Transcript View File - University of Engineering and Technology, Taxila

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
vS  iR1  iR2  i ( R1  R2 )
vS
i
R1  R2
R1
v1  iR1  vS
R1  R2
R2
v2  iR2  vS
R1  R2
Connect a “Load” Resistor in Parallel
Determine vo
vO 
Req
vS
R1  Req
R2 RL
Req 
R2  RL
vO 
R2
  R2
R1 1  
  RL

   R2

vS
vO 
Req
vS
R1  Req
R2 RL
Req 
R2  RL
vO 
R2
  R2
R1 1  
  RL

   R2

vS
RL  ,
R2
vO 
R1  R2
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:
Rx
Vx 
Vy
Ry
• The subscripts must match (x and y)
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)
b)
c)
d)
e)
f)
g)
h)
i)
Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases.
Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged.
Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases.
Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is
unchanged.
Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is
unchanged.
Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit
decreases.
Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit
decreases.
Bulb 1 is unaffected, but bulb 3 gets brighter. The total light emitted by the circuit
increases.
None of the above.
When the bulb #2 is
not burnt out:
R eq  R 
R 3
 R
2 2
Power , P  I 2 R
I
I2  1
2
V
R
I
For Bulb #1
V 2V
I1  3 
3R
2R
4V 2
V2
P1  I R 
 .44
9R
R
For Bulb #2
I
V
I2  1 
2 3R
V2
V2
P2  I R 
 .11
9R
R
2
1
2
2
For Bulb #3
I
V
I3  1 
2 3R
I1
V2
V2
P3  I R 
 .11
9R
R
2
3
I3 
I1
2
When the bulb #2 is
burnt out:
I1
I 3  I1
R eq  R  R  2R
Power , P  I 2 R
I
V
R
For Bulb #1
I1 
V
2R
V2
V2
P1  I R 
 .25
4R
R
2
1
For Bulb #2
I2  0
After total power is
V2 V2
V2
Pa 

 .50
R eq 2R
R
P2  I 22 R  0
So, Bulb #1 gets dimmer and bulb #3 gets
brighter. And the total power decreases.
For Bulb #3
V
I 3  I1 
2R
2
2
2
Before total power was Pb  V  V  .66 V
R eq 32 R
R
V2
V2
P3  I R 
 .25
4R
R
2
3
f) is the answer.
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 Bridge
An 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
Rg
Usually when R g  20
Ig  0  0.5milliAmp
Wheatstone Bridge
Analysis
• Identify the currents
Consider the bridge at “balance”, ig=0
i1  i3
i2  ix
i3 R3  ix Rx
i1 R1  i2 R2
Some Algebra i R  i R
1 3
2 x
i1 R1  i2 R2
R3 Rx

R1 R2
R2
Rx 
R3
R1
Use to Measure Resistance
Ratio Arms
R2
Rx 
R3
R1
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 Transformations
Also known as T-P 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?????????
I
9
10 
V
+
_
5
10 
Req
8
4
We cannot use resistors in parallel. We cannot use
resistors in series. If we knew V and I we could solve
Req =
V
I
There is another way to solve the problem without solving
for I (given, assume, V) and calculating Req for V/I.
Delta-to-Wye Transformations
Three 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
A
B
A
B
R1
RB
R2
RA
R3
C
C
Rest of Circuit
Rest of Circuit
Notes on Names
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.
RC
A
B
A
B
R1
RB
R2
RA
R3
C
C
Rest of Circuit
Rest of Circuit
• 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 Equations
When we perform the delta-to-wye transformation
(going from left to right) we use the equations given
below.
RC
A
B
A
B
R1
RB
R2
RA
R3
C
C
Rest of Circuit
Rest of Circuit
RB RC
R1 
RA  RB  RC
R2 
RA RC
RA  RB  RC
RA RB
R3 
RA  RB  RC
Wye-to-Delta Transformation Equations
When we perform the wye-to-delta transformation
(going from right to left) we use the equations given
below.
RC
A
B
A
B
R1
RB
R2
RA
R3
C
C
Rest of Circuit
Rest of Circuit
RA 
R1 R2  R2 R3  R1 R3
R1
RB 
R1 R2  R2 R3  R1 R3
R2
RC 
R1 R2  R2 R3  R1 R3
R3
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 you
name
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
10 k
R2
10 k
R3
5.1 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
 R1
I1  
 RT

 IT

 R2
b. I1  
 RT

 IT

a.

R2 
 IT
 R1  R2 
c. I1  

R1 
 IT
 R1  R2 
d. I1  
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 +
12 V
R1
10 k
R2
2.0 k
Quiz
8. The current in R1 is
a. 6.7 mA
b. 13.3 mA
I = 20 mA
c. 20 mA
d. 26.7 mA
R1
100 
R2
200 
Quiz
9. The voltage across R2 is
a. 0 V
b. 0.67 V
c. 1.33 V
d. 4.0 V
I = 20 mA
R1
100 
R2
200 
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
6. c
2. b
7. d
3. a
8. b
4. a
9. c
5. b
10. d
Q&A