electricity magnetism lecture 10: kirchhoff’s rules

Electricity & MagnetismLecture 10: Kirchhoff’s Rules

Today’sConcept:

Kirchhoff’sRules

Electricity&Magne7smLecture10,Slide1

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,baPeries,andresistors)aroundacircuitmustbezero.

WHY?Thepoten@aldifferencebetweenapointanditselfiszero!


Kirchhoff'sCurrentRulestatesthatthesumofallcurrentsenteringanygivenpointinacircuitmustequalthesumofallcurrentsleavingthesamepoint.

WHY?ElectricChargeisConserved

Kirchhoff’s Current Rule


Kirchhoff’s Laws

1)LabelallcurrentsChooseanydirec7on

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.Thedirec7onofthecurrentthrougheachresistorisindicatedbyblackarrows.

IfwearetowriteKirchoff'svoltageequa7onforthisloopintheclockwisedirec7onstar7ngfrompointa,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– WritedownLoopEqua7ons(KVR)– WritedownNodeEqua7ons(KCR)– Solve

I2

Calculation

Inthiscircuit,assumeViandRiareknown.

WhatisI2?


+ −

+ −

+ −

ThisiseasyforbaPeries

V1R1

R2


WhatisI2?

R3

V2

V3

I1

I3

I2

Labelandpickdirec7onsforeachcurrent

Labelthe+ and−sideofeachelement

− +

+ −

− +

Forresistors,the“upstream”sideis+

Nowwritedownloopandnodeequa7ons

Calculation


Howmanyequa7onsdoweneedtowritedowninordertosolveforI2?

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

Why?– Wehave3unknowns:I1,I2,andI3

– Weneed3independentequa7onstosolvefortheseunknowns

V1R1

R2

R3

V2

V3

+ −

+ −

+ −− +

+ −

− +

I1

I3

I2


WhatisI2?

Calculation


Whichofthefollowingequa7onsisNOTcorrect? A)I2 = I1 + I3 B)− V1 + I1R1 − I3R3 + V3 = 0C)− V3 + I3R3 + I2R2 + V2 = 0D) − V2 − I2R2 + I1R1 + V1 = 0

Why?– (D) isanaPempttowritedownKVRforthetoploop– Startatnega7veterminalofV2andgoclockwise

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

Wehavethefollowing4equa7ons:

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

– Weneed3INDEPENDENTequa7ons– Equa7ons2,3,and4areNOTINDEPENDENT

Eqn 2+Eqn 3= − Eqn 4 – WemustchooseEqua7on1andanytwooftheremaining(2,3,and4)

Weneed3equa7ons:Which3shouldweuse?

V1R1

R2

R3

V2

V3

I1

I3

I2


WhatisI2?

Calculation


V1R1

R2

R3

V2

V3

I1

I3

I2

Wehave3equa7onsand3unknowns.I2 = I1 + I3

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

Thesolu7onwillgetverymessy!Simplify:assumeV2 = V3 = V V1 = 2V R1 = R3 = R R2 = 2R

2VR

2R

R

V

V

I1

I3

I2

Calculation


WhatisI2?


Inthiscircuit,assumeVandRareknown.WhatisI2?

Withthissimplifica7on,youcanverify:I2 = ( 1/5) V/RI1 = ( 3/5) V/RI3 = (−2/5) V/R

Wehave3equa7onsand3unknowns.I2 = I1 + I3

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

2VR

2R

R

V

V

I1

I3

I2

currentdirec7on

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 = 0BoPomLoopEqua7on:

Follow Up

Vab = 0


V R R

a b

Isthereacurrentflowingbetweenaandb?

A)YesB)No

a & b havethesamepoten7al Nocurrentflowsbetweena&b

CurrentflowsfrombaPeryandsplitsataSomecurrentflowsdown

SomecurrentflowsrightElectricity&Magne7smLecture10,Slide16

Clicker Question

CheckPoint: Circuits w/ Resistors and a Battery 1


Considerthecircuitshownbelow.Whichofthefollowingstatementsbestdescribesthecurrentflowinginthebluewireconnec7ngpointsaandb?

A.Posi7vecurrentflowsfromatobB.Posi7vecurrentflowsfrombtoaC.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? CurrentflowinginandoutofthebaTery.

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.Inwhichcaseisthecurrentflowinginthebluewireconnec7ngpointsaandbbigger?

IA IB

Currentwillflowfromlemtorightinbothcases.

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

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.








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 10: kirchhoff’s rules

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