SOLUTION OF ELECTRIC CIRCUIT

ELECTRIC CIRCUIT

AN ELECTRIC CIRCUIT IS A CONFIGURATION OF ELECTRONIC COMPONENTS THROUGH WHICH ELECTRICITY IS MADE TO FLOW. THE FIGURE BELOW

SHOWS A TYPICAL EXAMPLE:

OBJECTIVE :

WE ARE GOING TO LEARN TO SOLVE SIMPLE ELECTRIC CIRCUITS USING

ONLY :

• THE CONCEPT OF “EQUIVALENT RESISTANCE”

• FIRST OHM’S LAW

S1 S2

R1

R2

R3

R4

R5R6

V1

V2

C1 C2

WE ARE GOING TO READ FOUR SHORT TEXTS ON THE FOLLOWING TOPICS:

• ELECTRIC CURRENT READING 1

• ELECTROMOTIVE FORCES READING 2

• EQUIVALENT RESISTANCE READING 3

• FIRST OHM’S LAW READING 4

THEY ARE ESSENTIAL TO SOLVE ELECTRIC CIRCUIT ( SIMPLEELECTRIC CIRCUIT )

WRITE THE ITALIAN TRANSLATION OF THE FOLLOWING WORDS ABOUT THE ELECTRIC CIRCUITS :• ELECTRIC CURRENT __________________________________________________________• CHARGE _____________________________________________________________________• ELECTROMOTIVE FORCE _____________________________________________________• CONDUCTOR ________________________________________________________________• VOLTAGE ____________________________________________________________________• WIRE ________________________________________________________________________• WIRES JOINED ______________________________________________________________• BRANCH _____________________________________________________________________• ARM ________________________________________________________________________• CAPACITOR __________________________________________________________________• AMMETER ___________________________________________________________________• RESISTANCE __________________________________________________________________• DIRECT CURRENT (DC) ________________________________________________________• VOLTMETER _________________________________________________________________

READINGREADING 11 ELECTRIC CURRENT

AN ELECTRIC CURRENT IS A FLOW OF ELECTRIC CHARGE AND IT IS DEFINED AS THE AMOUNT OF ELECTRIC CHARGE (MEASURED IN COULOMB) FLOWING THROUGH THE SURFACE IN THE TIME t :

THE S.I. UNIT OF ELECTRIC CURRENT IS THE AMPERE (A) , WHICH EQUALS A FLOW OF ONE COULOMB OF CHARGE PER SECOND. A DIRECT CURRENT (DC) IS A UNIDIRECTIONAL FLOW. CURRENT IS A SCALAR QUANTITY, BUT IN CIRCUIT ANALYSIS THE DIRECTION OF CURRENT IS RELEVANT AN IS INDICATED BY ARROWS.

CONVENTIONAL CURRENT : FOR HISTORICAL REASONS, ELECTRIC CURRENT IS SAID TO FLOW FROM THE POSITIVE PART OF A CIRCUIT TO THE MOST NEGATIVE PART ( THIS WAS GUESSED AT BEFORE THE ELECTRONS WERE DISCOVERED ). AN ELECTRIC CURRENT WILL ONLY FLOW WHEN :THERE IS A POTENTIAL DIFFERENCE BETWEEN TWO POINTS AND THE TWO POINTS ARE CONNECTED BY A CONDUCTOR.

IN SOLID METALS, LIKE WIRES, THE POSITIVE CHARGE CARRIERS ARE MOTIONLESS, AND ONLY THE NEGATIVELY CHARGED ELECTRONS FLOW. AS THE ELECTRONS CARRY NEGATIVE CHARGE, THE ELECTRON CURRENT IS IN THE DIRECTION WHICH IS OPPOSITE TO THAT OF THE CONVENTIONAL ( OR ELECTRIC )CURRENT. MOST FAMILIAR CONDUCTORS ARE METALLIC , HOWEVER, THERE ARE ALSO MANY NON-METALLIC CONDUCTORS, INCLUDING GRAPHITE, SOLUTIONS OF SALT, AND ALL PLASMAS. THE ISTRUMENT WHICH IS USED TO MEASURE THE CURRENT FLOWING IN A CONDUCTOR IS CALLED AN AMMETER. IN A CIRCUIT IT MUST BE PLACED IN SERIES WITH THE PARTS THROUGH WHICH THE CURRENT TO BE MEASURED IS PASSING.

t

Qi

READINGREADING 22 ELECTROMOTIVE FORCE (EMF)

THE ELECTROMOTIVE FORCE (OFTEN ABBREVIATE “EMF” AND DENOTED ) IS AN ARCHAIC TERM THAT INDICATES AN ELECTRIC POTENTIAL. WHEN WE COSIDER AN “IDEAL BATTERY” (‘ THE INTERNAL RESISTANCE IS ZERO’)

THE POTENTIAL DIFFERENCE ACROSS THE BATTERY EQUALS THE ELECTROMOTIVE FORCE (= ).THE ELECTROMOTIVE FORCE, WHICH TRIES TO MOVE A POSITIVE CHARGE FROM A HIGHER TO A LOWEL POTENTIAL, THERE MUST BE ANOTHER ‘FORCE’ TO MOVE CHARGE FROM A POTENTIAL TO A HIGHER INSIDE THE BATTERY. THIS SO-CALLED FORCE IS CALLED THE ELECTROMOTIVE FORCE, OR EMF. THE INSTRUMENT WHICH IS USED TO MEASURE THE POTENTIAL DIFFERENCE BETWEEN TWO POINTS IN A CIRCUIT IS THE VOLTMETER. IT IS CONNECTED IN PARALLEL WITH THE TWO POINTS WHOSE POTENTIAL IS BEING MEASURED.

READINGREADING 33 OHM’S LAW

IN 1826 A GERMAN SCIENTIST GEORG SIMON OHM, WHO EXPERIMENTED WITH CIRCUITS, FOUND OUT THE RELATIONSHIPS BETWEEN CURRENT AND VOLTAGE:

THE POTENTIAL DIFFERENCE BETWEEN THE ENDS OF A METALLIC CONDUCTOR IS DIRECTLY PROPORTIONAL TO THE CURRENT FLOWING. IT IS CALLED OHM’S LAW AND CAN BE FORMALLY DEFINED AS FOLLOWS (TEMPERATURE IS CONSTANT):

THE VALUE OF

GIVES AN INDICATION OF HOW A CURRENT CAN REALLY FLOW IN A PARTICULAR CONDUCTOR AND IT IS CALLED RESISTANCE.

HENCE

CAN BE WRITTEN

THIS FORMULA IS OFTEN EXPRESSED MATHEMATICALLY AS

WHERE : V IS THE APPLIED VOLTAGE, I IS THE CURRENT, R IS THE RESISTANCE

WITH OHM’S LAW IS POSSIBLE TO CALCULATE THE CURRENT IN AN (IDEAL) RESISTOR (OR OTHER OHMIC DEVICE) DIVIDING VOLTAGE BY RESISTANCE :

I

V

CONSTANTI

V

RI

V

iRV

R

VI

READINGREADING 44 EQUIVALENT RESISTANCEIF TWO ( OR MORE ) RESISTORS R1 AND R2 ( OR R1,……,RN ) ARE CONNECTED IN SERIES OR IN PARALLEL, WE CAN REPLACE THEM WITH THEIR EQUIVALENT RESISTANCE AND REDRAW THE CIRCUIT, IN THIS CASE WE GET A SEMLIFIED VERSION CIRCUIT AND WE CALL THIS EQUIVALENT RESISTANCE R12

IF R1 AND R2 ARE CONNECTED IN PARALLEL

IF R1 AND R2 ARE CONNECTED IN SERIES

THE VALUE OF THE EQUIVALENT RESISTANCE R12 IS GIVEN BY :

R1

R2

R12

R1 R2

R12

2112

111

RRR 2112 RRR

Nn RRR

1.........

11

1..1

NN RRR .........1..1

THE TOTAL CURRENT I THROUGH THE COMBINATION EQUALS THE SUM OF CURRENTS IN EACH BRANCH OF THE CIRCUIT

21 iii

SERIES CONNECTION

i i1

i2

PARALLEL CONNECTION

TEST 1 :TEST 1 :MATCH EACH CIRCUIT SYMBOL WITH ITS COMPONENT

WIRE RESISTOR VOLTMETER LAMP (LIGHTING)

WIRES JOINED (JUNCTION,

NODE)

AMMETER BATTERY SWITCH BRANC (ARM ) CAPACITOR

SWITCH

A V

TEST 2 :TEST 2 :

COMPLETE THE FOLLOWING SENTENCES: IN THE S.I. :

THE AMPERE IS THE UNIT _________________________ ITS SYMBOL IS __________________

THE VOLT IS THE UNIT ____________________________ ITS SYMBOL IS __________________

THE COULOMB IS THE UNIT _______________________ ITS SYMBOL IS __________________

THE ELECTRIC CURRENT IS GIVEN BY ______________________________________________

THE UNIT OF THE ELECTRICAL CURRENT IS ____________ DEFINED AS _________________

____________________________________________________________________________________

THE DIRECTION OF CURRENT IS RELEVANT AND IT IS INDICATED BY __________________

WHAT IS THE ELECTRIC CURRENT ? _________________________________________________

____________________________________________________________________________________

CHARGE, CURRENT AND TIME ARE RELATED BY _____________________________________

IF THE CHARGE ON 1 ELECTRON IS_____________________ , FIND HOW MANY ELECTRONS ARE INVOLVED IF A CURRENT FLOW RESULT IN THE MOVEMENT OF 3.60 105 OF CHARGE :

____________________________________________________________________________________

HOW MANY ELECTRONS FLOW THROUGH A BATTERY THAT DELIVERS A CURRENT OF

1.5 A FOR 10 s ? ______________________________________________________________________

THE POTENTIAL DIFFERENCE (VOLTAGE) ACROSS AN IDEAL CONDUCTOR IS ____________

________________TO THE CURRENT THROUGH IT.

USE OHM’S LAW TO FIND THE POYENTIAL DIFFERENCE BETWEEN TWO POINTS INCLUDING A RESISTANCE R= 8 WHEN THIS IS RUN THROUGH BY A CURRENT OF 0.25A?

EXERCISE 1 :EXERCISE 1 : FIND THE EQUIVALENT RESISTANCE USING THE RULES OF RESISTORS IN SERIES OR IN PRARALLEL R1 = 20 ; R2 = 40 ; R3 = 35 ; R4 = 15 . COMPLETE:

R1

R2 R3

R4

R1

R1

R23

R4

R234

R1234

CONNECTION EQUIVALENT RESISTANCE

R2 – R3 in series R23=R2+R3= ………….

EQUIVALENT RESISTANCE =

TEST 3 :TEST 3 :

CONSIDER THE CIRCUIT IN THE FOLLOWING FIGURE :

• HOW MANY NODES ARE THERE ? _________________________________

• HOW MANY BRANCHES ARE THERE ? _____________________________

• DRAW AN ARROW FOR EACH BRANCH TO INDICATE THE DIRECTION OF CONVENTIONAL CURRENT

• DRAW AGAIN THE CIRCUIT AND INSERT AN AMMETER AND A VOLTMETER IN THE CORRECT PLACE TO MEASURE THE CURRENT AND THE POTENTIAL DIFFERENCE ACROSS THE RESISTOR R5

R1

R2

R3

R4

+

R5

HOW CAN WE USE THE CONCEPTS • EQUIVALENT RESISTANCE• OHM’S LAWTO SOLVE THE ELECTRIC CIRCUITS ?• EQUIVALENT RESISTANCEREPLACING EACH GROUP OF RESISTORS WITH THEIR EQUIVALENT RESISTANCE AND REDRAWING THE CIRCUIT, WE GET A NEW SEMPLIFIED VERSION OF CIRCUIT. WE CONTINUE TO SIMPLIFY THE NEW VERSION OF THE CIRCUIT, AS FOR AS WE GET A SOURCE OF ELECTRICAL ENERGY THAT WILL PRODUCE A POTENTIAL DIFFERENCE BETWEEN TWO POINTS WHICH IS CONNECTED WITH ONLY ONE RESISTANCE

• FIRST OHM’S LAWTO DETERMINE :THE CURRENT IN A RESISTOR ( R ) WHICH IS GIVEN BY VOLTAGE ( V ) DIVIDED BY RESISTANCE:

THE POTENTIAL DIFFERENCE BETWEEN TWO POINTS WHICH INCLUDE A RESISTANCE ( R ) IS GIVEN BY THE PRODUCT OF THE RESISTANCE AND THE CURRENT FLOWING THROUGH THE RESISTANCE :

R

Vi

iRV

EXERCISE 2 :EXERCISE 2 :CONSIDER THE ELECTRIC CIRCUIT SHOWN IN THE DIAGRAM BELOW ( THE

INTERNAL RESISTANCE OF THE BATTERY CAN BE IGNORED )

IN THIS EXERCISE WE ARE GOING TO LEARN HOW TO CALCOLATE THE CURRENT FLOWING IN EACH ARM OF THE CIRCUIT. EXAMINE THE CIRCUIT DIAGRAM AND LIST ITS COMPONENTS: ONE BATTERY

_________________________

_________________________

_________________________

_________________________

LET’S REDRAW THE CIRCUIT DIAGRAM WITHOUT DRAWING :

THE INSTRUMENT USED TO MEASURE IT

THE BRANCHES WHERE THERE IS AN OPEN SWITCH (SINCE THEY

AREN’T RUN THROUGH BY ELECTRIC CURRENT )

A1

A2

S1 S2

R1

R2

R3

R4 EMF

+

EFM = 10 V

R1 = 2

R2 = 3

R3 = 4

R4= 6

THERE ARE FOUR POSSIBLE CIRCUIT DIAGRAMS :

(a) (b) (c ) (d)

SWITCHES CIRC.DIAGR CONNECTION

S1 AND S2 OPEN

S1 OPEN AND S2 CLOSED

S1 CLOSED AND S2 OPEN ( a ) R1-R2-R3 IN SERIES

S1 AND S2 CLOSED

COMPLETE :

(a)

USE

“EQUIVALENT

RESISTANCE”

V1+V2+V3=………….

V1= ………….=

V2=…………..=

V3= R3 i = ……………

R123 =

USE FIRST OHM’S

LAW

R1

R2

R3

R123

..........

.........i

(b)

USE

“EQUIVALENT

RESISTANCE”

V1+V2+V3=………….

V1= …………….=

V2=……………..=

V4= …………….=

………………….

USE FIRST OHM’S

LAW

R1

..........

.........i

R2

R4

(c)

USE

“EQUIVALENT

RESISTANCE”

USE FIRST OHM’S

LAW

R1

R2

R4

( d )

USE

“EQUIVALENT

RESISTANCE”

i12=

i4=

……………………

…………………….

i3=

V34=

………………………

V12=

i =

USE OHM’S LAW

R1

R2

R3

R4

WHAT DOES EACH VOLTMETER IN THE CIRCUIT BELOW INDICATE ?

V0

V2

V4

SWITCHES

S1 AND S2 OPEN

S1 OPEN AND S2 CLOSED

S1 CLOSED AND S2 OPEN

S1 AND S2 CLOSED

COMPLETE :

V4 V2 V0

+

S1 S2

R1

R2

R3

R4

EXERCISE 3 :EXERCISE 3 :THREE IDENTICAL LAMPS (EACH BULB HAS A RESISTANCE R ) ARE CONNECTED AS SHOWN IN THE CIRCUIT DIAGRAM BELOW. THE

POTENTIAL DIFFERENCE ACROSS THE BATTERY IS 5.7 V ( THE INTERNAL RESISTANCE OF THE BATTERY CAN BE IGNORED )

CONSIDER THE POSITION (OPEN/CLOSED) OF THE SWITCHES S1 AND S2 AND PUT (V) FOR EACH LIGHTED LAMP (L1,L2,L3)

POSITION SWITCHES L1 L2 L3

S1 AND S2 CLOSED

S1 CLOSED AND S2 OPEN

S1 OPEN AND S2 CLOSED

S1 AND S2 OPEN

+

S1

S2

L1

L2

L3

A

LET’S CONSIDER NOW THE SWITCH S1 CLOSED; DESCRIBE WHAT HAPPENS TO THE GROUP OF PARALLEL LAMPS L2 AND L3 WHEN WE CLOSE THE SWITCH S2 .

WHY ?

LABORATORY EXPERIMENTCHECK YOUR ANSWERS

THERE ARE MORE COMPLICATED CIRCUITS WHICH CANNOT BE SIMPLY

REDUCED TO A PARALLEL OR SERIES CIRCUIT USING EQUIVALENT

RESISTANCES. THESE ONES NEED TO BE SOLVED USING TWO LAWS :

• KIRCHHOFF’S CURRENT LAW ;

• KIRCHHOFF’S VOLTAGE LAW.

OFTEN, WHEN WE USE KIRCHHOFF’S LAWS, WE GET A LOT OF EQUATIONS

WHICH ARE COMPLICATED TO SOLVE. THE ANALISYS OF THIS CIRCUIT IS

MORE SIMPLE IF WE USE THE FOLLOWING LAWS:

• THEVENIN

• NORTON