<onforthisloopintheclockwisedirec<onstar<ngfrompointa,whatisthecorrectorderofvoltagegains/dropsthatwewillencounterforresistorsR1,R2andR3?

A.drop,drop,dropB.gain,gain,gainC.drop,gain,gainD.gain,drop,dropE.drop,drop,gain

Withthecurrent VOLTAGEDROP

DROP

Againstthecurrent VOLTAGEGAIN

GAIN

GAIN

2V

1V

1V

ConceptualAnalysis:– CircuitbehaviordescribedbyKirchhoff’sRules:

• KVR:Σ Vdrops = 0 • KCR:Σ Iin = Σ Iout

StrategicAnalysis– WritedownLoopEqua<ons(KVR)– WritedownNodeEqua<ons(KCR)– Solve

I2

Calculation

Inthiscircuit,assumeViandRiareknown.

WhatisI2?


+ −

+ −

+ −

ThisiseasyforbaKeries

V1R1

R2


WhatisI2?

R3

V2

V3

I1

I3

I2

Labelandpickdirec<onsforeachcurrent

Labelthe+ and−sideofeachelement

− +

+ −

− +

Forresistors,the“upstream”sideis+

Nowwritedownloopandnodeequa<ons

Calculation


Howmanyequa<onsdoweneedtowritedowninordertosolveforI2?

A)1B)2C)3D)4E)5

Why?– Wehave3unknowns:I1,I2,andI3

– Weneed3independentequa<onstosolvefortheseunknowns

V1R1

R2

R3

V2

V3

+ −

+ −

+ −− +

+ −

− +

I1

I3

I2


WhatisI2?

Calculation


Whichofthefollowingequa<onsisNOTcorrect? A)I2 = I1 + I3 B)− V1 + I1R1 − I3R3 + V3 = 0C)− V3 + I3R3 + I2R2 + V2 = 0D) − V2 − I2R2 + I1R1 + V1 = 0

Why?– (D) isanaKempttowritedownKVRforthetoploop– Startatnega<veterminalofV2andgoclockwise

Vgain (−V2) thenVgain (−I2R2) thenVgain(−I1R1)thenVdrop (+V1)

V1R1

R2

R3

V2

V3

+ −

+ −

+ −− +

+ −

− +

I1

I3

I2


WhatisI2?

Calculation


A)Any3willdoB)1,2,and4C)2,3,and4

Wehavethefollowing4equa<ons:

1. I2 = I1 + I3 2.− V1 + I1R1 − I3R3 + V3 = 03.− V3 + I3R3 + I2R2 + V2 = 04.− V2 − I2R2 − I1R1 + V1 = 0Why?

– Weneed3INDEPENDENTequa<ons– Equa<ons2,3,and4areNOTINDEPENDENT

Eqn 2+Eqn 3= − Eqn 4 – WemustchooseEqua<on1andanytwooftheremaining(2,3,and4)

Weneed3equa<ons:Which3shouldweuse?

V1R1

R2

R3

V2

V3

I1

I3

I2


WhatisI2?

Calculation


V1R1

R2

R3

V2

V3

I1

I3

I2

Wehave3equa<onsand3unknowns.I2 = I1 + I3

V1 + I1R1 − I3R3 + V3 = 0V2 − I2R2 − I1R1 + V1 = 0

Thesolu<onwillgetverymessy!Simplify:assumeV2 = V3 = V V1 = 2V R1 = R3 = R R2 = 2R

2VR

2R

R

V

V

I1

I3

I2

Calculation


WhatisI2?


Inthiscircuit,assumeVandRareknown.WhatisI2?

Withthissimplifica<on,youcanverify:I2 = ( 1/5) V/RI1 = ( 3/5) V/RI3 = (−2/5) V/R

Wehave3equa<onsand3unknowns.I2 = I1 + I3

−2V + I1R − I3R + V = 0 (outside)−V − I2(2R) − I1R + 2V = 0 (top)

2VR

2R

R

V

V

I1

I3

I2

currentdirec<on

Calculation: Simplify


Weknow:I2 = ( 1/5) V/RI1 = ( 3/5) V/RI3 = (−2/5) V/R

a b

SupposeweshortR3:WhathappenstoVab(voltageacrossR2?)

A)Vab remainsthesame

B)Vab changessign C)Vab increasesD)Vabgoestozero

Why?Redraw:

2VR

2R V

V

I1

I3

I2a b

c

d

2VR

2R

R

V

V

I1

I3

I2

Vab + V − V = 0BoKomLoopEqua<on:

Follow Up

Vab = 0


V R R

a b

Isthereacurrentflowingbetweenaandb?

A)YesB)No

a & b havethesamepoten<al Nocurrentflowsbetweena&b

CurrentflowsfrombaKeryandsplitsataSomecurrentflowsdown

SomecurrentflowsrightElectricity&Magne<smLecture10,Slide16

Clicker Question

CheckPoint: Circuits w/ Resistors and a Battery 1


Considerthecircuitshownbelow.Whichofthefollowingstatementsbestdescribesthecurrentflowinginthebluewireconnec<ngpointsaandb?

A.Posi<vecurrentflowsfromatobB.Posi<vecurrentflowsfrombtoaC.Nocurrentflowsbetweenaandb

I1R − I2 (2R) = 0

I4R − I3 (2R) = 0

I = I1 − I3

I + I2 = I4

I2 = ½ I1

I4 = 2 I3

I1 − I3 + ½ I1 = 2I3 I1 = 2I3 I = +I3

II1

I2

I3I4

Whatisthesame? CurrentflowinginandoutofthebaKery.

Whatisdifferent? Currentflowingfromatob.

2R3

2R3

Prelecture CheckPoint


2RI1/3R

2/3I

V

R 2R

a b

I2/3I

V/2

I

1/3

0

2/3I

2/3I

2/3I

1/3I1/3I

1/3I

2/3I1/3I


CheckPoint: Circuits w/ Resistors and a Battery 2


Considerthecircuitshownbelow.Inwhichcaseisthecurrentflowinginthebluewireconnec<ngpointsaandbbigger?

IA IB

Currentwillflowfromlentorightinbothcases.

CaseACaseBTheyarethesameA B C

Inbothcases,Vac = V/2

c c

IA = IR − I2R

= IR − 2I4R IB = IR − I4R

I2R = 2I4R

V0

r

R VL

r

V0

+

VLR

Usuallycan’tsupplytoomuchcurrenttotheloadwithoutvoltage“sagging”

Model for Real Battery: Internal Resistance


Using Breadboards (protoboards)

Original Breadboards

Circuit Technique

58 CHAPTER 6. INTRODUCTORY ELECTRONICS NOTES: PRACTICE

Figure 6.1: Bad and Good breadboarding technique.

• Try to build your circuit so that it looks like its circuit diagram:

– Let signal flow in from the left, exit on the right (in this case, the “signal” is justV ; the “output” is just I, read on the ammeter);

– Place ground on a horizontal breadboard bus strip below your circuit. When youreach circuits that include negative supply, place that on a bus strip below theground bus.

– Use colour coding to help you follow your own wiring: use black for ground, redfor the positive supply. Such colour coding helps a little now, a lot later, whenyou begin to lay out more complicated digital circuits.

Figure 6.2 shows bad and good examples of breadboard layouts. Figure 6.3 showsthe layout of a typical breadboard. Typically, one places components in the middlegroups with vertical interconnects and power lines and grounds in the horizontalinterconnects at top and bottom.

Figure 6.2: Bad and good breadboard layouts of a simple circuit

Bad

Goodugly!

Bad

ugly!

Good and Bad component layout








Figure 6.2: Bad and good breadboard layouts of a simple circuitConnections among pins in the breadboard.

Use horizontal rows for voltage busses: +5V, ±12V, gnd.

Use vertical rows for connecting components

together.

Good

`








Figure 6.2: Bad and good breadboard layouts of a simple circuit

+5V bus

gnd bus

to +5V ofpower supply

to gnd ofpower supply

to scope

connection

electricity magnetism lecture 8: kirchhoff’s rules

Documents