KIRCHHOFF VOLTAGE LAW ONE OF THE FUNDAMENTAL CONSERVATION LAWS IN ELECTRICAL ENGINERING

THIS IS A CONSERVATION OF ENERGY PRINCIPLE “ENERGY CANNOT BE CREATE NOR DESTROYED”

KIRCHHOFF VOLTAGE LAW (KVL) KVL IS A CONSERVATION OF ENERGY PRINCIPLE A POSITIVE CHARGE GAINS ENERGY AS IT MOVES TO A POINT WITH HIGHER VOLTAGE AND RELEASES ENERGY IF IT MOVES TO A POINT WITH LOWER VOLTAGE 

W



q

(

V B



V A

)

B V B q



V A

A “THOUGHT EXPERIMENT” 

W



qV AB



V AB



q



V A B V B



V BC

 

W



qV BC



V CA



W

 

qV CA

V C

IF THE CHARGE COMES BACK TO THE SAME INITIAL POINT THE NET ENERGY GAIN MUST BE ZERO (Conservative network) OTHERWISE THE CHARGE COULD END UP WITH INFINITE ENERGY, OR SUPPLY AN INFINITE AMOUNT OF ENERGY

q

(

V AB



V BC



V CD

)  0

q



q



a c



V ab

 

V cd



b d

LOSES 

W



qV ab

GAINS 

W



qV cd

KVL: THE ALGEBRAIC SUM OF VOLTAGE DROPS AROUND ANY LOOP MUST BE ZERO

A



V

   ( 

V B A

A VOLTAG E RISE IS ) 

B

A NEGATIVE DROP



V S



V R

1 

V R

2 

V R

3  0

V R

2  12

V V R

1  18

V

PROBLEM SOLVING TIP: KVL IS USEFUL TO DETERMINE A VOLTAGE - FIND A LOOP INCLUDING THE UNKNOWN VOLTAGE THE LOOP DOES NOT HAVE TO BE PHYSICAL 

V be

 EXAMPLE : V R1 , V R3 ARE KNOWN DETERMINE THE VOLTAGE V be

V R

1 

V be



V R

3  30 [

V

]  0 LOOP abcdefa

BACKGROUND: WHEN DISCUSSING KCL WE SAW THAT NOT ALL POSSIBLE KCL EQUATIONS ARE INDEPENDENT. WE SHALL SEE THAT THE SAME SITUATION ARISES WHEN USING KVL A SNEAK PREVIEW ON THE NUMBER OF LINEARLY INDEPENDENT EQUATIONS IN THE CIRCUIT DEFINE

N

NUMBER OF NODES

B

NUMBER OF BRANCHES

N

 1 LINEARLY INDEPENDEN KCL EQUATIONS T

B

 (

N

 1 ) LINEARLY INDEPENDEN T KVL EQUATIONS EXAMPLE: FOR THE CIRCUIT SHOWN WE HAVE N = 6, B = 7. HENCE THERE ARE ONLY TWO INDEPENDENT KVL EQUATIONS THE THIRD EQUATION IS THE SUM OF THE OTHER TWO!!

FIND THE VOLTAGES

V ae

,

V ec

DEPENDENT SOURCES ARE HANDLED WITH THE SAME EASE GIVEN THE CHOICE USE THE SIMPLEST LOOP

V ac

 4  6  0

V bd

 2  4  0

V ad



V ac V bd

  ______

V eb

 4  6  12  0

V ad

 12  8  6  0

V bd

 MUST FIND V R  12 

V R

1  1  1 FIRST 10

V R

1  0 

V R

1  1

V

DEPENDENT SOURCES ARE NOT REALLY DIFFICULT TO ANALYZE REMINDER: IN A RESISTOR THE VOLTAGE AND CURRENT DIRECTIONS MUST SATISFY THE PASSIVE SIGN CONVENTION

V ad

 _______,

V eb

 ________ 

V

 

V



SAMPLE PROBLEM 

V

1  +  4

V



R

 2

k



b



V x



V

1  12

V

,

V

2  4

V

 + -

a V

2 

DETERMINE V

x

V

ab 

4V



-8V

Power disipated on the 2k resistor P

2k  Remember past topics We need to find a closed path where only one voltage is unknown

FOR V V X X

 

V X V

4 2 

V

1  12   4 4   0 0

V X V ab



V

2   

V X V ab



V

2  0

10

k

 5

k



There are no loops with only one unknown!!!

 

V x

 

- Vx/2 +

25 

V

+ -

V

1  The current through the 5k and 10k resistors is the same. Hence the voltage drop across the 5k is one half of the drop across the 10k!!!

+ 

V X

25 [

V

] 

V X

  20 [

V

] 

V X

2 

V X

4  0

V

4

x V

1 

V X

4

V

1  

V X

4 

V X

2  0  5 [

V

]