notes_tee301_ac-dc network theorems concept notes

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Prepared by: Nafees Ahmed

AC-DC NETWORK THEOREM

Mesh Analysis:

Steps are

1. Identify the total meshes

2. Assume some mesh current in each mesh (clockwise or anticlockwise)

3. Apply KVL in each mesh.

4. Solve the above equations.

Node Analysis:

Steps are

1. Identify the total Principal Nodes

2. Assume one node as reference node (Voltage of this node = 0 Volts)

3. Assume some node voltages for other remaining nodes w.r.to

reference node. (V1, V2, V3 etc).

4. Assume some branch currents in different branches.

5. Apply KCL at different nodes and make the equations in terms of node

voltages and circuit elements.

6. Solve the above equations.

Superposition theorem:

“In a linear circuit, containing more then one independent

energy sources, the overall response (Voltage or current) in any branch of the circuit is

equal to sum of the response due to each independent source acting one at a time while

making other source in-operative.”

⇒ +

So according to superposition theorem I=I’+I”

(a)

V1

I

R2 I1 R1

(b)

V1

I’ R2 R1

(c)

I” R2

I1 R1

Fig.1



Thevenin’s theorem:

“Any linear two terminal circuits can be replaced by an

equivalent network consisting of a voltage source (VTh) in series with impedance ZTh (or

RTh for DC network).”

⇒

Where

VTh = Open circuit voltage at load terminals

ZTh (or RTh) = Equivalent Impedance (Resistance) of the network at load

terminals when the sources are made in-operative.

And LTh

ThL

ZZ

VI

+=

Steps to find out Thevenin’s equivalent circuit

Steps are

1. Remove the load.

2. Find out the open circuit voltage at load terminals.

3. Find out ZTh (or RTh).

� Case I: Circuit with independent sources only

o Make the source in-operative

o Find out the equivalent impedance (or resistance) between the load

terminals

� Case II: Circuit with independent + dependent sources

o Short circuit the load terminals and find out short circuit current Isc.

o sc

ThThTh

I

VRorZ =)(

� Case III: Circuit with dependent sources only

o In this case VTh =0 & Isc =0.

o Apply 1 A current source at load terminals and find out voltage (V) across

load terminals then VV

I

VRorZ ThTh ===

1)( .

OR

o Apply 1 V voltage source at load terminals and find out current (I)

provided by this source thenII

VRorZ ThTh

1)( == .

Fig.2

(a)

Any Linear

Two

terminals

Network

IL

ZL

A

B (b) Thevenin’s Equivalent

Circuit

IL

A

B

VTh

ZTh ZL



Norton’s theorem:

“Any linear two terminal circuits can be replaced by an

equivalent network consisting of a current source (IN) in series with impedance ZN (or RN

for DC network).”

⇒

Where

IN = Short circuit current at load terminals

ZN (or RN) = Equivalent Impedance (or Resistance) of the network at load

terminals when the sources are made in-operative = ZTh (or RTh)

And N

LTh

ThL I

ZZ

ZI

+=

Steps to find out Thevenin’s equivalent circuit

Steps are

4. Remove the load and short circuit the load terminals.

5. Find out the short circuit current Isc (= IN) at load terminals.

6. Find out ZN (or RN).

� Case I: Circuit with independent sources only

o Make the source in-operative

o Find out the equivalent impedance (or resistance) between the load

terminals

� Case II: Circuit with independent + dependent sources

o Open circuit the load terminals and find out open circuit voltage Voc

(=VTh).

o sc

ThNN

I

VRorZ =)(

� Case III: Circuit with dependent sources only

o In this case VTh =0 & Isc =0.

o Apply 1 A current source at load terminals and find out voltage (V) across

load terminals then VV

I

VRorZ NN ===

1)( .

OR

o Apply 1 V voltage source at load terminals and find out current (I)

provided by this source thenII

VRorZ NN

1)( == .

Fig.3

(a)

Any Linear

Two

terminals

Network

IL

ZL

A

B (b) Norton’s Equivalent

Circuit

IL

A

B

IN ZN ZL



Maximum Power Transfer Theorem:

For DC Netwrok:

“Maximum power is transferred by a circuit to a load resistance (RL) when RL is

equal to Thevenin’s equivalent resistance (RTh) of the network.”

So for maximum power

RL = RTh

And maximum power will be

L

Th

R

VP

4

2

max =

Proof:

Load current will be

)1(−−−−−+

=LTh

ThL

RR

VI

Power

LLRIP 2=

L

LTh

Th RRR

V2

+=

)2()( 2

2 −−−−−+

=LTh

LTh

RR

RV

Differentiating equation equa (2) w. r. t. RL and put dP/dRL = 0

)3()(

)(1)(4

22 −−−−−

++×−×+

=LTh

LThLLThTh

L RR

RRRRRV

dR

dP

RL-RTh = 0

Or RL = RTh Put this in equation (2)

L

Th

R

VP

4

2

max =

For AC Network:

“Maximum power is transferred by a circuit to load impedance (ZL) when ZL is

equal to complex conjugate of Thevenin’s equivalent impedance (ZTh) of the network.”

So for maximum power

ThL ZZ =

And maximum power will be

L

Th

R

VP

4

2

max =

IL

RL

A

B

VTh

RTh

Fig.4a Thevenin’s Equivalent

Circuit

Fig.4b Thevenin’s Equivalent

Circuit

IL

A

B

VTh

ZTh ZL



Proof:

ThThTh jXRZ +=

LLL jXRZ +=

Load current will be

)4()()(

−−−−−+++

=+

=LThLTh

Th

LTh

ThL

XXjRR

V

ZZ

VI

Power

LL RIP2

=

L

LThLTh

Th RXXjRR

V2

22 )()(

+++=

)5()()( 22

2 −−−−−+++

=LThLTh

LTh

XXjRR

RV

Differentiating equation equa (5) w. r. t. XL and put dP/dXL = 0

[ ]

)6()()(

)(20222

2 −−−−−

+++

+×−=

LThLTh

LThLTh

L XXjRR

XXRV

dR

dP

XL+XTh = 0

Or XL = -XTh Put this in equation (5)

)6()( 2

2 −−−−−−+

=LTh

LTh

RR

RVP

Differentiating equation equa (6) w. r. t. RL and put dP/dRL = 0

0)(

)(1)(4

2

2 =

+

+×−×+=

LTh

LThLLTh

Th

L RR

RRRRRV

dR

dP

RL-RTh = 0

Or RL = RTh Put this in equation (6) L

Th

R

VP

4

2

max =

So for maximum power transfer to a load ZL

ThL ZZ = and L

Th

R

VP

4

2

max =



Tellegen’s Theorem:

“In any liner/non linear, active/passive, time variant/invariant network, the

summation of power of each branch (instantaneous power in case of AC network) is

equal to zero”

So for a circuit having “n” no of branches

∑=

=n

K

KK iv1

0

Proof:

Consider Kth branch of a network

pqpqKK iviv =

Or )1()( −−−−−−−= pqqpKK ivviv

And )2()( −−−−−−−= qppqKK ivviv )( qppq ii −=∴

Adding equations (1) & (2)

qppqpqqpKK ivvivviv )()(2 −+−=

[ ]

)(2

)(

)()(2

1

qppq

qp

qppqpqqpKK

iivv

ivvivviv

−−

=

−+−=

For “n” braches

( )

∑ ∑∑ ∑

∑ ∑∑

= == =

= ==

−=

−=

n

q

n

p

pqq

n

p

n

q

pqp

n

K

n

q

pqqp

n

p

KK

iviv

ivviv

1 11 1

1 11

2

1

2

1

2

1

According to KCL at any node

∑∑==

==n

q

pq

n

p

pq ii11

0&0

Hence

∑=

=n

K

KK iv1

0

vp vq

vpq= vK

ipq= iK

Fig.5



Millman’s Theorem:

Statement 1: For Voltage sources in parallel

“If V1, V2, V3……… Vn voltage sources with internal impedances Z1, Z2,

Z3 ….Zn are connected in parallel; they can be replaced by an equivalent circuit

consisting of a voltage source Veq in series with impedance Zeq.”

Where

∑

∑

=

==+++

+++=

n

i i

n

i i

i

n

n

n

eq

Z

Z

V

ZZZZ

Z

V

Z

V

Z

V

Z

V

V

1

1

321

3

3

2

2

1

1

11........

111

........

∑=

=+++

=n

i in

eq

ZZZZZ

Z

1321

1

1

1........

111

1

Proof:

Apply source transformation to fig.6a

Where

1

11Z

VI =

2

22Z

VI =

3

33Z

VI = …….

n

nnZ

VI =

1

1

1

ZY =

2

2

1

ZY =

3

3

1

ZY = …….

n

nZ

Y1

=

Above figure can be replaced by fig.6d, see bellow

Z2

V2

+

-

~

Z3

V3

+

-

~

Zn

Vn

+

-

~

Z1

V1

+

-

~

ZL

IL

(a)

ZL

IL

(b)

Zeq

Veq +

-

~

Fig.6

ZL

IL

(c)

Yn

InY3

I3Y2

I2Y1

I1



Where

n

n

neq

Z

V

Z

V

Z

V

Z

V

IIIII

.....

.......

3

3

2

2

1

1

321

+++=

+++=

n

neq

ZZZZ

YYYYY

1......

111

.....

321

321

+++=

+++=

Apply source transformation to fig.6d, it will be converted to fig.6e, see above.

Where

∑

∑

=

==+++

+++==

n

i i

n

i i

i

n

n

n

eq

eq

eq

Z

Z

V

ZZZZ

Z

V

Z

V

Z

V

Z

V

Y

IV

1

1

321

3

3

2

2

1

1

11........

111

........

∑=

=+++

==n

i in

eq

eq

ZZZZZ

YZ

1321

1

1

1........

111

11 Proved

Statement 1: For current sources in series

“If I1, I2, I3……… In voltage sources with internal impedances Y1, Y2, Y3

….Yn are connected in series; they can be replaced by an equivalent circuit consisting of

a current source Ieq in parallel with admittance Yeq.”

Where

∑

∑

=

==+++

+++=

n

i i

n

i i

i

n

n

n

eq

Y

Y

I

YYYY

Y

I

Y

I

Y

I

Y

I

I

1

1

321

3

3

2

2

1

1

11........

111

........

ZL

IL

(d)

Yeq

Ieq ZL

IL

(e)

Zeq

Veq +

-

~

I2

Y2

I3

Y3

I1

Y1

In

Yn

(a)

Ieq

Yeq

(b)Fig.7



∑=

=+++

=n

i in

eq

YYYYY

Y

1321

1

1

1........

111

1

Proof:

Apply source transformation to fig.7a, see fig.7c

Where

1

11Y

IV =

2

22Y

IV =

3

33Y

IV = ……

n

nnY

IV =

1

1

1

YZ =

2

2

1

YZ =

3

3

1

YZ = ……

n

nY

Z1

=

Above figure can be replaced by fig.7d, see bellow

Where

n

n

neq

Y

I

Y

I

Y

I

Y

I

VVVVV

......

......

3

3

2

2

1

1

321

+++=

+++=

n

neq

YYYY

ZZZZZ

1......

111

....

321

321

+++=

+++=

Apply source transformation to fig.7d, it will be converted to fig.7e, see above.

Where

∑

∑

=

==+++

+++==

n

i i

n

i i

i

n

n

n

eq

eq

eq

Y

Y

I

YYYY

Y

I

Y

I

Y

I

Y

I

Z

VI

1

1

321

3

3

2

2

1

1

11........

111

........

∑=

=+++

==n

i in

eq

eq

YYYYY

ZY

1321

1

1

1........

111

11 Proved

(c)

Z2 V2

-

~ Z3

V3

-

~ Z1

V1

-

~ Zn

Vn

-

~

(d)

ZeqVeq

-

~

Ieq

Yeq

(e)



Reciprocity Theorem: “The ration of voltage and current remains same with respective to an interchange

between the point of application of source and the measurement of response.”

→ If excitation is a voltage source

Above network is reciprocal if

1

2

2

1

I

V

I

V=

→ If excitation is a current source

Above network is reciprocal if

2

1

1

2

I

V

I

V=

Compensation Theorem:

Consider a linear circuit with a voltage source “V” and its internal impedance ZTh

(Thevenin’s Equivalent impedance) delivering a current “I” to an impedance “Z” as

shown in following figure.

If “Z” changes to "" ZZ ∆+ , the current I will change to I’ as shown in following figure.

Network V1

± I2

Network I1 V1

±

Network V2I1

Network V1 I2

Fig.8

(a) (b)

(c) (d)

(a)

V

ZTh

Z I+

-

~



Now change in current I∆ ( )'III −=∆ can be find out by replacing the voltage source

with its internal impedance (ZTh) and placing a compensating voltage VC ( )ZIVc ∆×= in

series with impedance ZZ ∆+ such that polarity of VC opposes the flow of current.

)( ZZZ

VI

Th

C

∆++−=∆

Proof:

From fig.9a

)1(−−−−−−+

=ZZ

VI

Th

From fig.9b

)2(' −−−−−−∆++

=ZZZ

VI

Th

)()(

'

ZZZZZ

ZV

III

ThTh +×∆++

∆×−=

−=∆

)()( ZZZ

Z

ZZ

V

ThTh ∆++∆

×+

−=

)(

)(

ZZZ

V

ZZZ

ZI

Th

c

Th

∆++−=

∆++∆×

−=

Use equation (1)

Where

ZIVC ∆×=

So )( ZZZ

VI

Th

C

∆++−=∆

(b)

V

ZTh

ZZ ∆+I’+

-

~

(c)

VC

ZTH

ZZ ∆+I∆

+

-

~

Fig.9

notes_tee301_ac-dc network theorems concept notes

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