BASICS OF ELECTRICAL ENGINEERING

UNDERSTANDING BASICS OF ELECTRICITY:

THE GREATEST INVENTION OF MANKIND IS AN ELECTRICITY. THERE IS NO FUTURE FOR HUMANITY WITHOUT AN ELECTRICITY. MOST OF THE EQUIPMENTS IN DAY TO DAY LIFE REQUIRES AN ELECTRICITY AT SOME POINT IN ORDER TO RUN. MACHINES ARE THE MOST IMPORTANT PART OF HUMAN LIFE AS THEY HELP SAVING HUGE AMOUNT OF HUMAN EFFORTS AND TIME.

AN ELECTRICITY CAN BE CREATED PHYSICALLY (FOR EX. FRICTION OF CARPET AND SOCKS) AND/OR CHEMICALLY (LIKE IN BATTERIES) IN THE FORM OF CHARGE. NOW, ELECTRICITY CAN BE EXPRESSED SIMPLY AS MOVEMENT OF CHARGE FROM, POSITIVE POINT TO A NEGATIVE POINT. NO MATTER HOW THE ELECTRICITY IS CREATED THE MOVEMENT OF DISCHARGE IS CALLED ELECTRICITY.

UNDERSTANDING CURRENT:

THE RATE/MEASURE OF FLOW OF ELECTRICAL CHARGE IS CALLED AS "ELECTRICAL CURRENT". THE UNIT OF CURRENT IS "AMPERES". THERE ARE TWO TYPES OF ELECTRICAL CURRENT, ALTERNATING CURRENT(AC) AND DIRECT CURRENT(DC).
DIRECT CURRENT(DC) - DC IS THE CONSTANT CURRENT THAT FLOWS IN THE SAME DIRECTION TILL THE VOLTAGE POLARITIES ARE CONSTANT. IN ORDER TO CHANGE THE DIRECTION OF FLOW OF DC CURRENT WE NEED TO CHANGE THE VOLTAGE POLARITIES OF THE CIRCUIT.

DC SUPPLY

ALTERNATING CURRENT(AC) - AC IS THE CURRENT THAT HAS A NATURAL ZERO PERIOD. THAT IS, IT CHANGES ITS DIRECTION ALONG WITH VOLTAGE POLARITIES AT A CERTAIN PERIOD. THIS PERIOD IS DEPEND UPON THE FREQUENCY.

AC SUPPLY

SCIENTIST SUCH AS, THOMAS EDISON AND A. VOLTA HAS CREATED THE THEORY OF DIRECT CURRENT WHILE NICOLA TESLA HAS INVENTED AN AC CURRENT THEORY. THE DC IS MOSTLY USED AT PORTABLE APPLICATIONS AS THE TRANSMISSION OF DC IS TOO INEFFICIENT.

HOWEVER THE AC CURRENT THEORY OF N. TESLA HAS CHANGE THE LIFESTYLE OF HUMANITY. WITH THE USE OF AC CURRENT THEORY IT BECAME POSSIBLE TO TRANSMIT THE ELECTRICITY AT LONGER DISTANCE AND ALSO TO PRODUCE HIGH AC VOLTAGE IN POWER PLANTS.

    

OHM'S LAW:

THE MOST BASIC LAW AND IMPORTANT LAW OF ELECTRICITY IS OHM'S LAW. THIS LAW STATES THAT "THE CURRENT THROUGH A CONDUCTOR BETWEEN TWO POINTS IS DIRECTLY PROPORTIONAL TO THE VOLTAGE ACROSS THE TWO POINTS". NOW, THE RESISTANCE IS TERMED AS A COMPONENT OF CIRCUIT WHICH OPPOSES THE FLOW OF CURRENT. UNIT OF RESISTANCE IS "OHM". AS RESISTANCE RESISTS THE FLOW OF CURRENT FRICTION OCCURS AND HEAT GETS PRODUCED. THIS HEAT THEN WHICH CAUSES THE VOLTAGE DROP IN CIRCUIT. VOLTAGE IS TERMED AS THE MEASURE OF WORK(PRESSURE) REQUIRED TO MOVE THE CURRENT FROM ONE POINT OF CIRCUIT TO THE OTHER. UNIT OF VOLTAGE IS "VOLTS". FROM THE TWO POINTS OF CIRCUIT ONE POINT HAS HIGHER VOLTAGE VALUE AND OTHER POINT HAS LOWER VOLTAGE VALUE FOR EX. POSITIVE POINT HAS +5V AND NEGATIVE POINT HAS 0V OR CALLED AS "NEUTRAL POINT". TO ALLOW THE FLOW OF CURRENT IN CIRCUIT IT IS NECESSARY TO HAVE NON-ZERO THE POTENTIAL DIFFERENCE BETWEEN POSITIVE AND NEGATIVE POINT. SOMETIME THE NEGATIVE POINT COULD BE NON-ZERO VOLTAGE FOR EX. IF POSITIVE POINT HAS +20V THEN NEGATIVE POINT COULD BE -20V, +10V. AS LONG AS THERE IS A NON-ZERO POTENTIAL DIFFERENCE BETWEEN THE TWO POINT THE CURRENT WILL KEEP FLOWING THROUGH THE CIRCUIT. THE DIRECTION OF THIS CURRENT COULD BE CHANGE IF THE POTENTIAL DIFFERENCE BECOMES NEGATIVE. FOR EX. POSITIVE POINT NOW HAS +10V AND NEGATIVE POINT CHANGED TO -20V THE DIFFERENCE WILL BE 10-20=-10V. THIS WILL CAUSE THE CURRENT TO FLOW IN OPPOSITE DIRECTION.

THE PARAMETERS OF OHM'S LAW -

V = I * R
"V" IS THE APPLIED VOLTAGE IN "VOLTS".
"I" IS THE CURRENT FLOWING THROUGH CIRCUIT  IN "AMPERES".
"R" IS THE RESISTANCE OF THE CIRCUIT IN "OHMS" or "Ω".

BASIC TYPES OF ELECTRICAL CIRCUIT:

MULTIPLE RESISTANCE IN THE CIRCUIT USED TO DIVIDE VOLTAGE OR ALTER THE CURRENT PATH. THERE ARE TWO TYPES OF CIRCUIT DIVISION POSSIBLE USING RESISTANCES, "SERIES CIRCUIT" AND "PARALLEL CIRCUIT".

SERIES CIRCUIT AND PARALLEL CIRCUIT -

THE SERIES CIRCUIT IS IN-LINE WITH POWER SOURCE. THAT IS RESISTANCES HERE ARE CONNECTED IN SERIES WITH VOLTAGE SOURCE. THE CURRENT IS THE SERIES CIRCUIT IS CONSTANT. HOWEVER, THE VOLTAGE NOW GETS DIVIDED BY THE NUMBER OF RESISTANCES. HENCE, IN SERIES CIRCUIT NUMBER OF RESISTANCES CAUSES REDUCTION IN VOLTAGE DEPENDING UPON THE VALUES OF RESISTANCES.

PARALLEL CIRCUIT IS THE CIRCUIT THAT IS DIVIDED INTO DIFFERENT LOOPS OR BRANCHES WITH COMPONENTS ARE ON THE DIFFERENT BRANCH AND POWER SOURCE IS ON THE DIFFERENT BRANCH. IN PARALLEL CIRCUIT THE VOLTAGE IS CONSTANT. HOWEVER, DUE TO THE PRESENCE OF DIFFERENT LOOPS OR BRANCHES IN THE CIRCUIT IT WILL CAUSE CURRENT TO DIVIDE IN EACH BRANCH.

SERIES AND PARALLEL AC CIRCUIT

THE BEST EXAMPLE OF SERIES CIRCUIT IS DECORATION LIGHTS. IT BECOME SO ANNOYING TO SEE WHEN THE DECORATIVE LIGHTS SHUTS DOWN COMPLETELY ONLY BECAUSE OF ONE LAMP IN HUNDREDS IS BROKEN. THIS IS MAINLY BECAUSE ALL THE LAMP NEEDS SAME AMOUNT OF CURRENT TO LIGHT UP AND THE CURRENT NEEDS A CLOSED LOOP TO FLOW THROUGH. BECAUSE OF ONE FAULTY LAMP CIRCUIT TENDS TO GET OPEN AND CURRENTS WILL BECOME ZERO. THIS WILL CAUSE WHOLE SERIES OF LIGHTS TO SHUT DOWN.

HEALTHY SERIES DECORATION LIGHTS

FAULTY SERIES DECORATION LIGHTS

NOWADAYS, PARALLEL CONNECTED DECORATIVE LIGHTS ARE AVAILABLE IN THE MARKET. THIS USUALLY WORKS AS, LAMPS WILL NOW BE DIVIDED INTO DIFFERENT BRANCHES IF ONE LAMP BREAKS DOWN THE ONLY ONE BRANCH THAT OF THE SAME LAMP WILL SHUT DOWN BUT  REST OF THE BRANCHES AND LAMPS WILL REMAIN HEALTHY.

HEALTHY PARALLEL BRANCHES OF DECORATION LIGHTS

ONE FAULTY AND ONE HEALTHY PARALLEL BRANCH OF DECORATION LIGHTS

HOW TO APPLY OHM'S LAW:

WE WILL TAKE COMPONENTS WITH VALUES IN CONSIDERATION AS,

R1 = 300 Ω
R2 = 250 Ω
I  = 5 AMPS (BATTERY CURRENT) DC
V  = 5 VOLTS (BATTERY VOLTAGE) DC
S1 = SWITCH

WE KNOW OHM'S LAW IS V = I * R IT CAN BE USED TO FIND "I",

AS THE CIRCUIT HAS TWO RESISTANCES CONNECTED IN SERIES WE CAN ADD THEIR VALUES TO FIND COMMON VALUE THIS IS CALLED EQUIVALENT RESISTANCE "Req".

NOW,

Req = R1 + R2 ;
    = 300 + 250 ;
Req = 550 Ω.

NOW, WE CAN CALCULATE CURRENT - 

I = V / Req
  = 5 / 550
  = 9 mAMPS

APPLYING OHM'S LAW APPLIED TO SERIES CIRCUIT - 

IN ORDER TO APPLY OHM'S LAW WE WILL TAKE A CIRCUIT IN CONSIDERATION THAT HAS TWO LED LIGHTS AS OUR TWO RESISTANCES "R1" AND "R2" RESPECTIVELY. "S1" WILL BE THE SWITCH OF THE CIRCUIT.
ONCE THE SWITCH IS CLOSED, CURRENT WILL START FLOWING THROUGH THE LOOP. BATTERY SUPPLIES CURRENT OF 5 AMPS HOWEVER THE CIRCUIT LOADS ONLY USES NECESSARY AMOUNT OF IT.

DC SERIES CIRCUIT SUPPLYING TWO LED LAMP LOADS

WE KNOW THE BATTERY VOLTAGE WHICH IS Vs = 5 VDC.

NOW, WE NEED TO CALCULATE THE VOLTAGE DROP ACROSS LED1 AND LED2 THAT IS V1 AND V2 RESPECTIVELY.
WE KNOW OHM'S LAW IS V = I * R ;
AS,

I  = 9 mAMPS
R1 = 300 Ω
R2 = 250 Ω
V1 = I * R1 ; & V2 = I * R2 ;
   = 0.009 * 300 ;    = 0.009 * 250 ;
V1 = 2.7 VOLTS V2 = 2.3 VOLTS

KIRCHHOFF'S VOLTAGE LAW:

THE VOLTAGE DROPS IN SERIES CIRCUIT BRINGS A NEW LAW CALLED KIRCHHOFF'S VOLTAGE LAW IN CONSIDERATION. KIRCHHOFF'S VOLTAGE LAW STATES THAT AN ALGEBRAIC SUM OF VOLTAGES IN CLOSED LOOP IS EQUAL TO ZERO.
KIRCHHOFF'S VOLTAGE LAW USED IN SERIES CIRCUITS AS ONLY SERIES CIRCUITS HAS VOLTAGE DROPS NOT PARALLEL CIRCUITS. IF WE KNOW THE VOLTAGE DROP IN R1 ONLY WE CAN FIND THE VOLTAGE DROP OF R2 WITH THE HELP OF KIRCHHOFF'S VOLTAGE LAW (KVL).
IN KVL WE NEED TO ASSUME ONE CURRENT DIRECTION, WE WILL ASSUME IT AS CLOCKWISE(+VE TO -VE), ALONG WITH POLARITIES OF THE COMPONENTS R1 AND R2 AS SHOWN.

KVL APPLIED TO DC SERIES CIRCUIT

NOW,

Vs = V1 + V2 ; REARRANGING EQUATION ACCORDING TO KIRCHHOFF'S VOLTAGE LAW
Vs - V1 - V2 = 0 ;

        (5) - (I * R1) - (I * R2) = 0 ;

5 - (0.009 * 300) - (0.009 * 250) = 0 ;
5 - (2.7 - 2.3) = 0;

        5 - 5 = 0.

THUS, Vs - (V1 + V2) = 0, HENCE THEORY PROVED.

OHM'S LAW APPLIED IN PARALLEL CIRCUIT -

NOW TO APPLY OHM'S LAW TO A PARALLEL CIRCUIT WE WILL TAKE COMPONENTS AS SERIES CIRCUIT AND CONNECT THEM PARALLEL WITH THE SAME SOURCE.
THIS WILL DIVIDE THE CIRCUIT IN DIFFERENT BRANCHES:

BRANCH 1 HAS R1, 
BRANCH 2 HAS SUPPLY SOURCE AND SWITCH, 
BRANCH 3 HAS R2.

AS WE ALREADY KNOW THE VOLTAGE IN PARALLEL CIRCUIT IS CONSTANT IN ALL BRANCHES. BUT THE CURRENT IN PARALLEL CIRCUIT IS DIVIDED IN EVERY BRANCH WITH VALUE DEPENDING UPON THE RESISTANCE OFFERED BY THE BRANCH.

DC PARALLEL CIRCUIT SUPPLYING TWO LED LAMPS

CALCULATE CURRENT FLOWING THROUGH BRANCH 1 -

I1 = Vs / R1 ;
   = 5 / 300 ;
I1 = 0.017 AMP.

FOR BRANCH 2 THE TOTAL CURRENT USED BY THE CIRCUIT -

It = Vs / Req ;

FOR PARALLEL CIRCUIT  Req CAN BE FOUND OUT BY TAKING ADDITION OF RECIPROCALS OF THE RESISTANCES ;

(1 / Req) = (1 / R1) + (1 / R2) ;
(1 / Req) = (1 / 300) + (1 / 250) ;
(1 / Req) = (0.00733) ;
Req   = (1 / 0.00733) ;
Req   = 136.4 Ω.

NOW,

        It = (5 / 136.4) ;
It = 0.037 AMPS.

CALCULATE CURRENT FLOWING THROUGH BRANCH 3 -

I2 = Vs / R2 ;
   = 5 / 250 ;
I2 = 0.02 AMP.

THE CURRENT GETS DIVIDED IN EACH BRANCH BUT BRANCH WITH LOWER RESISTANCE WILL HAVE HIGHER VALUE OF CURRENT FLOWING THROUGH IT, HENCE THEORY PROVED.

KIRCHHOFF'S CURRENT LAW:

AS WE KNOW, CURRENT IN SERIES CIRCUITS ARE CONSTANT BUT CURRENT IN PARALLEL CIRCUIT IS VARYING, DUE TO THIS ANOTHER KIRCHHOFF'S LAW BROUGHT OUT.
THIS LAW STATES THAT CURRENT ENTERING IN THE NODE WILL BE EQUAL TO CURRENT OUT-COMING FROM THE NODE. THAT IS THE ACTUAL CURRENT IN NODE WILL BE ZERO.

i.e. Iin - Iout = 0.

BEFORE WE SOLVE THIS WITH KCL WE HAVE ALREADY VALUES DETERMINED FROM PARALLEL CIRCUIT ABOVE SO, 

        It = I1 + I2 ;
HERE, It = Iin ;
(I1 + I2) = Iout ;

SO, THE ACCORDING TO THEORY WE CAN SAY THAT, 

It - (I1 + I2) = 0 ;
PUT THE VALUES DETERMINED BY SOLVING PARALLEL CIRCUIT ABOVE,
0.037 - (0.017 + 0.02) = 0 , HENCE THEORY PROVED.

NOW LET US PROVE THIS THEORY WITH KCL -

KCL APPLIED TO DC PARALLEL CIRCUIT

WE ALREADY KNOW THAT,

It = (I1 + I2) ;

WHERE It IS INCOMING CURRENT TO THE NODE AND (I1 + I2) ARE THE CURRENTS OUT-COMING FROM NODE.

FIND THE VALUES OF CURRENTS USING OHM'S LAW ;

It = (Vs / Req) ;
I1 = (Vs / R1) ;
I2 = (Vs / R2)

PUTTING VALUES OF CURRENTS IN EQUATION OF KCL;

It - (I1 + I2) = 0 ;
(Vs / Req) - [(Vs / R1) + (Vs / R2)] = 0 ;
(5 / 136.4) - [(5 / 300) + (5 / 250)] = 0 ; 
(0.036) - [0.036] = 0 ; HENCE THEORY PROVED.

ELECTRICAL POWER EQUATION:

ELECTRICAL POWER IS MEASURE OF ELECTRICAL ENERGY IN WATTS. THE POWER EQUATION IS AS FOLLOWS,

P = V * I ;
WHERE, P = POWER IN WATTS ;
V = VOLTAGE IN VOLTS ;
I = CURRENT ;

THUS DEFINITION OF POWER IS PRODUCT OF VOLTAGE APPLIED TO THE CIRCUIT AND CURRENT FLOWING THROUGH THE CIRCUIT. HOWEVER, FOR AN AC CURRENT CIRCUIT, THERE ARE SOME LOSSES ALSO INVOLVED IN THE ELECTRICAL SYSTEM THAT IS WHEN THE POWER FACTOR TERM COMBINES WITH ELECTRICAL POWER EQUATION.

THE POWER EQUATION FOR AC CIRCUITS IS AS FOLLOWS,

P = V * I * COSФ, 
WHERE, COSФ = POWER FACTOR.

WE CAN USE OHM'S LAW WITH POWER EQUATION TO FIND OUT UNKNOWN VALUES OF VARIABLES. FOR EXAMPLE,

I = V / R ; 

COMBINE THIS WITH POWER EQUATION WE GET,

P = V * (V / R) ;
P = (V^2 / R).

OR

V = I * R ;

COMBINE THIS WITH POWER EQUATION WE GET,

P = (I * R) * I ;
P = (I^2 * R).

USING THIS EQUATION WE CAN FIND THE POWER CONSUMED BY PARALLEL CIRCUIT,

P = (I^2 * R) ;
HERE,
I^2 = It^2 ;
R = Req ;
P = [(It^2) * Req] ;
P = [(0.036^2) * 136.4] ;
P = 183 mWATTS.

SIMILARLY USING OTHER EQUATION,

P = (V^2 / R) ;
HERE,
V^2 = Vs^2 ;
R = Req ;
P = [(Vs^2) / Req] ;
P = [(5^2) / 136.4] ; 
P = 183 mWATTS.