Download FIND THE EQUIVALENT RESISTANCE USING THE RULES OF

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:
R3
R1
R4
V2
S1
V1
R2
C1
S2
R6
C2
R5
OBJECTIVE :
WE ARE GOING TO LEARN TO SOLVE SIMPLE ELECTRIC CIRCUITS USING
ONLY :
• THE CONCEPT OF “EQUIVALENT RESISTANCE”
• FIRST OHM’S LAW
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 ( SIMPLE
ELECTRIC 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 _________________________________________________________________
READING 1
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
Q
SURFACE IN THE TIME t :
i
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 NONMETALLIC 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.
READING 2
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.
OHM’S LAW
READING 3
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 V
I
GIVES AN INDICATION OF HOW A CURRENT CAN REALLY FLOW IN A PARTICULAR
CONDUCTOR AND IT IS CALLED RESISTANCE.
HENCE V
I
 CONSTANT
CAN BE WRITTEN
V
R
I
THIS FORMULA IS OFTEN EXPRESSED MATHEMATICALLY AS V  R  i
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
R
EQUIVALENT RESISTANCE
IF 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
READING 4
PARALLEL CONNECTION
SERIES CONNECTION
R1
R1
R2
R12
R2
R12
THE VALUE OF THE EQUIVALENT RESISTANCE R12 IS GIVEN BY :
1
1
1


R12 R1 R2
R12  R1  R2
IF R1 AND R2 ARE CONNECTED IN
1
1
PARALLEL 1
R1.. n

R1
 ......... 
R1.. N  R1  .........  RN
RN
THE TOTAL CURRENT I THROUGH THE
COMBINATION EQUALS THE SUM OF
CURRENTS IN EACH BRANCH OF THE
CIRCUIT
1
2
i  i i
IF R1 AND R2 ARE CONNECTED IN
SERIES
i1
i
i2
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 :
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 : 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
CONNECTION
R4
R2 – R3 in series
EQUIVALENT RESISTANCE
R23=R2+R3= ………….
R23
R1
R4
R1
R234
R1234
EQUIVALENT RESISTANCE =

TEST 3 :
CONSIDER THE CIRCUIT IN THE FOLLOWING FIGURE :
R4
+
R1
R3
R5
R2
•
•
•
•
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
HOW CAN WE USE THE CONCEPTS
•
EQUIVALENT RESISTANCE
•
OHM’S LAW
TO SOLVE THE ELECTRIC CIRCUITS ?
•
EQUIVALENT RESISTANCE
REPLACING 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 LAW
TO DETERMINE :
THE CURRENT IN A RESISTOR ( R ) WHICH IS GIVEN BY VOLTAGE ( V ) DIVIDED
BY RESISTANCE:
V
i
R
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 :
V  Ri
EXERCISE 2 :
CONSIDER THE ELECTRIC CIRCUIT SHOWN IN THE DIAGRAM BELOW ( THE
INTERNAL RESISTANCE OF THE BATTERY CAN BE IGNORED )
+
A1
EMF
EFM = 10 V
R4
R3
R1
R2
S1
S2
R1 = 2 
A2
R2 = 3 
R3 = 4 
R4 = 6 
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 )
THERE ARE FOUR POSSIBLE CIRCUIT DIAGRAMS :
(a)
(b)
(c )
(d)
COMPLETE :
SWITCHES
CIRC.DIAGR
CONNECTION
S1 AND S2 OPEN
S1 OPEN AND S2 CLOSED
S1 CLOSED AND S2 OPEN
S1 AND S2 CLOSED
(a)
R1-R2-R3 IN SERIES
(a)
V1+V2+V3=………….
USE
“EQUIVALENT
RESISTANCE”
R3
R1
R2
V1= ………….=
V2=…………..=
V3= R3 i = ……………
i
.........
..........
USE FIRST OHM’S
LAW
R123 =
R123
(b)
V1+V2+V3=………….
USE
“EQUIVALENT
RESISTANCE”
R4
R1
V1= …………….=
V2=……………..=
V4= …………….=
R2
………………….
i
.........
..........
USE FIRST OHM’S
LAW
(c)
USE
“EQUIVALENT
RESISTANCE”
R4
R1
R2
USE FIRST OHM’S
LAW
(d)
USE
“EQUIVALENT
RESISTANCE”
R4
i12=
R3
R1
i4=
R2
i3=
……………………
…………………….
V34=
V12=
………………………
i=
USE OHM’S LAW
WHAT DOES EACH VOLTMETER IN THE CIRCUIT BELOW INDICATE ?
V4
+
V0
R3
R4
R1
R2
S1
S2
V2
COMPLETE :
SWITCHES
S1 AND S2 OPEN
S1 OPEN AND S2 CLOSED
S1 CLOSED AND S2 OPEN
S1 AND S2 CLOSED
V0
V2
V4
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 )
S1
L1
+
A
L2
L3
S2
CONSIDER THE POSITION (OPEN/CLOSED) OF THE SWITCHES S1 AND S2 AND PUT
(V) FOR EACH LIGHTED LAMP (L1,L2,L3)
POSITION SWITCHES
S1 AND S2 CLOSED
S1 CLOSED AND S2 OPEN
S1 OPEN AND S2 CLOSED
S1 AND S2 OPEN
L1
L2
L3
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 EXPERIMENT
CHECK 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