ECA - Lecture 03

Electric Circuit Analysis

Basic analysis techniques

Ahsan [email protected] 102Department of Electrical Engineering

Circuit Abstraction

Question: What is the current through the bulb?

Concept of Abstraction

Solution:

In order to calculate the current, we can replace the bulb with a resistor.

R is the only subject of interest, which serves as an abstraction of the bulb.

Voltage (Potential)

a

b

VVab 5 a 、 b, which point’s potential is higher ？

b

a

V6aV V4bV Vab = ?

a b +Q from point b to point a get energy ， Point a is

Positive? or negative ?

Basic Quantities

Example

Voltage (Potential)

ab

c´

c d

d´

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

,

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

Basic Quantities

Example

I

Open Circuit R=

I=0, V=E , P=0E

R0

Short Circuit R=0

E

R0

R ＝ 0 0R

EI 00 IREV

02RIPE

Basic Quantities

Nodes, Branches, Loops, mesh

• A circuit containing three nodes and five branches.• Node 1 is redrawn to look like two nodes; it is still one nodes.

Kirchhoff's Current and Voltage Laws

• sum of all currents entering a node is zero• sum of currents entering node is equal to sum of currents leaving node

KCL

KCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

Kirchhoff's Current and Voltage Laws

• sum of all currents entering a node is zero• sum of currents entering node is equal to sum of currents

leaving node

KCL

P1.9

DCBA iiii

Kirchhoff's Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+-120V

50* 1W Bulbs

Is

• Find currents through each light bulb:

IB = 1W/120V = 8.3mA

• Apply KCL to the top node:

IS - 50IB = 0

• Solve for IS: IS = 50 IB = 417mA

KCL-Christmas Lights

Kirchhoff's Current and Voltage Laws

KCL

Current divider

N VG1

G2

I+

-

I1 I2

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

Kirchhoff's Current and Voltage Laws

In case of parallel : 1 21 2

1 1 1 , , V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero.

KVL

• A voltage encountered + to - is positive.• A voltage encountered - to + is negative.

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

Kirchhoff's Current and Voltage Laws

KVL

Determine the voltages Vae and Vec.

Kirchhoff's Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

N

Kirchhoff's Current and Voltage Laws

KVLVoltage divider

kR 151

Volume control?

Example: Vs = 9V, R1 = 90kΩ, R2 = 30kΩ

Kirchhoff's Current and Voltage Laws

Mesh Analysis• Technique to find voltage drops around a

loop using the currents that flow within the loop, Kirchoff’s Voltage Law, and Ohm’s Law– First result is the calculation of the mesh currents• Which can be used to calculate the current flowing

through each component

– Second result is a calculation of the voltages across the components• Which can be used to calculate the voltage at the

nodes.

Mesh Analysis

What is a mesh: It is a loop which does not

contain any other loops within it.

V1

10V

R1

4ohmR26ohm

R3

3.0ohm i5A

i1i2

This circuit has two meshes

Steps in Mesh Analysis

Vin

Step 1

• Identify all of the meshes in the circuit

Vin

Step 2

• Label the currents flowing in each mesh

i1

i2

Vin

Step 3

• Label the voltage across each component in the circuit

i1

i2

+ V1

_

Vin

+ V3

_

+ V5

_

+ V6

_

+ V2 -

+ V4 -

Step 4

• Use Kirchoff’s Voltage Law

i1

i2

+ V1

_

Vin

+ V3

_

+ V5

_

+ V6

_

+ V2 -

+ V4 -

0

0

543

6321

VVV

VVVVVin

Step 5

• Use Ohm’s Law to relate the voltage drops across each component to the sum of the currents flowing through them.– Follow the sign convention on the resistor’s voltage.

RIIV baR

Step 5

i1

i2

+ V1

_

Vin

+ V3

_

+ V5

_

+ V6

_

+ V2 -

+ V4 -

616

525

424

3213

212

111

RiV

RiV

RiV

RiiV

RiV

RiV

Step 6

• Solve for the mesh currents, i1 and i2

– These currents are related to the currents found during the nodal analysis.

213

542

62171

iiI

IIi

IIIIi

Step 7

• Once the mesh currents are known, calculate the voltage across all of the components.

12V

From Previous Slides

616

525

424

3213

212

111

RiV

RiV

RiV

RiiV

RiV

RiV

0

0

543

6321

VVV

VVVVVin

Substituting in Numbers

kiV

kiV

kiV

kiiV

kiV

kiV

1

3

6

5

8

4

16

25

24

213

12

11

0

012

543

6321

VVV

VVVVV

Substituting the results from Ohm’s Law into the KVL equations

0365

0158412

2221

12111

kikikii

kikiikikiV

Results

• One or more of the mesh currents may have a negative sign.

Mesh Currents (mA)i1 740i2 264

Results

Currents (mA)IR1 = i1 740IR2 = i1 740

IR3 = i1- i2 476IR4 = i2 264IR5 = i2 264IR6 = i1 740I Vin = i1 740

The currents through each component in the circuit.

Example

Using Mesh analysis find the currents through each resistor

KVL in loop 1:

-2+2i1+4(i1-i2)= 0

6i1-4i2-2= 0

6i1-4i2= 2 …………..1)

KVL in loop 2:

6+4(i2-i1)+i2= 0

-4i1+5i2= -6 ………..2)

From eqn 1)&2)

Loop currents: i1= -1A, i2= -2A

Currents through each resistor:

i2= i1= -1A

i1 = i2 = -2A

i4= i1-i2

= -1A-(-2A)

= 1A

V12V

R12ohm

R24ohm

R31ohm

V26V

i1i2

Example

Use the mesh current method to determine the power associated with each voltage source in the circuit. Calculate the voltage vo across the 8 resistor.

KVL in Loop 1:

-40+2i1+8(i1-i2)=0

KVL in Loop 2:

8(i2-i1)+6i2+6(i2-i3)=0

KVL in Loop 3:

6(i3-i2)+4i3+20=0

V140V

V220V

R1

2.0ohm

R28ohm

R3

6ohm

R46ohm

R5

4ohm

i1 i2 i3

10i1-8i2+0i3=40

-8i1+20i2-6i3=0

0i1-6i2+10i3= -20

By solving the matrix

i1=5.6A

i2= 2A

i3= -0.8A

P40v= -40i1= -40(5.6) = -224W

P20v = 20i3= 20(-0.8) = -16W

vo= 8(i1-i2)= 8(5.6-2) = 8(3.6)=28.8V

35

Example

Apply KVL to each mesh

2 1 7 5 0sV v v v

2 6 7 0v v v

15 3 0sv v v

Mesh 1:

Mesh 2:

Mesh 3:

14 8 6 0sv v V v Mesh 4:

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1v 2v

3v 4v

5v6v

7v

8v

+ +

+ +

++

+

+

-

-- -

-

-

-

-

1sV

2sV 1i 2i

3i 4i

36

DC

DC

1R

3R5R

7R

2R

6R

8R

4R

1v 2v

3v 4v

5v6v

7v

8v

+ +

+ +

++

+

+

-

-- -

-

-

-

-

1sV

2sV 1i 2i

3i 4i

2 1 7 5 0sV v v v

2 6 7 0v v v

15 3 0sv v v

Mesh 1:

Mesh 2:

Mesh 3:

14 8 6 0sv v V v Mesh 4:

2 1 1 1 2 7 1 3 5( ) ( ) 0sV i R i i R i i R

2 2 2 4 6 2 1 7( ) ( ) 0i R i i R i i R

13 1 5 3 3( ) 0si i R V i R

Mesh 1:

Mesh 2:

Mesh 3:

14 4 4 8 4 2 6( ) 0si R i R V i i R Mesh 4:

Express the voltage in terms of the mesh currents:

37

2 1 1 1 2 7 1 3 5( ) ( ) 0sV i R i i R i i R

2 2 2 4 6 2 1 7( ) ( ) 0i R i i R i i R

13 1 5 3 3( ) 0si i R V i R

Mesh 1:

Mesh 2:

Mesh 3:

14 4 4 8 4 2 6( ) 0si R i R V i i R Mesh 4:

Mesh 1:

Mesh 2:

Mesh 3:

Mesh 4:

21 5 7 1 7 2 5 3( ) sR R R i R i R i V

7 1 2 6 7 2 6 4( ) 0R i R R R i R i

15 1 3 5 3( ) sR i R R i V

16 2 4 6 8 4( ) sR i R R R i V

38

Mesh 1:

Mesh 2:

Mesh 3:

Mesh 4:

21 5 7 1 7 2 5 3( ) sR R R i R i R i V

7 1 2 6 7 2 6 4( ) 0R i R R R i R i

15 1 3 5 3( ) sR i R R i V

16 2 4 6 8 4( ) sR i R R R i V

2

1

1

1 5 7 7 5 1

7 2 6 7 6 2

5 3 5 3

6 4 6 8 4

0

00

0 0

0 0

s

s

s

VR R R R R i

R R R R R i

VR R R i

R R R R i V