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AMIE(I) STUDY CIRCLE(REGD.) A Focused Approach

Dated: Feb 06, 2010

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Additional Course Material

NETWORK THEOREMS Additional theorems

SUPERPOSITION THEOREMThis theorem finds use in solving a network where two or more sources are present andconnected not in series or in parallel.

Statement If a number of voltage or current sources are ecting simultaneously in a linear network, theresultant current in any branch is the algebraic sum of the currents that would be produced in it, when each source acts alone replacing all other independent sources by their internalresistances.

Step 1: Take only one independent source of voltage/current and deactivate the otherindependent voltage/current sources. (For voltage sources, remove the source and shortcircuit the respective circuit terminals and for current sources, just delete the source keeping

the respective circuit terminals open). Obtain branch currents.Step 2. Repeat the above step for each of the independent sources.

Step 3. To determine the net branch current utilising superposition theorem, just add thecurrents obtained in step 1 and step 2 for each branch. If the currents obtained in step 1 andstep 2 are in same direction, just -add them; on the other hand, if the respective currents aredirected opposite in each step, assume the direction of the clockwise current to be +ve andsubtract the current obtained in the next step from the original current. The net current in eachbranch is then obtained.

Example

Find v in the circuit of given figure using superposition theorem.

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Solution

Let us first take 2 V source deactivating the current sources (see figure below)

1

21

2 21

2 2

i A x

v1 (drop across r L due to 2 V source) = 1 x 1 = 1 V

Next, taking the lower current source only (see figure below)

2

1 3 15( 5) ( 5)

1 1 (2 / 3) 8 8i A

3 15 2 15 2 5.8 1 2 8 3 4i A

This gives 25

(1) (5 / 4)4

v V

In following figure

4

15 15 / 5 3

(2 / 3) 1i A

This gives2

3. 22 1rL

i A

3 2 1 2v x V

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By superposition

1 2 3 1 5 / 4 2 7 / 4v v v v V

Example

Find v by superposition theorem.

Solution

Taking the 10 V source only, with reference to following figure.

1

1010 10 6.67

5 10v xi V

Taking the 5A source only, with reference to figure given below

10

55. 1.67

5 10i A

2 1.67 10 16.70v x V

By superposition theorem

1 2 6.67 16.70 23.37v v v V

Example

Find i 0 and i from the circuit given below using superposition theorem.

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Solution

Assuming only 6 V source to be active, with reference to figure given below

0 0 06 (1 5) ' 2 ' 0 ' 3 / 4i i i A

0 ' ' (3 / 4)i i A

Next, assuming 1 A source active only, with reference to figure given below

00 0

" 2 ""1 " " 1.2 " 0.4 "

1 5v iv

i i v i

But 0 " "/1i v

We finally get

0 0 01 1.2 " 0.4 " 0.8 "i i i

i.e. 0 " 1.25i A

and 0 0" 2 "

" "/ 5 0.255

v ii i A

Using the principle of superposition

0 0 0' " (3 / 4) 1.25 0.5i i i A

and 0 ' " (3 / 4) 0.25 0.5i i i A

RECIPROCITY THEOREM

Statement In any branch of a network, the current (1) due to a single source of voltage N) elsewl!ere inthe network is equal to the current through the branch in which the source was originally

placed when the source is placed in the branch in which the current (1) was originallyobtained.

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ExplanationIn simple sense, the location of the voltage source and the through current may beinterchanged without a change in current. However, the. polarity of the voltage source shouldhave the identicality with the direction of branch current in each position.

The limitation of this theorem is that it is applicable only to single source networks and not inmulti source network. Moreover, the network where reciprocity theorem is applied should bea linear one and containing of resistors, inductors, capacitors and coupled circuits. Thenetwork should not have any time varying element.

Steps for Solving a Network Utilising Reciprocity TheoremStep 1. The braches between which reciprocity is to be established are to be selected first.

Step 2 . The current in the branch is obtained using conventional network analysis.

Step 3. The voltage source is interchanged between the branches concerned.Step 4. The current in the branch where the voltage source was existing earlier is calculated.

Example

Show the application of reciprocity theorem in the network of given figure.

Solution

With reference to given figure

[(2 1) ||3] 2 3.5eq R

110

2.863.5

I A

2

32.86 1.43

3 3 I x A

3 2.86 1.43 1.43 I A

With reference to following figure

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(2 || 3) 1 2 (6 / 5) 3 21/ 5 4.2eq R

210

2.3814.2

V I A

This gives 1 23 3

2.381 1.433 2 5

I I x A

TELLEGEN'S THEOREM

For any given time, the sum of power delivered to each branch of any electric network is zero.

Thus for Kth branch, this theorem states that1

0n

K K K

v i .

Here n is number of branches, v K is the drop in the branch and i K the through current.

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