Outline

• Resistances in Series and Parallel• Network Analysis by Using Series and Parallel Equivalents• Voltage-divider and Current-Divider Circuits• Node-Voltage Analysis• Mesh-Current Analysis•Thevenin and Norton Equivalent Circuits

In a parallel ckt, the voltage across each element is the same.Applying Ohm’s law, we can write

vRRR

i

R

v

R

v

R

viiii

R

vi

R

vi

R

vi

)111

(

,,

321

321321

33

22

11

For your own practice!

Voltage-divider and current-divider ckts

When a voltage is applied to a seriescombination of resistances, a fractionof a voltage appears across each ofthe resistances.

total

total

total

total

eq

total

vRRR

RiRv

vRRR

RiRv

vRRR

RiRv

RRR

v

R

vi

321

333

321

222

321

111

321

Voltage Division

Current DivisionThe total current flowing intoa parallel combination of resistances divides, and a fraction of the total current flows through each resistance.

total

total

totaltotaleq

iRR

R

R

vi

iRR

R

R

vi

iRR

RRiRv

21

1

22

21

2

11

21

21

Node-voltage analysis

Although they are very important concepts, series/parallelequivalents and the current/voltage division principles are notsufficient to solve all ckt problems. That’s how the node-voltageanalysis comes in.

A node is a point at which two or more ckt elements are joined together. In node-voltage analysis, we first select one of the nodesas the reference node. In principle, any node can be picked to be the reference node.

Next, we label the voltages at each of the other nodes.

Write equations to solve for the voltages, then the current.

Note: the positive reference is at the head of the arrow.

Writing KCL equations in terms of the node voltage:to find the current flowing out of node n through a resistance toward node k, we subtract the voltage at node k from the voltageat node n and divide the difference by the resistance. For example,If vn and vk are the node voltages and R is the resistance connectedbetween the nodes, the current flowing from node n toward node kis given by

R

vv kn

Apply KCL at node 2, we get 03

32

4

2

2

12

R

vv

R

v

R

vv

Apply KCL at node 3, we get 03

23

5

3

1

13

R

vv

R

v

R

vv

Thevenin and Norton Equivalent ckts

In this section, we learn how to replace two-terminal ckts Containing resistances and sources by simple equivalent ckts.

Thevenin Equivalent circuits

One type of equivalent ckts is the Thevenin equivalent, which Consists of an independent voltage source in series with a resistance.

Thevenizing Procedure

1. Calculate the open-ckt voltage (Vth)across the network terminals.2. Redraw the network with each independent source replaced by its

internal resistance. This is called “deactivation of the sources”.3. Calculate the resistance (RTH ) of the redrawn network as seenfrom the output terminals.

Different methods of finding RTH

(a) For independent sources: Deactivate the sources, i.e. for independent current source, deactivate it by open circuiting itsterminals and for voltage source, deactivate it by shorting it. To In case of non-ideal sources, the internal resistance will remainConnected across the deactivated source terminals.

(b) For dependent sources in addition or in absence of independentsource.(i) Find open circuit voltage Voc across the open circuited loadterminals. Next short circuit the load terminals and find the shortckt current (Isc) through the shorted terminals.The Thevenin’s equivalent resistance is then obtained as RTh = Voc/Isc

Norton’s Theorem

According to this theorem, any two-terminal active network, containing voltage sources and resistance when viewed from its output terminals is equivalent to a constant current source and an internal (parallel) resistance. The constant current source (known as Norton’s equivalent currentsource) is of the magnitude of the short circuit current at theterminals.

Nortonizing Procedure

1. Remove RL, and short circuit the terminals a and b. Thecurrent through the short circuited path is Isc or In.

2. Finding internal resistance of the network by deactivatingall the sources and replace them by its internal resistance. ThenCalculate RN of the redrawn network as seen from the outputTerminals.3. Draw the Norton’s equivalent ckt.

Source Transformation

A voltage source in series with a resistance is externallyEquivalent to a current source in parallel with the resistance,Provided that In = Vt/Rt

Maximum Power Transfer

Suppose that we have a two-terminal ckt and we want to connect aLoad resistance RL such that the maximum possible power is Delivered to the load.

To analyze this problem, we replace the original ckt by its TheveninEquivalent as shown

tL

Lt

LtLtLtt

L

L

Lt

LtL

LLL

Lt

tL

RR

RR

RRRVRRV

dR

dp

RR

RVp

Rip

RR

vi

0)(

)(2)(

)(

4

222

2

2

2

Thus, the load resistance that absorbs the maximum power from aTwo-terminal ckt is equal to the Thevenin resistance.

The maximum power is given by

t

tL R

VP

4

2

max