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.
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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.
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HEALTHY SERIES DECORATION LIGHTS |
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FAULTY SERIES DECORATION LIGHTS
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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.
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HEALTHY PARALLEL BRANCHES OF DECORATION LIGHTS
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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.
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DC SERIES CIRCUIT SUPPLYING TWO LED LAMP LOADS
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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.
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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.
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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 -
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KCL APPLIED TO DC PARALLEL CIRCUIT
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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.