basically kvl and kcl for thoes who study in high school and superpostion theorem for higher studies which we learn when we do graduation or engineering.

1. B U D G E B U D G E I N S T I T U T E
O F T E C H N O L O G Y
NISCHINTAPUR, BUDGE BUDGE, KOLKATA- 700137
DEPARTMENT:___________________________________
ASSIGNMENT ON CA1
PAPER NAME :_BASIC ELECTRICAL ENGINEERING
PAPER CODE:_ES-EE101_________________________
NAME: ADITYA SINGH__________________________
YEAR : YEAR 1st SEMESTER: SEM-1st SEASON: 2023-2024
UNIVERSITY ROLL NO:_
UNIVERSITY REGISTRATION NO:________________________________

2. Today’s Focus
• KIRCHHOFF’S CURRENT LAW (KCL)
• KIRCHHOFF’S VOLTAGE LAW (KVL)
• SUPERPOSTION THEOREM

3. KIRCHHOFF’S CURRENT LAW
KIRCHHOFF’S CURRENT LAW IS ALSO KNOWN AS KIRCHHOFF’S FIRST LAW OR KIRCHHOFF’S LAW OF THE
JUNCTION, BUT THE MOST USED TERM IS KIRCHHOFF’S CURRENT LAW OR KCL. KCL IS BASED ON THE LAW
OF CONSERVATION OF CHARGE.
DEFINE KIRCHHOFF’S CURRENT LAW
KIRCHHOFF’S CURRENT LAW STATES THAT THE ALGEBRAIC SUM OF CURRENTS ENTERING A NODE OR A
CLOSED BOUNDARY EQUALS ZERO.
IF THERE ARE N NUMBER OF BRANCHES CONNECTED TO A NODE AND IT IS THE CURRENT OF THE NTH
BRANCH, THEN MATHEMATICALLY, KCL STATES,
N∑N=1IN = 0
KCL

4. HERE, THE THREE CURRENTS ENTERING THE NODE, I1, I2, I3 ARE ALL POSITIVE IN VALUE AND THE
TWO CURRENTS LEAVING THE NODE, I4 AND I5 ARE NEGATIVE IN VALUE. THEN THIS MEANS WE
CAN ALSO REWRITE THE EQUATION AS;
I1 + I2 + I3 – I4 – I5 = 0
THE TERM NODE IN AN ELECTRICAL CIRCUIT GENERALLY REFERS TO A CONNECTION OR JUNCTION OF
TWO OR MORE CURRENT CARRYING PATHS OR ELEMENTS SUCH AS CABLES AND COMPONENTS. ALSO FOR
CURRENT TO FLOW EITHER IN OR OUT OF A NODE A CLOSED CIRCUIT PATH MUST EXIST. WE CAN USE
KIRCHHOFF’S CURRENT LAW WHEN ANALYSING PARALLEL CIRCUITS.
KCL

5. APPLICATION OF KIRCHHOFF’S CURRENT LAW
KCL IS USED TO COMBINE THE CURRENT SOURCES PRESENT IN PARALLEL. THE OVERALL
EQUIVALENT CURRENT IS THE ALGEBRAIC SUM OF INDIVIDUAL CURRENTS PRESENT IN PARALLEL,
AS SHOWN BELOW:
APPLYING KCL AT NODE A,
KCL

6. VALIDITY OF KIRCHHOFF’S CURRENT LAW
THERE EXIST SOME CONDITIONS WHERE KCL IS VALID, AND IN SOME CASES, IT IS NOT VALID. THOSE
CONDITIONS ARE:
KCL IS INDEPENDENT OF THE VARIATION IN TEMPERATURE IN THE CIRCUIT.
KCL IS VALID FOR LINEAR, NON-LINEAR, BILATERAL, UNILATERAL, PASSIVE, AND ACTIVE ELEMENTS.
KCL IS VALID FOR LUMPED ELECTRICAL NETWORKS ONLY, NOT FOR DISTRIBUTED ELECTRICAL
NETWORKS. AT HIGH FREQUENCIES, THE CIRCUIT IS TREATED AS DISTRIBUTED AND NOT LUMPED,
AND THE EFFECT OF PARASITIC RESISTANCE CANNOT BE IGNORED, SO KCL IS INVALID AT HIGH
FREQUENCIES.
KIRCHHOFF’S LAW IS NOT VALID FOR TIME-VARYING MAGNETIC FIELDS.
KCL

7. 1: FIND CURRENT I0 AND VOLTAGE V0 IN THE
CIRCUIT SHOWN BELOW
YOUR PARAGRAPH TEXT
KCL

8. KIRCHHOFF’S VOLTAGE LAW
KIRCHHOFF’S SECOND LAW OR THE VOLTAGE LAW STATES THAT
THE NET ELECTROMOTIVE FORCE AROUND A CLOSED CIRCUIT LOOP IS EQUAL TO THE SUM OF
POTENTIAL DROPS AROUND THE LOOP
IT IS TERMED KIRCHHOFF’S LOOP RULE, WHICH IS AN OUTCOME OF AN ELECTROSTATIC FIELD
THAT IS CONSERVATIVE.
HENCE,
IF A CHARGE MOVES AROUND A CLOSED LOOP IN A CIRCUIT, IT MUST GAIN AS MUCH ENERGY AS
IT LOSES.
THE ABOVE CAN BE SUMMARIZED AS THE GAIN IN ENERGY BY THE CHARGE = CORRESPONDING
LOSSES THROUGH RESISTANCES
KVL DEPENDS UPON THE CONCEPT OF A LOOP. A LOOP IS ANY CLOSED PATH THROUGH THE CIRCUIT
WHICH ENCOUNTERS NO NODE MORE THAN ONCE. ESSENTIALLY, TO CREATE A LOOP, START AT ANY NODE
IN THE CIRCUIT AND TRACE A PATH THROUGH THE CIRCUIT UNTIL YOU GET BACK TO YOUR ORIGINAL
NODE.
KVL

9. Formulations of Kirchhoff’s Voltage Law
(CONSERVATION OF ENERGY)
FORMULATION 1:
SUM OF VOLTAGE DROPS AROUND LOOP
= SUM OF VOLTAGE RISES AROUND LOOP
FORMULATION 2:
ALGEBRAIC SUM OF VOLTAGE DROPS AROUND LOOP = 0
VOLTAGE RISES ARE INCLUDED WITH A MINUS SIGN.
FORMULATION 3:
ALGEBRAIC SUM OF VOLTAGE RISES AROUND LOOP = 0
VOLTAGE DROPS ARE INCLUDED WITH A MINUS SIGN.
(HANDY TRICK: LOOK AT THE FIRST SIGN YOU ENCOUNTER ON EACH ELEMENT WHEN TRACING THE
LOOP)
KVL

10. PATH 1:
PATH 2:
PATH 3:
THREE CLOSED
PATHS:
KVL EXAMPLE
KVL

11. IN LOOP AXBC, KVL GIVES
-5+4I_1+2I_1=0
OR, I_1= DFRAC{5}{6}A
V_{BX}= -V_{XB}=4I_1 (DROP ACROSS 4Ω RESISTOR)
=3.33V (X TERMINAL –VE AS THE CURRENT I1 FLOWS FROM B TO X)
SIMILARLY, IN LOOP DEFY,
-10 + 2I_2 + 5I_2 = 0 I_2 = DFRAC{10}{7}A
V_{DY}=DFRAC{10}{7} TIMES 2=2.857V(D TERMINAL +VE)
THE VOLTAGE BETWEEN THE TERMINAL X AND Y IS THEN
V_{XB} + V + V_{DY} = (-3.333 + 2 + 2.857) V = 1.524V V_{XY} = 1.524V
VXB IS –VE (NEGATIVE) BECAUSE POLARITY OF TERMINAL X IS –VE (NEGATIVE) AND THE
EQUIVALENT CIRCUIT OF THE NETWORK FOR THE PART X-Y OF THE CIRCUIT IS AS FOLLOW
IN FIGURE 5 FIND VOLTAGE DROP ACROSS X-Y
TERMINALS.
KVL

12. SUPERPOSITION THEOREM
THE SUPERPOSITION THEOREM IS A CIRCUIT ANALYSIS THEOREM USED TO SOLVE THE NETWORK WHERE TWO OR MORE
SOURCES ARE PRESENT AND CONNECTED.
SUPERPOSITION THEOREM STATES THE FOLLOWING:
“IN ANY LINEAR AND BILATERAL NETWORK OR CIRCUIT HAVING MULTIPLE INDEPENDENT SOURCES, THE RESPONSE OF AN
ELEMENT WILL BE EQUAL TO THE ALGEBRAIC SUM OF THE RESPONSES OF THAT ELEMENT BY CONSIDERING ONE SOURCE
AT A TIME.”
TO CALCULATE THE INDIVIDUAL CONTRIBUTION OF EACH SOURCE IN A CIRCUIT, THE OTHER SOURCE MUST BE REPLACED
OR REMOVED WITHOUT AFFECTING THE FINAL RESULT. THIS IS DONE BY REPLACING THE VOLTAGE SOURCE WITH A SHORT
CIRCUIT. WHILE REMOVING A VOLTAGE SOURCE, ITS VALUE IS SET TO ZERO. WHEN REMOVING A CURRENT SOURCE, ITS
VALUE IS SET TO INFINITE. THIS IS DONE BY REPLACING THE CURRENT SOURCE WITH AN OPEN CIRCUIT.
THE SUPERPOSITION THEOREM IS VERY IMPORTANT IN CIRCUIT ANALYSIS BECAUSE IT CONVERTS A COMPLEX CIRCUIT
INTO A NORTON OR THEVENIN EQUIVALENT CIRCUIT.
GUIDELINES TO KEEP IN MIND WHILE USING THE SUPERPOSITION THEOREM
WHEN YOU SUM THE INDIVIDUAL CONTRIBUTIONS OF EACH SOURCE, YOU SHOULD BE CAREFUL WHILE ASSIGNING SIGNS
TO THE QUANTITIES. IT IS SUGGESTED TO ASSIGN A REFERENCE DIRECTION TO EACH UNKNOWN QUANTITY. IF A
CONTRIBUTION FROM A SOURCE HAS THE SAME DIRECTION AS THE REFERENCE DIRECTION, IT HAS A POSITIVE SIGN IN THE
SUM; IF IT HAS THE OPPOSITE DIRECTION, THEN A NEGATIVE SIGN.
ALL THE COMPONENTS MUST BE LINEAR TO USE THE SUPERPOSITION THEOREM WITH CIRCUIT CURRENTS AND VOLTAGES.
IT SHOULD BE NOTED THAT THE SUPERPOSITION THEOREM DOES NOT APPLY TO POWER, AS POWER IS NOT A LINEAR
QUANTITY.
S.T

13. HOW TO APPLY SUPERPOSITION THEOREM?
*THE FIRST STEP IS TO SELECT ONE AMONG THE MULTIPLE SOURCES PRESENT IN THE
BILATERAL NETWORK.
*AMONG THE VARIOUS SOURCES IN THE CIRCUIT, ANY ONE OF THE SOURCES CAN BE CONSIDERED
FIRST.
*EXCEPT FOR THE SELECTED SOURCE, ALL THE SOURCES MUST BE REPLACED BY THEIR INTERNAL
IMPEDANCE.
*USING A NETWORK SIMPLIFICATION APPROACH, EVALUATE THE CURRENT FLOWING THROUGH OR
THE VOLTAGE DROP ACROSS A PARTICULAR ELEMENT IN THE NETWORK.
*THE SAME CONSIDERING A SINGLE SOURCE IS REPEATED FOR ALL THE OTHER SOURCES IN THE
CIRCUIT.
*UPON OBTAINING THE RESPECTIVE RESPONSE FOR INDIVIDUAL SOURCE, PERFORM THE
SUMMATION OF ALL RESPONSES TO GET THE OVERALL VOLTAGE DROP OR CURRENT THROUGH THE
CIRCUIT ELEMENT
S.T

14. REVIEW OF THE SUPERPOSITION THEOREM
THE SUPERPOSITION THEOREM STATES THAT A CIRCUIT WITH MULTIPLE POWER SOURCES CAN BE
ANALYZED BY EVALUATING ONLY ONE POWER SOURCE AT A TIME. THEN, THE COMPONENT
VOLTAGES AND CURRENTS ARE ADDED ALGEBRAICALLY TO DETERMINE THE CIRCUIT RESPONSE
WITH ALL POWER SOURCES IN EFFECT.
STEP 1: REPLACE ALL OF THE POWER SOURCES EXCEPT ONE. REPLACE VOLTAGE SOURCES WITH A
SHORT CIRCUIT (WIRE) AND CURRENT SOURCES WITH AN OPEN CIRCUIT (BREAK).
STEP 2: CALCULATE THE VOLTAGES AND CURRENTS DUE TO EACH INDIVIDUAL SOURCE.
STEP 3: REPEAT STEPS 1 AND 2 FOR EACH POWER SUPPLY.
STEP 4: SUPERIMPOSE THE INDIVIDUAL VOLTAGES AND CURRENTS. ALGEBRAICALLY ADD THE
COMPONENT VOLTAGES AND CURRENTS; PAYING PARTICULAR ATTENTION TO THE DIRECTION OF
THE VOLTAGE DROPS AND CURRENT FLOWS.
THE SUPERPOSITION THEOREM IS LIMITED TO USE WITH LINEAR, BILATERAL CIRCUITS.
THE SUPERPOSITION THEOREM CAN BE APPLIED TO DC, AC, AND COMBINED AC/DC CIRCUITS.
THE SUPERPOSITION THEOREM CANNOT BE USED TO ADD POWER.
S.T

15. EXAMPLE: FIND I IN THE CIRCUIT
SHOWN IN FIGURE 1
NEXT, LET US ASSUME THE CURRENT SOURCE ONLY
SOLUTION
[BY CURRENT DIVISION FORMULA]
[BY CURRENT DIVISION FORMULA]
THE CURRENT THROUGH 2Ω RESISTOR IS OBTAINED
AS S.T

16. THANK
YOU