KVL and KCL examples

Mesh Analysis

Nodal Analysis

Ohms alw • A voltage of 20V is applied across AB shown in

the Figure 1(b).Calculate the total current, the current in each resistor and power dissipated in each resistor and the value of the series resistance to have the total current.( All resistors are measured in ohms.)

Definitions • Electric circuit: A combination of various electric

elements connected in any manner is called an electric circuit.

• Passive network: is one which contains no source of emf. In it.

• Active network: is one which contains one of more source of emf. In it.

• Node: is a junction in a circuit where two or more circuit elements are connected together

• Branch: is a part of a network which lies between two junctions.

Definitions

• Loop : It is close path in a circuit in which no elements or node is encountered more than once.

• Mesh: It is loop that contains no other loop within it.

Network Analysis Direct methods : Determining different voltages and currents in the original circuit.

• KCL and KVL

• Nodal Analysis

• Superposition theorem

Network Reduction Method : the original circuit converted into a much simpler circuit.

• Thevenin’s theorem

• Norton’s theorem

Polarity • Sign of battery e.m.f:

Rise in voltage should be given a +ve sign and fall in voltage a –ve sign

• Sign of IR drop

FOR A GIVEN CIRCUIT LET B NUMBER OF BRANCHES N NUMBER OF NODES

THE MINIMUM REQUIRED NUMBER OF LOOP CURRENTS IS

)1( NBL

MESH CURRENTS ARE ALWAYS INDEPENDENT

AN EXAMPLE

2)16(7

6

7

L

N

BTWO LOOP CURRENTS ARE REQUIRED. THE CURRENTS SHOWN ARE MESH CURRENTS. HENCE THEY ARE INDEPENDENT AND FORM A MINIMAL SET

KVL ON LEFT MESH

REPLACING AND REARRANGING

DETERMINATION OF LOOP CURRENTS

KVL ON RIGHT MESH

2 4 5 30

Sv v v v

USING OHM’S LAW

1 1 1 2 1 2 3 1 2 3

4 2 4 5 2 5

, , ( )

,

v i R v i R v i i R

v i R v i R

SHORTCUT: POLARITIES ARE NOT NEEDED. APPLY OHM’S LAW TO EACH ELEMENT AS KVL IS BEING WRITTEN

1I @KVL

2I @KVL

396

12612

21

21

kIkI

kIkIREARRANGE

add and 2/*

mAIkI 5.0612 22

mAIkIkI4

561212 121

EXPRESS VARIABLE OF INTEREST AS FUNCTION OF LOOP CURRENTS

21 IIIO

1I @KVL

1IIO NOW

THIS SELECTION IS MORE EFFICIENT

996

12612

21

21

kIkI

kIkIREARRANGE

substract and 2/*

3/*

mAIkI4

31824 11

EXAMPLE: FIND Io

AN ALTERNATIVE SELECTION OF LOOP CURRENTS

2I @KVL

Mesh Analysis: KVL

Applying KVL to the three loops. We get

V1-I1R1-(I1-I2)R4=0 or I1(R1+R4)-I2R4=V1 -I2R2—(I2-I3)R5—(I2-I1)R4=0 or I1R4+I2(R2+R4+R5)+I3R5=0 -I3R3—V2-(I3-I2)R5=0 or I2R5-I3(R3+R5)=V2

Take any resistatnce values, write equations and solve the current

Solve this example

Answer: I1=5A I2: -1A Current through r2=4A

Exercise: Try • Determine the loop currents and current

supplied by each battery in the circuit shown in Figure

Answer: I2=542/299A , I3=-1875/598A , II1= 765/299A.

Discharge current of V1=765/299A Discharge current of V2=I1-I2=220/299A. Discharge current of V3= I2+I3=2965/598A Discharge current of V4 =I2=545/299A. Discharge current of V5= 1875/598A.

Nodal Analysis

Applying KCL at each node

THE STRATEGY FOR NODE ANALYSIS 1. IDENTIFY ALL NODES AND SELECT A REFERENCE NODE

2. IDENTIFY KNOWN NODE VOLTAGES

3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION (e.g.,SUM OF CURRENT LEAVING =0)

0:@321 IIIV

a

4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES

0369

k

VV

k

V

k

VV baasa AND GET ALGEBRAIC EQUATIONS IN THE NODE VOLTAGES ...

REFERENCE

SV

aV

bV

cV

0:@543 IIIV

b

0:@65 IIV

c

0943

k

VV

k

V

k

VVcbbab

039

k

V

k

VVcbc

SHORTCUT: SKIP WRITING THESE EQUATIONS...

AND PRACTICE WRITING

THESE DIRECTLY

EXAMPLE

@ NODE 1 WE VISUALIZE THE CURRENTS LEAVING AND WRITE THE KCL EQUATION

REPEAT THE PROCESS AT NODE 2

03

12

4

122

R

vv

R

vvi

OR VISUALIZE CURRENTS GOING INTO NODE

WRITE THE KCL EQUATIONS

mA6

1I

2I

3I

036

6:@ 222

k

V

k

VmAV 2

12V V

CURRENTS COULD BE COMPUTED DIRECTLY USING KCL AND CURRENT DIVIDER!!

1

2

3

8

3(6 ) 2

3 6

6(6 ) 4

3 6

I mA

kI mA mA

k k

kI mA mA

k k

IN MOST CASES THERE ARE SEVERAL DIFFERENT WAYS OF SOLVING A PROBLEM

NODE EQS. BY INSPECTION

mAVVk

6202

121

mAVkk

V 63

1

6

10 21

116V V

k

VI

k

VI

k

VI

362

23

22

11

Once node voltages are known

1

1@ : 2 6 0

2

VV mA mA

k

Node analysis

VV

VV

4

6

4

1

SOURCES CONNECTED TO THE REFERENCE

SUPERNODE

CONSTRAINT EQUATION VVV 1223

KCL @ SUPERNODE

02

)4(

212

6 3322

k

V

k

V

k

V

k

Vk2/*

VVV 22332

VVV 1232 add and 3/*

VV 385 3

mAk

VI

O8.3

2

3 LAW SOHM'

OIV FOR NEEDED NOT IS

2

LEARNING EXAMPLE

1 1 21@ : 4 0

6 12

V V VV mA

k k

USING KCL

2 2 12@ : 2 0

6 12

V V VV mA

k k

BY “INSPECTION”

1 2

1 1 14

6 12 12V V mA

k k k

1 2

1 1 12

12 6 12V V mA

k k k

LEARNING EXAMPLE

3 nodes plus the reference. In principle one needs 3 equations...

…but two nodes are connected to the reference through voltage sources. Hence those node voltages are known!!!

…Only one KCL is necessary

012126

12322

k

VV

k

VV

k

V

][5.1][64

0)()(2

22

12322

VVVV

VVVVV

EQUATIONS THE SOLVING

Hint: Each voltage source connected to the reference node saves one node equation

][6

][12

3

1

VV

VV

THESE ARE THE REMAINING TWO NODE EQUATIONS

Nodal voltage (Current source)-Try

Use the nodal analysis to find the current in 5 ohm resistance

210

1

1

2

4

11

10

13

10

13

1

23

4

32

:3

01

3

2

11

5

1

2

12

1

32

5

2

2

21

:2

2810

3

2

2

10

1

2

1

2

11

10

31

2

21

2

128

:1

VVV

VVVVV

Node

VVV

VVVVV

Node

VVV

VVVVV

Node

Solve equation to find v1,V2 and V3 and find the current.

Nodal analysis

Dependent sources

VOLTAGE SOURCE CONNECTED TO REFERENCE

VV 31

0263

: 212

xI

k

V

k

VV VKCL@

2

CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES k

VI

x

6

2

REPLACE

06

263

2212

k

V

k

V

k

VVk6/*

VVVV 602212

mAk

VVI

O1

3

21

OI FIND

LEARNING EXAMPLE

Try these examples

Ans: I2=V1/R1 I3=(V1-V2) / R3 I4=V2/R2