Basic Electrical Circuits & Machines (EE-107)

Course TeacherShaheena Noor

Assistant ProfessorComputer Engineering Department

Sir Syed University of Engineering & Technology.

Useful Circuit analysis Techniques

The basic goals of this is learning methods of simplifying the analysis of more complicated circuits.

We are interested only in the detailed performance of an isolated portion of a complex circuit; a method of

replacing the remainder of the circuit by a greatly simplified equivalent is then very desirable.

Superposition

The superposition Principle• It states that “ the response (a desired current

or voltage) in a linear circuit having more than one independent source can be obtained by adding the responses caused by the separate independent sources acting alone”

Superposition

No voltage dro acrossterminals but currentcan flow

I

0 V

I

• A voltage source set to zero acts like a short circuit.

• A current source set to zero acts like an open circuit.

+ V -

+ V -

0 A

No current flows,but a voltage canappear across theterminals

Example 5.1 (page 104)• Use superposition to write an expression for

the unknown branch current ix.

is = 2AVs = 3V 9 Ohm

6 Ohm

ix

Source Transformations• A real voltage source can be converted to an

equivalent real current source and vice versa.• For Example:

R1

V I = ?

R1

2 Ohm

R210 V iL

• Compute the current through the 4.7kΩ resistor after transforming the 9mA source into an equivalent voltage source.

Example 5.4 (page 113)

9 mA5k Ohm

3 V

3k Ohm4.7k Ohm

I

• For the circuit, compute the voltage V across 1MΩ resistor using repeated source transformations.

Drill Problem 5.4 (page 115)

6M Ohm40µ A

+ V -

75 µA3 V

1M Ohm

200k Ohm4M Ohm

Thevenin’s Theorem

• It states that “ any linear circuit is equivalent to a single voltage source in series with a single resistance.”

Procedure:1. Open circuit the terminals with respect to which Thevenin

equivalent circuit is desired.

2. The Thevenin equivalent resistance RTH is the total resistance at the open circuited terminals when all voltage sources are replaced by short circuits and all current are replaced by open circuits.

3. The Thevenin equivalent voltage VTH (or ETH) is the voltage across the open circuited terminals.

4. Replace the original circuitry by its Thevenin equivalent circuit with the Thevenin terminals occupying the same position as the original terminal.

Thevenin’s Theorem

• Find the Thevenin equivalent to the left of terminal x – y

Thevenin’s Theorem (Example)

120 Ohm

x

y

100 Ohm60 Ohm

24 V50 Ohm

Drill Problem 5.6 (page 119)• Use Thevenin’s theorem to find the current

through 2Ω resistor.

9 V

5 Ohm4 Ohm

6 Ohm

4 Ohm

2 Ohm

I2Ω

Norton’s Theorem

• It states that “any linear circuit is equivalent to a real current source at a selected set of terminals.”

Procedure:• First find the Thevenin’s equivalent circuit and

then convert it to an equivalent current source.

Example• Find the Norton equivalent current source at

terminals x - y 20 Ohm 10 Ohm

30 Vx y

18 V

Drill Problem 5.5 (page 118)• Determine the Norton equivalent of the high

lighted network.

2 Ohm

8 Ohm

5 A 10 OhmRL

Maximum Power Transfer• An independent voltage source in series with

a resistance RS, or an independent current source in parallel with a resistance RS, delivers a maximum power to that load resistance RL for which RL = RS.

RL

VS

RS

+

VL

-

iL

A voltage source connected to a load resistor RL

Delta-Wye (∆ -Y) Conversion• Some electrical circuits have no components in

series and in parallel.• So they can not be reduced to simpler circuits

containing equivalent resistance of series or parallel combination.

• However in many cases it is possible to transform a portion of the circuit in such a way that the resulting configuration does contain series and parallel connected components.

Delta-Wye (∆ -Y) Conversion• The transformation produces an equivalent circuit

in the sense that voltages and current in the other (untransformed) components remain the same.

• Therefore, once the circuit has been transformed, voltages and current in the unaffected components can be determined using series-parallel analysis methods.

Delta-Wye (∆ -Y) Conversionb d

a c

RB

RCRA

RB

RCRA

a

c d

b

(a) ∏ network consisting of three resistors and three unique

connections

(b) Same network drawn as a Δ network

b d

a c

(c) A T network consisting of three resistors

(d) Same network drawn as a Y network

R1

R3

R2

R3

R2

R1

To convert from a Y network to a ∆ network, the new resistor values are calculated using the following relations:

• RA = R1R2 + R2R3 + R3R1

R2

• RB = R1R2 + R2R3 + R3R1

R3

• RC = R1R2 + R2R3 + R3R1

R1

Delta-Wye (∆ -Y) Conversion

To convert from a ∆ network to a Y network.• R1 = RARB

RA+ RB + RC

• R2 = RBRC

RA+ RB + RC

• R3 = RCRA

RA+ RB + RC

Delta-Wye (∆ -Y) Conversion

Drill Problem (page 129)• Use the technique of Y- Δ conversion to find

the Thevenin equivalent resistance of the circuit given below.

Each R is 10 Ω

Example (page 128)• Use the technique of Δ-Y conversion to find

the Thevenin equivalent resistance of the circuit given below.

1Ω 4Ω

3Ω

2Ω5Ω