1

EE211: NOTE # 7

Voltage Divider Rule (VDR) and Current Divider Rule (CDR)

A: Revisiting Resistors in Series and Parallel

EXAMPLE 7.1

2

SOLUTION

Mho or Siemen [S]

3

EXAMPLE 7.2

Remember that current flow in relation to voltage a passive element is defined

as

(i : is positive) ( i :is negative)

4

I. VOLTAGE DIVIDER RULE (VDR) is used to compute voltage across ith passive

element in N series connected passive elements.

EXAMPLE 7.3

For the circuit shown below:

5

Solution:

EXAMPLE 7.4

II. CURRENT DIVIDER RULE (CDR) is used to calculate current flowing in the kth

passive element of N passive elements connected in parallel.

6

Consider a special case of R1 & R2 in parallel as shown below:

I1 I2

I

IN

G1

G2 GN

7

EXAMPLE 7.5

Note that it easier to convert resistances into conductances for the calculation

of currents flowing in parallel combination of resistances if the parallel

elements are more than two.

Example 7.5

8

Solution:

Repeat solution with conductances instead of resistances in the calculations of

currents flowing in the resistors.

9

EXERCISES

1.

Answer: i = 0.2A & i2 = 0.6A

2.

Answers: &

3. Determine current I1 and I2 in the circuit.

Answer: I1 =-0.098A & I2 = 0.031A

10

EE211: NOTE # 8 SOURCE TRANSFORMATION & SUPERPOSITION PRINCIPLE

8.1 SOURCE TRANSFORMATION (ST)

This involves transforming an ideal voltage source, Vs in series with resistor,

Rs into an ideal current source, Is in parallel with resistor Rp or vice versa.

Connecting same load resistor RL across terminals a – b of circuits 1 and 2 then

the circuits are said to be equivalent as seen from terminals if and only if I1=I2

and V1 = V2. Applying Ohm’s law and CDR to circuits 1 &2 yields the following:

Example 8.1

Convert the following circuits using source transformation (ST).

Solution

11

For circuit 1, and

Thus:

For circuit 2, and

Example 8.2

Convert the following circuits using source transformation.

Solution:

For circuit 1:

12

For circuit 2:

Cases for which source transformation do not apply are:

1. Ideal voltage source in parallel with resistor; and

2. Ideal current source in series with resistor.

Their respective equivalences are summarised as follows.

Example 8.3

Use source transformation to calculate i1 and i2 in the following circuit.

13

Solution

Apply source transformation to 8V source in series with 2Ω into current

source of (8V/2Ω) = 4A in parallel with 2Ω.

Combine the resulting two current sources into an equivalent current source

of 6A in parallel with two resistors of 1Ω and 2Ω also in parallel. Apply CDR

to determine current through 1Ω.

i2 = 6x 1/(1+0.5) = 4A ; Apply CDR to the original circuit to

calculate i1 thus: -i2 + i1 + 2=0 i1=4-2 = 2A.

As an exercise solve this problem by applying ST to the current source in

parallel with 1Ω.

Exercises

1.

2. Using ST, calculate i1 and i2 in the following circuit.

Answer : V =3.232V

Answer: i1 = 1.277A & i2 = 0.615A

14

8.2 SUPERPOSITION PRINCIPLE

The superposition principle (SP) is a very useful tool for analyzing linear circuits

with several active sources. In general, a circuit can be represented by a block

diagram having multiple inputs and multiple outputs as shown below.

Where

S1, S2, S3 ...SN denote inputs representing either independent voltage sources or

current sources; and

U1, U2, U3 ...UM denote outputs comprising voltages across passive elements

and/or currents flowing in various elements. Without any loss of generality, let

us consider the special case of multi-input single-output (MISO) circuit in which

all the inputs contribute to the single output as shown in this figure.

The superposition principle states that if the contribution of S1 to the output is

U1 with all other inputs removed, U2 is contributed to the output by S2 with all

Multi-Input

Multi-Output

(MIMO)

Linear Circuit

Multi-Input

Single-Output

(MISO)

linear Circuit

15

other inputs removed and so on until the last contribution of UN by SN then the

single output U with all inputs S1, S2, S3...SN is given by:

U = U1 +U2 +U3.....+UN

The same concept can also be extended to multi-outputs. Note that the

Superposition is applicable only to calculations of voltages and currents with

respect to linear circuits. To apply superposition principle to a given linear

with several input sources, the removal of source follows this procedure;

1. To remove a voltage source simply replace it with a short circuit branch;

2. To remove a current source, replace it with an open circuit branch as

illustrated in the following figures.

The following examples illustrate the application of superposition principle to

circuit analysis.

Example 8.4

Apply superposition principle to determine I in the following circuit.

Solution

The block diagram for the circuit comprises two-input (1 voltage source + 1

current source) and single output (current flowing in 10Ω resistor).

Voltage source Current source

Short

circuit

Open

Circuit

16

Apply Ohm’s to calculate I’ with respect to (wrt) voltage source only circuit and

apply CDR wrt to current source only thus:

I = I’ + I’’ =0.25A + 0.75A = 1.00A

Example 8.5

Using Superposition principle, calculate I1 and I2 in the following circuit.

Solution

The block diagram corresponding to the circuit is shown below.

Apply superposition principle to the circuit as follows:

= +

Voltage Source only Current source only

I1

↓

I2 →

2A

8V

3-Input 2-output

linear circuit 4A

I1

I2

= + +

17

Apply Ohm’s law to Circuit 1 yields: I1’ = 8/ (10+6) = 0.5A; I2

’ = 0A;

Apply CDR to circuit 2 yields: I1’’ = -4x6/ (10+6) =-1.5A; I2

’’ = 4A; and

Apply CDR to circuit 3 yields: I1’’’ = 2X6/(10+6) = 0.75A; I2

’’ = -2A.

Now apply SP as follows:

I1 = I1’ +I1’’ +I1’’’ = 0.5 + (-1.5) + 0.75 = -0.25A

I2 = I2’ +I2’’ +I2’’’ = 0.0+4+ (-2) = 2A

Exercises

1. Apply Superposition principle to determine currents I1 and I2 in the

following circuit.

Answer: I1= -0.314A & I2 = 2.468A

2. Using Superposition principle, to determine V and I shown in the following

circuit.

EE211: NOTE # 9

I1

I2

+

V

_

I

Answer: V=6⅔V ; I=-2/9A

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THEVENIN’S & NORTON’S THEOREMS AND MAXIMUM

POWER TRANSFER

9.1 Thevenin and Norton Equivalent Circuits (TEC & NEC)

An electrical circuit can, in general, be split into two sub-circuits labelled A

and B and linked via a-b terminals as shown in Figure 9.1. It is desired to

replace A as seen from a-b terminals with either Thevenin Equivalent

Circuit (TEC) or Norton Equivalent Circuit (NEC). The development of any

of these equivalent circuits is possible for a linear circuit composed of

active ideal voltage or current sources and passive elements as formally

presented by the following theorems.

Fig. 9.1(a) Development of Thevenin Equivalent Circuit (TEC)

Fig. 9.1(b) Development of Norton Equivalent Circuit (NEC)

A. Definition of Thevenin’s Theorem:

It states that any linear passive or active two-terminal network may be

modelled by an ideal voltage in series with a resistor.

Original circuit Split into two sub-circuits Replacement of A with (NEC)

Original circuit Split into two sub-circuits Replacement of A with (TEC)

Sub –

Circuit

(A)

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B. Definition of Norton’s Theorem:

It states that any linear passive or active two-terminal network may be

modelled by an ideal current in parallel with a resistor.

Figure 9.2:

Vth : Is Thevenin voltage source;

Rth : Is Thevenin resistance;

Isc : Is Norton Current source; and

Rn: Is Norton Parallel Resistance.

Recall that from Source Transformation earlier discussed:

If Rn = Rth then

.

This implies that if the Thevenin equivalent circuit is determined for a given

linear network, the Norton equivalent circuit is known via source

transformation (ST) or vice versa.

The task is to determine Vth and Rth for Thevenin equivalent circuit or Isc and

Rn for Norton equivalent circuit. There are several procedures available for

achieving this task but only two procedures will be taught in this course.

Procedure 1:

Step 1: Remove the load (B) such that a-b terminals constitute open circuit

looking into sub-circuit (A).

Step 2: Calculate Voc across terminals (a-b) using a combination of circuit

laws which is equal to Vth.

(ST)

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Step 3 : Short circuit terminals a-b and use superposition principle as well as

circuit laws (KVL, KCL, Ohm’s law) to determine Isc. Calculate Rth as

follows: Rth = Voc/Isc.

Procedure 2:

Step 1: Same as procedure 1.

Step 2: Same as procedure 1.

Step 3: Replace all active sources of sub-circuit (A) with their respective

internal resistances (short circuits for all voltage sources and open

determine equivalent resistance, Rth seen between a-b terminals.

Example 9.1

Determine the Thevenin equivalent circuit for the following linear circuit.

Solution: Use Procedure 1

Note the circuit given has already removed the load.

Applying superposition principle to the circuit (remove 3A source then put back current source and remove 25V source) yields: Voc = 25x (20/25) + 20x (5/25) x3 = 32V.

Calculate Isc = (3x(¼))/((1/5)+(1/20)+(1/4))+5x(0.25/0.5)=4A

Rth = Voc/Isc = 32/4=8Ω

Thevenin Equivalent Circuit

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Example 9.2

Determine Norton equivalent circuit for the following circuit with respect to

terminals a-b.

Solution:

Apply ST to 9A//12Ω→84V in series with 12Ω and apply again ST to 108V in

series with (4+12+2) Ω→108/18 A =6A//18Ω →Isc=6A & Rn =18Ω//9Ω=6Ω.

In = 6A & Rn = 6Ω.

C. Maximum Power Transfer Theorem:

It states that an independent source in series with a resistance, Rth or an

independent current source in parallel with Rth , delivers maximum power to a

load RL , if and only if RL = Rth . As corollary the following can also be deduced.

Maximum voltage transfer occurs if and only if RL Rth.

Maximum current transfer occurs if and only if Rth RL.

Exercise

1. Determine the Thevenin and Norton equivalent circuits for the following

circuit with respect to terminals a-b and hence calculate the current i.

What is the power delivered to the 12Ω? What is the maximum power

transferrable?

Answer TEC: Rth = 4Ω & Vth = -8V; NEC: Rn = 4Ω & Isc = -2A;

P 12Ω =3W; Pmax = 4W.

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EE211: NOTE # 10

MESH AND NODAL ANALYSES APPLIED TO LINEAR NETWORKS

10.1 MESH ANALYSIS

Recall from earlier lectures that mesh was defined as the simplest type of loop

comprising a number of connected branches which form a closed path. In

addition, circuit graph, graph tree, trivial and non-trivial loops were

introduced. For a given network, the number of meshes needed to determine

currents flowing in all branches equal the number of branches removed to

create its graph tree. For the sake of clarity, this is illustrated again by the

following sample network and its corresponding circuit graph. It is possible to

select the number of meshes by choosing all non overlapping closed paths.

The procedure for applying mesh analysis is summarized below:

1. For a given circuit, establish all the non overlapping meshes and assign

directions (clockwise (CW) or counter clockwise (CCW)) to all labelled

circulating mesh currents as shown in the figure below.

Circuit Graph =Graph tree +

removed branches (graph co-tree)

1 2 3

4

Graph tree branches=4

Removed branches=4

Number of meshes=4 (2 trivial +2 non-trivial)

Sample Network

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2. Apply KVL to each mesh and observe consistently the KVL convention with

respect to voltage drop across each passive element. Where there are

shared current sources in any mesh closed path, assign to them unknown

voltages across them and apply KCL to each current source.

3. Solve the resulting set of linear equations for the unknown mesh currents.

We will now demonstrate the application of mesh analysis to solve circuit

problems in what follows.

Example 10.1

Calculate mesh currents i1 and i2 shown in the following circuit.

Solution

Apply KVL around the closed path of mesh 1 -10+2xi1+6x (i1-i2) =0

Simplify 8i1 - 6i2 = 10 (1)

Apply KVL around the closed path of mesh 2 6x (i2-i1) +4xi2-5=0

Simplify 10i2 - 6i1 = 5 (2)

Solve equations (1) & (2) yield i1=2.96A and i2 = 2.27A

Example 10.2

Calculate mesh currents i1, i2 and i3 of the following circuit.

Solution:

Mesh 1: 4xi1-8+6x (i1-i2) = 0 10i1-6i2 =8 (1)

Mesh 2: 6x (i2-i1) +8x (i2-i3) +12 = 0 -6i1+14i2-8i3 =-12 (2)

24

Mesh 3: 8+ 2xi3+8x (i3-i2) = 0 -8i2+10i3 = -8 (3)

Solving equations 1-3 yield i1 = -1.24A, i2 = -3.40A & i3 = -3.52A

Example 10.3

Calculate mesh currents i1, i2 and i3 of the following circuit which comprises a

voltage source and current source.

Solution

Mesh 1: i1+2x (i1-i3) +v1 =0 3i1-2i3 = -v1 (1)

Mesh 2: 4x (i2-i3) +5xi2-v1=0 9i2-4i3 = v1 (2)

Meshes 1&2: i2-i1 = 3 (3)

Mesh 3: 4x (i3-i2) +2x (i3-i1) +4+3xi3=0 -2i1-4i2+9i3 = -4 (4)

Equation1+Equation 2 3i1+9i2-6i3 =0 i1+3i2-2i3= 0 (5)

Solving eqns 3 to 5 yield i1=-2.708A i2 = 0.292A & i3 = -0.917A

10.2 NODAL ANALYSIS

The nodal analysis is used to determine the voltage at node relative to the

reference node (or ground). The procedure of nodal analysis is summarised as

follows:

1. Identify all the nodes (trivial and non-trivial) with respect to an electrical

circuit and number them sequentially with unknown voltages: V1, V2, V3...VN.

Select a reference node amongst the nodes identified and assign it 0V.

2. Apply KCL at each node which does not include the reference node. Simplify

each resulting equation. Where a node is connected directly to a voltage

source onto the reference node; the node voltage is that of the source

voltage. Current sources connected to nodes are easily amendable to KCL.

3. Collect all equations for the nodes and solve the set of linear equations. The

number of equations must equal the remaining of unknown node voltages.

25

Example 10.4

Calculate the nodal voltages V1 and V2 of the following circuit.

Solution:

Node 1: Apply KCL:

)V1 -

V2 = 3 (1)

Node 2: Apply KCL:

V1 +(

)V2 =-1 (2)

Solving equations 1 &2 yield V1 = 4.25V & V2 = -1V

Example 10.5

Calculate nodal voltages V1, V2 and V3 shown in the following circuit.

Solution:

Node 1: Apply KCL

V1 - ⅟2V3 = -2 (1)

Node 2: Apply KCL

V2 -⅟4V3 = 2 (2)

Node 3: Node 3 connected to voltage source onto ground. V3 = 5V (3)

Solving the eqns 1, 2 & 3 yield: V1 = 0.6V; V2 =2.6V& V3 = 5V

Reference node symbol

26

Example 10.6

Calculate the nodal voltages V1, V2 &V3 shown in the following circuit with

floating voltage source (i.e. voltage source not connected to the reference

node).

Solution

Node 1: Apply KCL

V1 -

V2 = i (1)

Node 2: Apply KCL

V2 -

V3 = -i (2)

Node 3: Apply KCL

V1 -

V2 + (

)V3 = -3 (3)

Floating voltage source: Apply KVL V1 – V2 = 2 (4)

Eliminate i from eqns 1 & 2 to yield

V1 +

V2 -

V3 = 0 (5)

Solve eqns 3, 4 & 5 to yield: V1= 0.25V; V2 = -1.75V & V3 = -4.4167V

EXERCISES ON MESH & NODAL ANALYSES

1. Calculate the mesh currents, i1, i2 & i3 shown in the following circuit.

Answer: i1= 0.955A , i2= 1.864A & i3 =0.591A

I

4V

10V 8

4

4

5

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2. Determine the mesh currents shown in the following circuits.

Answer: i1=-⅓ A , i2= -3⅓A & i3 =-4⅓ A

3. Determine the unknown nodal voltages shown in the following circuit.

Answer: V1=25V , V2=6V & V3 =1V

4. Calculate the unknown nodal voltages shown in the following circuit.

Answer: V1=

, V2=

& V3 =

4A

10V

10 Ω

5 2

5 5 2

10V

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EE211: NOTE # 11

DEPENDENT SOURCES

DEFINITIONS OF DEPENDENT SOURCES

A dependent voltage or current source is one whose value is determined by

voltages or currents elsewhere in the circuit. Dependent sources are

commonly used to model the behaviour of transistors and op-amps which will

be taught in higher level courses. The symbols for dependent sources are

shown in the following figures.

(i) Dependent Voltage Source Symbol (ii) Dependent Current Source Symbol

The presence of any dependent sources in a given circuit can be treated in

similar manner as done for independent sources using all circuit laws and

network analyses so far presented. The only exception is in the application of

superposition principle wherein all dependent sources are left untouched.

Some illustrative examples will now be presented on circuits with dependent

sources.

Example 11.1

Consider the circuit shown in the figure below in which the dependent (or

controlled) current source derives from the current i1 flowing in the 5Ω

resistor. This is known as current dependent current source with specific

value of 4i1. Note that the constant 4, which is the same as β in the symbol

definition, is dimensionless.

i1 + 4i1 i2

2A 5Ω 4Ω

-

Circuit with current dependent current source

+

-

29

Applying KCL to the circuit yields: 2 + 4i1 = i1 + i2 → 2 = -3i1 + i2 (1) Apply Ohm’s law to the circuit yields: i1 = ѵ/5; i2 = ѵ/4 (2)

Substituting eqn(1) into eqn(2) yields 2 =

→ ѵ =

(3)

i1 =

& i2 =

Example 11.2

The circuit shown below contains a dependent voltage source. Apply mesh analysis to determine the mesh currents i1, i2 and i3.

4Ω 2Ω 1Ω

- ѵ +

1.5V I1 1Ω I2 5Ω I3 2ѵ V

Solution

Apply KVL around mesh I1 4i1 + 1(i1-i2) – 1.5 = 0 5i1 –i2 =1.5 (1)

Apply KVL around mesh I2 2I1 + 5(I2-I3) + 1(I2-I1) = 0 -I1+8I2-5i3 = 0 (2)

Apply KVL around mesh I3 1I3 + 2ѵ+5(I3-I2)=0 -8I1-5I2+6I3=0; Note ѵ=-4I1 (3)

Put eqns 1-3 in matrix form:

Apply Cramer’s rule,

∆=69; ∆1 =34.5; ∆2 = 69 & ∆3 = 103.5 I1 =∆1/∆ =34.5/69= 0.5A; I2 = ∆2/∆ =69/69 = 1A & I3 = ∆3/∆ = 103.5/69 = 1.5A.

Example 11.3

Determine the node voltages ѵ1, ѵ2 and ѵ3 shown in the circuit below:

4Ω

1Ω 2i A

2A 1A 3Ω 2Ω

+

-

+

- +

-

+

-

i Ѵ3

Ѵ2 Ѵ1

30

Solution

Apply KCL to node 1: V1/2 +(V1-V2)/1 +(V1-V3)/4 -2 =0 1.75V1-V2-0.25V3=2 (1) Apply KCL to node 2: V2/3+(V2-V1)/1=2i=2(V1-V2)/1 3V1-10V2/3=0 (2) Apply KCL to node 3: (V3-V1)/4 +2i-1=0 1.75V1 -2V2+0.25V3 =1 (3) From eqn 2: V1 = 10V2/9 Adding eqns 1 & 3 yield: 3.5V1-3V2 = 3 V2=27/8V ; V1=15/4V & V3 = 4((15/4)(7/4)-27/8-2)=19/4V

V1=15/4V; V2= 27/8V & V3 = 19/4V

Example 11.4

Determine the Thevenin and Norton equivalents for the circuit shown below.

(i) Apply ST to Dependent current source to obtain (ii)

Solution:

Apply KVL to mesh 1: -3 + I1 +5I1 I1 = 0.5A ѵ1 = 0.5x5=2.5V (1) Apply KVL to mesh 2: I2 = 0 - ѵ1 + 4ѵ1 +Voc = 0 Voc =-3ѵ1 = -7.5V (2)

Apply short circuit between terminals a & b.

Now apply KVL to mesh 1: -3+I1+5(I1-Ish) = 0; ѵ1 = 5(I1 – Ish); (3)

Apply KVL to mesh 2: 5(Ish-I1) + 4ѵ1+ 5Ish =0 5(Ish-I1) + 20(I1-Ish)+ 5Ish =0 (4)

Solving eqns 3 & 4 yield: Ish = 1.5I1; I1 = -2A; Ish = -3A ;

Rth = Voc/Ish = (-7.5/-3)Ω= 2.5Ω.

Thevenin Equivalent Norton Equivalent

2 1

b

a

+

-

Ѵ1 3V +

- 5Ω

2ѵ1 A

1Ω 3Ω

2Ω

Voc I1

+ - 2 1

b

a

+

-

Ѵ1 3V +

- 5Ω

4ѵ1 V

AV

1Ω 3Ω 2Ω

VOC

I2

I1

+ - 2 1

b

a

+

-

Ѵ1 3V +

- 5Ω

4ѵ1 V

AV

1Ω 3Ω 2Ω

Ish

2.5Ω

7.5V -

+ 2.5Ω

3A