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 S1S1 S2S2 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 V1V1 V2V2 C1C1 C2C2

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 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 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.

READING 2 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.

READING 3 READING 3 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 :

READING 4 READING 4 EQUIVALENT RESISTANCE IF TWO ( OR MORE ) RESISTORS R 1 AND R 2 ( OR R 1,……,R N ) 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 R 12 IF R 1 AND R 2 ARE CONNECTED IN PARALLEL IF R 1 AND R 2 ARE CONNECTED IN SERIES THE VALUE OF THE EQUIVALENT RESISTANCE R 12 IS GIVEN BY : R1R1 R2R2 R 12 R1R1 R2R2 THE TOTAL CURRENT I THROUGH THE COMBINATION EQUALS THE SUM OF CURRENTS IN EACH BRANCH OF THE CIRCUIT SERIES CONNECTION i i1i1 i2i2 PARALLEL CONNECTION

TEST 1 : TEST 1 : MATCH EACH CIRCUIT SYMBOL WITH ITS COMPONENT WIRERESISTORVOLTMETERLAMP (LIGHTING) WIRES JOINED (JUNCTION, NODE) AMMETERBATTERYSWITCHBRANC (ARM )CAPACITOR SWITCH AV

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 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 R 1 = 20  ; R 2 = 40  ; R 3 = 35  ; R 4 = 15 . COMPLETE: R1R1 R2R2 R3R3 R4R4 R1R1 R1R1 R 23 R4R4 R 234 R 1234 CONNECTION EQUIVALENT RESISTANCE R 2 – R 3 in series R 23 =R 2 +R 3 = ………….  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 R 5 R1R1 R2R2 R3R3 R4R4 + R5R5

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: 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 :

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 ) A1A1 A2A2 S1S1 S2S2 R1R1 R2R2 R3R3 R4R4 EMF + EFM = 10 V R 1 = 2  R 2 = 3  R 3 = 4  R 4 = 6 

THERE ARE FOUR POSSIBLE CIRCUIT DIAGRAMS : (a)(b) (c )(d) SWITCHESCIRC.DIAGRCONNECTION S 1 AND S 2 OPEN S 1 OPEN AND S 2 CLOSED S 1 CLOSED AND S 2 OPEN ( a )R 1 -R 2 -R 3 IN SERIES S 1 AND S 2 CLOSED COMPLETE :

(a) USE “EQUIVALENT RESISTANCE” V 1 +V 2 +V 3 =…………. V 1 = ………….= V 2 =…………..= V 3 = R 3 i = …………… R 123 = USE FIRST OHM’S LAW R1R1 R2R2 R3R3 R 123

(b) USE “EQUIVALENT RESISTANCE” V 1 +V 2 +V 3 =…………. V 1 = …………….= V 2 =……………..= V 4 = …………….= …………………. USE FIRST OHM’S LAW R1R1 R2R2 R4R4

(c) USE “EQUIVALENT RESISTANCE” USE FIRST OHM’S LAW R1R1 R2R2 R4R4

( d ) USE “EQUIVALENT RESISTANCE” i 12 = i 4= …………………… ……………………. i 3 = V 34 = ……………………… V 12 = i = USE OHM’S LAW R1R1 R2R2 R3R3 R4R4

WHAT DOES EACH VOLTMETER IN THE CIRCUIT BELOW INDICATE ? V0V0 V2V2 V4V4 SWITCHES S 1 AND S 2 OPEN S 1 OPEN AND S 2 CLOSED S 1 CLOSED AND S 2 OPEN S 1 AND S 2 CLOSED COMPLETE : V4V4 V2V2 V0V0 + S1S1 S2S2 R1R1 R2R2 R3R3 R4R4

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 S 1 AND S 2 AND PUT (V) FOR EACH LIGHTED LAMP (L1,L2,L3) POSITION SWITCHES L1 L2 L3 S 1 AND S 2 CLOSED S 1 CLOSED AND S 2 OPEN S 1 OPEN AND S 2 CLOSED S 1 AND S 2 OPEN + S1S1 S2S2 L1L1 L2L2 L3L3 A

LET’S CONSIDER NOW THE SWITCH S 1 CLOSED; DESCRIBE WHAT HAPPENS TO THE GROUP OF PARALLEL LAMPS L 2 AND L 3 WHEN WE CLOSE THE SWITCH S 2. 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