What are the advantages of Network Theorems ? What is the

What are the advantages of Network Theorems ?

What is the main difference between Thevenin’sTheorem and Norton’s Theorem ?

Why the load current is reduced in practical current source.

We increase the shunt resistance in practical current source

DC Voltage

SourceDC current

source

Ideal Sources

Ideal Current Source

+

-

VIs

Is

V

Ideal Current Source

+

-

VIs RL

V = Is RL

As RL V

Practical Current Source

+

-

VIs

Is

V

RSh

Rsh

VIsI −=

I = Load currentIs = Supply Current from SourceV = Load VoltageRsh = Shunt Resistance

IIR

IIIs R +=

Practical Current Source

Using Current Divider Rule

mAI

KM

MmAI

L

L

990.9

)11(

110

=

+

=

10 mA1MΩ 1KΩ

10 mA10 MΩ 1KΩ

mAI

KM

MmAI

L

L

999.9

)11(

110

=

+

=

Using Current Divider Rule

ideal current source Vs Practical Current Source

Rs is infinite Have some Rs

ideal current source Vs Practical Current Source

Rs is infinite

Supplies constant current to the load Whatever voltage is across load

Have some Rs

Is changes with the voltagei.e. V increases Is Decreases

ideal current source Vs Practical Current Source

Rs is infinite

Supplies constant current to the load Whatever voltage is across load

Can supply any amount of Voltage across load, Voltage depends only on RL

Have some Rs

Is changes with the voltagei.e. V increases Is Decreases

Supply limited voltage across load

Ideal Voltage Source

VVs

V

I

+- RL

Ideal Voltage Source

VVs

V

I

+- RL

Apply KVL-Vs + V = 0

V = VsLoad voltage V independent of load Current

Practical Voltage Source

V

I

Practical Voltage Source

V

IApply KVL-Vs + I Rs + V = 0

V = Vs – I RsLoad voltage V is depending on load Current

Ideal

Practical

Ideal Voltage Source Vs Practical Voltage Source

Internal Resistance Rs should be zero Internal Resistance Rs should be minimum

Ideal Voltage Source Vs Practical Voltage Source

Internal Resistance Rs should be zero

Supplies constant voltage to the load Whatever current is across load

Internal Resistance Rs should be minimum

Vs changes with the currenti.e. As I increases Vs Decreases

Ideal Voltage Source Vs Practical Voltage Source

Internal Resistance Rs should be zero

Supplies constant voltage to the load Whatever current is across load

Can supply any amount of Voltage across load,

Voltage depends only on RL

Internal Resistance Rs should be minimum

Vs changes with the currenti.e. As I increases Vs Decreases

Supply limited voltage across load

KVL

•The sum of voltage drops around a closed path must equal zero.

KCL

•The sum of currents leaving a closed surface or point must equal zero.

Review of Kirchhoff’s Current Law& Kirchhoff’s Voltage Law

KVL Tricks

+

-V2

Path

+

-V1

Path

+-V3

Path

+ V1

- V2

+ V3

Path 1: 0vvv b2a =++−

vcva

+

−

+

−

3

21

+ −

vb

v3v2

+ −

+

-

ab c

Writing KVL Equations

Path 1: 0vvv b2a =++−

Path 2: 0vvv c3b =+−−

vcva

+

−

+

−

3

21

+ −

vb

v3v2

+ −

+

-

ab c

Writing KVL Equations

Path 1: 0vvv b2a =++−

Path 2: 0vvv c3b =+−−

Path 3: 0vvvv c32a =+−+−

vcva

+

−

+

−

3

21

+ −

vb

v3v2

+ −

+

-

ab c

Writing KVL Equations

Elements in Parallel

KVL clockwise, start at top:

Vb – Va = 0

Va = Vb

In order to satisfy KCL, what is the value of i?

KCL says:

24 μA + -10 μA + (-)-4 μA + -i =0

18 μA – i = 0

i = 18 μA

i10 mA

24 mA -4 mA

Kirchhoff’s Current Law (KCL)

Elements in Series

i1 = i2

i1 – i2 = 0

Mesh Current Analysis Nodal Voltage Analysis

Using KCL

Using Kirchhoff's Current Law

Equation No

1 : 10 = 50I1 + 40I2

Using Kirchhoff's Current Law

Equation No

1 : 10 = 50I1 + 40I2Equation No

2 : 20 = 40I1 + 60I2

Using Kirchhoff's Current Law

I3 = I1 + I2

I1 = -0.142 AmpsI2 = 0.428 AmpsI3 = 0.286 Amps

Equation No

1 : 10 = 50I1 + 40I2Equation No

2 : 20 = 40I1 + 60I2

Mesh Current Analysis

Only label inside loops in a clockwise direction with circulating currents

i1 = I1 , i2 = -I2 and I3 = I1 – I2

10 = 10 I1 + 40 I1 – 40 I2

-20 = 20 I2 + 40 I2 – 40 I1

1

2

Mesh Current Analysis

[ V ] gives the total battery voltage for loop 1 and then loop 2

[ I ] loop currents which we are trying to find

[ R ] resistance matrix

[ R-1 ] inverse of the [ R ] matrix

Mesh Current Analysis

I3 = I1 – I2 -0.143 – (-0.429) = 0.286 Amps

Same as found by Kirchhoff’s circuit law

Nodal Voltage Analysis

Reference node = D

Nodal Voltage Analysis

Reference node = D

Va = 10v

Vc = 20v

Vb = ?

Nodal Voltage Analysis

Reference node = D