java class

MiMe

Lecture3-TheoremsTheveninNortonMaximumPower2.pdf

Home >Engineering homework help >java class

EEE: 2220

ELECTRICAL & ELECTRONIC

ENGINEERING PRINCIPLES

Lecture 3

LECTURE ROADMAP

• Circuit Analysis-Theorems

– Thevenin’s Theorem

– Norton’s Theorem

– Maximum Power Theorem

– Examples

Thevenin’s Theorem

• Thevenin’s theorem as applied to d.c. circuits is stated below :

• Any linear, bilateral network having terminals A and B can be replaced by a

single source of e.m.f. VTh in series with a single resistance RTh.

• (i) The e.m.f. VTh is the voltage obtained across terminals A and B with load,

if any removed i.e. it is open-circuited voltage between terminals A and B.

• (ii) The resistance RTh is the resistance of the network measured between

terminals A and B with load removed and sources of e.m.f. replaced by their

internal resistances. Ideal voltage sources are replaced with short circuits and

ideal current sources are replaced with open circuits. 3

Thevenin’s Theorem

• Consider the circuit shown in Fig. (i). As far as the circuit behind terminals AB

is concerned, it can be replaced by a single source of e.m.f. VTh in series with a

single resistance RTh as shown in Fig. b (ii).

• (i) Finding VTh. The e.m.f. VTh is the voltage across terminals AB with load

(i.e. RL) removed as shown in Fig. (ii).

• With RL disconnected, there is no current in R2 and VTh is the voltage appearing

across R3

• (ii) Finding RTh. To find RTh, remove the load RL and replace the battery by a

short-circuit because its internal resistance is assumed zero. Then resistance

between terminals A and B is equal to RTh as shown in Fig. 3.84 (i). Obviously,

at the terminals AB in Fig. 3.84 (i), R1 and R3 are in parallel and this parallel

combination is in series with R2.

Procedure for Finding Thevenin Equivalent Circuit

• (i) Open the two terminals (i.e., remove any load) between which you want to

find Thevenin equivalent circuit.

• (ii) Find the open-circuit voltage between the two open terminals. It is called

Thevenin voltage VTh.

• (iii) Determine the resistance between the two open terminals with all ideal

voltage sources shorted and all ideal current sources opened (a non-ideal source

is replaced by its internal resistance). It is called Thevenin resistance RTh.

• (iv) Connect VTh and RTh in series to produce Thevenin equivalent circuit

between the two terminals under consideration.

• (v) Place the load resistor removed in step (i) across the terminals of the

Thevenin equivalent circuit. The load current can now be calculated using only

Ohm’s law and it has the same value as the load current in the original circuit.

• Note. Thevenin’s theorem is sometimes called Helmholtz’s theorem

Practice Problems

• Use Thevenin's theorem to find the current flowing in the 10Ω resistor for the

circuit shown in Fig below.

Solution

Practice Problems

• Using Thevenin’s theorem, find the current in 6 Ω resistor in Fig.2 (i).

Solution

Practice Problem

• Using Thevenin’s theorem, find the current through resistance R connected

between points a and b in Fig.(i). [4.3A]

Solution

Solution…

Practice Question

• Calculate the power which would be dissipated in a 50 Ω resistor connected

across xy in the network shown in Fig. below. [2·08 W]

Advantages of Thevenin’s Theorem

• It reduces a complex circuit to a simple circuit viz. a single source of e.m.f. VTh in

series with a single resistance RTh.

• It greatly simplifies the portion of the circuit of lesser interest and enables us to

view the action of the output part directly.

• This theorem is particularly useful to find current in a particular branch of a

network as the resistance of that branch is varied while all other resistances and

sources remain constant.

• Thevenin’s theorem can be applied in successive steps. Any two points in a circuit

can be chosen and all the components to one side of these points can be reduced to

Thevenin’s equivalent circuit. 16

Norton’s Theorem

• Any linear, bilateral network having two terminals A and B can be replaced by

a current source of current output IN in parallel with a resistance RN.

a) The output IN of the current source is equal to the current that would flow

through AB when A and B are short-circuited.

b) The resistance RN is the resistance of the network measured between A and B

with load removed and the sources of e.m.f. replaced by their internal

resistances. Ideal voltage sources are replaced with short circuits and ideal

current sources are replaced with open circuits.

Procedure for Finding Norton Equivalent Circuit

• To determine the current flowing in a resistance R of a branch AB of an active

network:

i. Open the two terminals (i.e. remove any load) between which we want to find

Norton equivalent circuit.

ii. short-circuit branch AB and determine the short-circuit current ISC (IN) flowing in

the branch

iii. remove all sources of e.m.f. and replace them by their internal resistance (or, if a

current source exists, replace with an open-circuit), then determine the resistance r,

(RN) ‘looking-in’ at a break made between A and B

iv. Connect IN and RN in parallel to produce Norton equivalent circuit between the

two terminals under consideration.

v. Place the load resistor removed in step (i) across the terminals of the Norton

equivalent circuit. The load current can now be calculated by using current-divider

rule. 18

Practice Problems

• Use Norton’s theorem to determine the current flowing in the 10 Ω resistance

for the circuit shown in Fig (a) and in 4Ω in Fig (b)

Solution

Practice Question

• Two batteries, each of e.m.f. 12 V, are connected in parallel to supply a

resistive load of 0.5 Ω. The internal resistances of the batteries are 0.12 Ω

and 0.08 Ω. Calculate the current in the load and the current supplied by

each battery using Norton’s theorem.

Solution

Solution…

Practice

• Using Norton’s theorem, calculate the current in the 5 Ω resistor in the

circuit shown in Fig. [4A]

Practice

• Find Norton equivalent circuit for Fig below. Also solve for load current

and load voltage.[0.24A; 2.4V]

Maximum Power Transfer Theorem

• This theorem deals with transfer of maximum power from a source to load and may

be stated as under :

• In d.c. circuits, maximum power is transferred from a source to load when the load

resistance is made equal to the internal resistance of the source as viewed from the

load terminals with load removed and all e.m.f. sources replaced by their internal

resistances.

• According to maximum power transfer theorem, maximum power will be

transferred from the circuit to the load when RL is made equal to Ri, the Thevenin

resistance at terminals AB. 26

Proof of Maximum Power Transfer Theorem

Important Points

• The circuit efficiency at maximum power transfer is only 50% as one-half of the

total power generated is dissipated in the internal resistance Ri of the source.

• Under the conditions of maximum power transfer, the load voltage is one-half

of the open circuited voltage at the load terminals.

Practice Problem

• Find the value of the load resistor RL shown in Fig. that gives maximum power

dissipation and determine the value of this power.

Solution

Problem

• Calculate the value of R which will absorb maximum power from the circuit of

Fig.(i). Also find the value of maximum power.

Solution

Practice

• Find the value of RL in Fig. below necessary to obtain maximum power in RL.

Also find the maximum power in RL.[150Ω ; 1.042W]