Basic Electrical Engineering

Lecture # 05 & 06

Course Instructor:Engr. Sana Ziafat

Agenda

•Voltage division•Current division•Wheatstone bridge

Voltage Divider Circuit

•Developing more than one voltage level from a single voltage supply

Identify the current and Apply KVL

1 2 1 2

1 2

11 1

1 2

22 2

1 2

( )S

S

S

S

v iR iR i R R

viR R

Rv iR v

R R

Rv iR v

R R

Connect a “Load” Resistor in Parallel

1

2

2

2

21 21

eqO S

eq

Leq

L

O S

L

Rv v

R R

R RR

R R

Rv v

RR R

R

Determine vo

2

1 2

,L

O

R

Rv

R R

The Voltage Divider Rule•Voltage dropped across each resistor may

be determined by the voltage across any other resistor (or combination of resistors) by using the voltage divider rule expressed as:

•The subscripts must match (x and y)

yy

x

x VR

RV

Voltage Divider Rule Application

•If a single resistor is very large compared to the other series resistors, the voltage across that resistor will be the source voltage

•If the resistor is very small, the voltage across it will be essentially zero

Voltage Divider Rule Application•If a resistor is more than 100 times larger

than another resistor▫Smaller resistor can be neglected

Voltage Division

•Voltage drop across a single resistor is proportional to the voltage drop across series connected resistance

•Constant of proportionality is ratio of specific resistance to equivalent resistance.

Current division

•Current at specific resistance is proportional to the total current applied to set of parallel connected resistance.

•Constant of proportionality is ratio of equivalent resistance to parallel resistance.

3 bulb question

The circuit above shows three identical light bulbs attached to an ideal battery. If the bulb#2 burns out, which of the following will occur?

a) Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases.b) Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged.c) Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases.d) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is

unchanged.e) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is

unchanged.f) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit

decreases.g) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit

decreases.h) Bulb 1 is unaffected, but bulb 3 gets brighter. The total light emitted by the circuit

increases.i) None of the above.

When the bulb #2 is not burnt out:

R23

2RRR eq

R3V2

RVI

231

RIP,Power 2RVI

For Bulb #1

For Bulb #2

For Bulb #3

RV44.

R9V4RIP

22211

R3V

2I

I 12

RV11.

R9VRIP

22222

R3V

2I

I 13 R

V11.R9

VRIP22

233

1I

2I

I 12 2

II 1

3

R2RRR eq

R2VI1

RIP,Power 2RVI

For Bulb #1

For Bulb #2

For Bulb #3

RV25.

R4VRIP

22211

0I2 0RIP 222

R2VII 13 R

V25.R4

VRIP22

233

1I

13 II

So, Bulb #1 gets dimmer and bulb #3 gets brighter. And the total power decreases.

f) is the answer.

Before total power wasRV66.

RV

RVP

2

23

2

eq

2

b

RV50.

R2V

RVP

22

eq

2

a After total power is

When the bulb #2 is burnt out:

Measuring Resistance

Classification of Resistance

•Low resistance is the range of .1 to 1 ohm.

•Medium resistance is the range of 1 to low megha ohm.

•High resistance is .1M to higher range.

Wheatstone BridgeAn Overview

History

•The Wheatstone Bridge was invented in 1833 by Samuel Hunter Christie

•Later named after Sir Charles Wheatstone for his many applications of the circuit through the 1840s

•The most common procedure for the bridge remains the testing of unknown electrical resistance

How Does it Work?

•Uses ratio of 3 known resistors•Measures fourth unknown resistance

How Does it Work? (cont.)

•By changing resistors to adjusting variable resistors to balance the device, the mathematical ratio is used to calculate the fourth (unknown) resistance

Impact of the Wheatstone Bridge•The Wheatstone Bridge is a very simple

design, although there are more complex versions of achieving the same outcome

•Can be adjusted easily•Fairly inexpensive to produce•Also indirectly measures any variable that

would change the resistance of a material▫Ex: temperature, force, pressure

Little about Instruments………..Galvanometers: a coil in a magnetic field that senses current.

Ammeters: measures current.

Voltmeter: measures voltage.

Ohmmeters: measures resistance.

Multimeters: one device that does all the above.

Galvanometer is a needle mounted to a coil that rotates in a magnetic field. The amount of rotation is proportional to the current that flows

through the coil.

Symbolically we write

gRUsually when 20Rg

milliAmp5.00Ig

Wheatstone Bridge

Analysis

•Identify the currents

Consider the bridge at “balance”, ig=0

1 3

2

3 3

1 1 2 2

x

x x

i i

i i

i R i R

i R i R

Some Algebra1 3 2

1 1 2 2

3

1 2

23

1

x

x

x

i R i R

i R i R

R R

R R

RR R

R

Use to Measure Resistance

23

1x

RR R

R

Ratio Arms

Review of few last slides

• Created in 1833, popularized in 1840s• Wheatstone bridges are one of the best methods of

measuring resistance due to the basic mathematical ratio involved.

• Accurate standards with sensitive enough voltmeter, measurements of resistance within .05% can be reached.

• Many calibration laboratories still use this method today.• The Wheatstone Bridge are replaceable; however, for its

simplicity and versatility the circuit is an indispensible piece of technology

Wye – Delta TransformationsAlso known as T- transformations

Delta-to-Wye Transformations• Delta-to-wye, or wye-to-delta are also

sometimes called pi-to-tee or tee-to-pi transformations.

• These are equivalent circuit pairs. They apply for parts of circuits that have three terminals. Each version of the equivalent circuit has three resistors.

WHY Wye to Delta Transformation?????????

9

1 0 5

8 4

V+_

R eq 1 0

I

We cannot use resistors in parallel. We cannot useresistors in series. If we knew V and I we could solve

Req =V

I

There is another way to solve the problem without solvingfor I (given, assume, V) and calculating Req for V/I.

Delta-to-Wye TransformationsThree resistors in a part of a circuit with three terminals can be

replaced with another version, also with three resistors. The two versions are shown here. Note that none of these resistors is in series with any other resistor, nor in parallel with any other resistor. The three terminals in this example are labeled A, B, and C.

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

Notes on Names

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

When we go from the delta connection (on the left) to the wye connection (on the right), we call this the delta-to-wye transformation. Going in the other direction is called the wye-to-delta transformation. One can go in either direction, as needed. These are equivalent circuits.

•Each resistor in the Y network is the product of the resistors in the two adjacent ∆ branches, divided by the sum of the three ∆ resistors.

•Each resistor in the ∆ network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor.

Delta-to-Wye Transformation EquationsWhen we perform the delta-to-wye

transformation (going from left to right) we use the equations given below.

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

1

2

3

B C

A B C

A C

A B C

A B

A B C

R RR

R R R

R RR

R R R

R RR

R R R

Wye-to-Delta Transformation EquationsWhen we perform the wye-to-delta transformation

(going from right to left) we use the equations given below.

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

1 2 2 3 1 3

1

1 2 2 3 1 3

2

1 2 2 3 1 3

3

A

B

C

R R R R R RR

R

R R R R R RR

R

R R R R R RR

R

Why Are Delta-to-Wye Transformations Needed?• This is a good question. In fact, it should be pointed

out that these transformations are not necessary. Rather, they are like many other aspects of circuit analysis in that they allow us to solve circuits more quickly and more easily. They are used in cases where the resistors are neither in series nor parallel, so to simplify the circuit requires something more.

• One key in applying these equivalents is to get the proper resistors in the proper place in the equivalents and equations. We recommend that youname the terminals each time, on the circuit diagrams, to help you get these things in the right places.

Simplification

•If R1 = R2 = R3 = R, then Ra = Rb =Rc = 3R

•If Ra = Rb = Rc = R’, then R1 = R2 = R3 = R’/3

Readings

•Chapter 3: 3.4, 3.5, 3.6, 3.7 (Electric Circuits)▫By James W. Nilson

Quiz

1. The total resistance of parallel resistors is equal to

a. the sum of the resistances

b. the sum of the reciprocals of the resistances

c. the sum of the conductances

d. none of the above

Quiz

2. The number of nodes in a parallel circuit is

a. one

b. two

c. three

d. can be any number

Quiz

3. The total resistance of the parallel resistors is

a. 2.52 k

b. 3.35 k

c. 5.1 k

d. 25.1 k

R1 R2 R310 k 5.1 k10 k

Quiz

4. If three equal resistors are in parallel, the total resistance is

a. one third the value of one resistor

b. the same as one resistor

c. three times the value of one resistor

d. there is not enough information to say

Quiz

5. In any circuit the total current entering a junction is

a. less than the total current leaving the junction

b. equal to the total current leaving the junction

c. greater than the total current leaving the junction

d. can be any of the above, depending on the circuit

Quiz

6. The current divider formula to find I1 for the special case of two resistors is

a.

b.

c.

d.

11 T

T

RI I

R

21 T

T

RI I

R

11 T

1 2

RI I

R R

21 T

1 2

RI I

R R

Quiz

7. The total current leaving the source is

a. 1.0 mA

b. 1.2 mA

c. 6.0 mA

d. 7.2 mA

VS R1+ R2

2.0 k10 k12 V

Quiz

8. The current in R1 is

a. 6.7 mA

b. 13.3 mA

c. 20 mA

d. 26.7 mA

R1 R2

200 100 I = 20 mA

Quiz

9. The voltage across R2 is

a. 0 V

b. 0.67 V

c. 1.33 V

d. 4.0 V

R1 R2

200 100 I = 20 mA

Quiz

10. The total power dissipated in a parallel circuit is equal to the

a. power in the largest resistor

b. power in the smallest resistor

c. average of the power in all resistors

d. sum of the power in all resistors

Quiz

Answers:

1. d

2. b

3. a

4. a

5. b

6. c

7. d

8. b

9. c

10. d

Q & A