# Network Theorems. Mesh analysis Nodal analysis Superposition Thevenin’s Theorem Norton’s Theorem Delta-star transformation.

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Network Theorems

Mesh analysis Nodal analysis Superposition Thevenin’s Theorem Norton’s Theorem Delta-star transformation

Direct application in conjunction with Ohm’s law Indirect application in conjunction with resistance Simultaneous equations

Since R 3 and R 4 are in parallel Determine current and source e.m.f Therefore By Kirchoff’s first law Also By Kirchoff’s second law

Determine I 1, E, I 3 and I By Kirchoff’s second law Also By Kirchoff’s first law

Power dissipated in R 3 is 20W. Calculate I 3, R 1,I 1, I 2 and E By Kirchoff’s first law in node b By Kirchoff’s first law in node a By Kirchoff’s second law in loop 2 P.D across 1  is 5 X 1=5V 12 3 ab c d

First find the total effective resistance Determine current I and I 4 Then Using current division

Effective resistance for parallel resistor 10  // 15  and 16  //16  Determine V AB Then Using voltage division

Applying Kirchoff’s 2nd law for loop 1 Calculate the current in each resistor Applying Kirchoff’s 2nd law for loop 2 But Thus ---(b) ---(a)

continue Solving the 2 simultaneous equations Then (c) + (d) (a) X 4 (b) X 7 ---(c) ---(d) Substitute in (b) In 14  resistor In 4  resistor In 28  resistor In 3  resistor In 8  resistor

Calculate the current in the network (a)x10 (b)X 9 ---(b) ---(d) ---(a) Applying Kirchoff’s 2nd law for loop 1 Applying Kirchoff’s 2nd law for loop 2 ---(c) (d)-(c) we get Substitute I 1 in(a) Current in 18  resistor

Calculate the current in the network Current in 18  resistor Applying Kirchoff’s 2nd law for outside loop Applying Kirchoff’s 2nd law for loop 2 Current in 1  resistor

The network shown is a 3 cells having an internal resistance of 30 . Calculate the current in the network Applying Kirchoff’s 2nd law The voltage drop due to internal resistor is 0.05 x30=1.5V Thus there is no potential different between two terminals

Create loop’s current rather than branch current Use Kirchoff’s second (voltage ) law Ohm’s law to calculate p.d Branch is calculated by taking the algebraic sum of the loop currents

Calculate the current in each branch First create loop current,i.e I 1, I 2, I 3 as shown

continue ---(b) ---(a) In loop 1 In loop 2 ---(c) In loop 3

continue Solving these equations Current in 60  In direction of I 1 Current in 30  In direction of I 1 Current in 50  Current in 40  Current in 10  Current in 20  In direction of I 2 In direction of I 3

Choose reference node where all nodes can refer Assign currents going to/out the nodes Assign voltage at nodes as V 1, V 2, V 3 ….which refer to reference node Apply Kirchoff’s current law at each node Relate the voltage, resistance andcurrent using ohm’s law Solve the equations obtained

Simplified At node 1 ….(a) Calculate V 1 and V 2 At node 2

continue Simplified …..(b) Solve for equations (a) and (b) (b) X 21 (a) X 15 Substitute V 2 we have

Simplified At node 1 ….(a) Calculate V 1 and V 2 and current in 8 

continue Simplified At node 2 ….(b) Solving the simultaneous equations (a) and (b)

The superposition states that in any network containing more than one source, the current in, or the p.d. across in any branch can be found by considering each source separately and adding their effects: omitted sources of e.m.f are replaced by resistance equal to their internal resistances.

Separating the network into several circuit contenting only one source Original network Separating into 2 networks

Network 1 Total resistance Also Thus and

Network 2 Total resistance Also Thus and

combination Also Thus and

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