Download BaS_06b [Compatibility Mode]

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Spark-gap transmitter wikipedia, lookup

Immunity-aware programming wikipedia, lookup

Integrated circuit wikipedia, lookup

Regenerative circuit wikipedia, lookup

Negative resistance wikipedia, lookup

Test probe wikipedia, lookup

Transistor–transistor logic wikipedia, lookup

Integrating ADC wikipedia, lookup

TRIAC wikipedia, lookup

Josephson voltage standard wikipedia, lookup

Multimeter wikipedia, lookup

Valve RF amplifier wikipedia, lookup

Operational amplifier wikipedia, lookup

CMOS wikipedia, lookup

Electrical ballast wikipedia, lookup

Two-port network wikipedia, lookup

RLC circuit wikipedia, lookup

Power electronics wikipedia, lookup

Schmitt trigger wikipedia, lookup

Ohm's law wikipedia, lookup

Voltage regulator wikipedia, lookup

Power MOSFET wikipedia, lookup

Switched-mode power supply wikipedia, lookup

Current source wikipedia, lookup

Resistive opto-isolator wikipedia, lookup

Surge protector wikipedia, lookup

Opto-isolator wikipedia, lookup

Current mirror wikipedia, lookup

Network analysis (electrical circuits) wikipedia, lookup

Rectiverter wikipedia, lookup

Transcript
07-Dec-11
Summary
Series circuits
All circuits have three common attributes. These are:
Basics of electrical systems
R1
1. A source of voltage.
2. A load.
R2
VS +
3. A complete path.
R3
Summary
Summary
Series circuits
Series circuit rule for current:
A series circuit is one that has only one current path.
R1
VS
Because there is only one path, the current everywhere
is the same.
For example, the reading on the first ammeter
is 2.0 mA, What do the other meters read?
R1
R2
VS
R1
R2
R3
VS
R3
R3
+ 2.0 mA _
R2
R1
+ 2.0 mA _
R2
VS
_
_
2.0 mA +
Summary
2.0 mA +
R1
Summary
Series circuits
VS
12 V
Series circuit
680 Ω
R2
1.5 kΩ
R3
The total resistance of resistors in series is
the sum of the individual resistors.
For example, the resistors in a series circuit are 680 Ω,
1.5 kΩ, and 2.2 kΩ. What is the total resistance?
R1
VS
12 V
680 Ω
R3
2.2 kΩ
R2
1.5 kΩ
4.38 kΩ
2.2 kΩ
Tabulating current, resistance, voltage and power is a
useful way to summarize parameters in a series circuit.
Continuing with the previous example, complete the
parameters listed in the Table.
I1= 2.74 mA
I2= 2.74 mA
I3= 2.74 mA
IT= 2.74 mA
R1= 0.68 kΩ
R2= 1.50 kΩ
R3= 2.20 kΩ
RT= 4.38 kΩ
V1= 1.86 V
V2= 4.11 V
V3= 6.03 V
VS= 12 V
P1=
P2=
P3=
PT=
5.1 mW
11.3 mW
16.5 mW
32.9 mW
1
07-Dec-11
Summary
Summary
Voltage sources in series
Kirchhoff’s voltage law
Voltage sources in series add algebraically.
For example, the total voltage of the sources
shown is 27 V
Kirchhoff’s voltage law (KVL) is generally stated as:
+
9V
The sum of all the voltage drops around a single closed
path in a circuit is equal to the total source voltage in that
closed path.
+
9V
+
What is the total voltage if one battery is
accidentally reversed? 9 V
KVL applies to all circuits, but you must apply it to only one
closed path. In a series circuit, this is (of course) the entire
circuit.
+
9V
n
A mathematical shorthand way of writing KVL is
∑V
i =1
R1
Summary
Kirchhoff’s voltage law
=0
Summary
680 Ω
VS
12 V
i
R2
1.5 kΩ
Voltage divider rule
R3
2.2 kΩ
The voltage drop across any given resistor in a series
circuit is equal to the ratio of that resistor to the total
resistance, multiplied by source voltage.
Notice in the series example given earlier that the sum
of the resistor voltages is equal to the source voltage.
I1= 2.74 mA
I2= 2.74 mA
I3= 2.74 mA
IT= 2.74 mA
R1= 0.68 kΩ
R2= 1.50 kΩ
R3= 2.20 kΩ
RT= 4.38 kΩ
V1= 1.86 V
V2= 4.11 V
V3= 6.03 V
VS= 12 V
P1=
P2=
P3=
PT=
5.1 mW
11.3 mW
16.5 mW
32.9 mW
VS
Assume R1 is twice the size of
R2. What is the voltage across
R1? 8 V
Summary
R1
12 V
R2
Summary
R1
Voltage divider
Voltage divider
15 kΩ
VS +
20 V
R2
10 kΩ
What is the voltage across R2?
The total resistance is 25 kΩ.
Applying the voltage divider formula:
R 
 10 kΩ 
V2 =  2  VS = 
 20 V = 8.0 V
 25 kΩ 
 RT 
Notice that 40% of
the source voltage is
across R2, which
represents 40% of
the total resistance.
Voltage dividers can be set up for a variable output using
a potentiometer. In the circuit shown, the output voltage
is variable.
VS +
15 V
What is the largest output
voltage available? 5.0 V
R1
20 kΩ
R2
10 kΩ
VOUT
2
07-Dec-11
Summary
Summary
A
Power in Series Circuits
Voltage measurements
R1
470 Ω
Use the voltage divider rule to
find V1 and V2. Then find the
power in R1 and R2 and PT.
Applying the voltage
divider rule:
 470 Ω 
V1 = 
 20 V = 11.75 V
 800 Ω 
 330 Ω 
V2 = 
 20 V = 8.25 V
 800 Ω 
R2
330 Ω
VS +
20 V
The power dissipated by each
resistor is:
P1 =
(11.75 V )
P2 =
2
470 Ω
2
(8.25 V )
330 Ω
= 0.29 W
} P0.5=W
Voltage is relative and is measured with
respect to another point in the circuit.
B
R2
10 kΩ
C
What are VA, VB, and VAB for the circuit shown?
VA = 12 V VB = 8 V VAB = 4 V
Summary
Summary
Voltage measurements
Ground reference is not always at the
lowest point in a circuit. Assume the
ground is moved to B as shown.
VS +
12 V
Voltages that are given with respect to
ground are shown with a single subscript. For
example, VA means the voltage at point A with
respect to ground (called reference ground). VB
means the voltage at point B with respect to ground.
VAB means the voltage between points A and B.
T
= 0.21 W
R1
5.0 kΩ
A
VS +
12 V
Voltage measurements
R1
5.0 kΩ
B
R2
10 kΩ
C
What are VA, VB, and VC for the circuit?
VA = 4 V VB = 0 V VC = −8 V
Has VAB changed from the previous circuit?
No, it is still 4 V
Assume that R2 is open. For this case,
what are VA, VB, and VC for the circuit?
A
VS +
12 V
R1
5.0 kΩ
B
R2
10 kΩ
C
If R2 is open, there is no current. Notice that VB = 0 V
because it is ground and VA = 0 V because it has the same
potential as VB. VC = −12 V because the source voltage is
across the open.
Selected Key Terms
Selected Key Terms
Series In an electric circuit, a relationship of
components in which the components are
connected such that they provide a single path
between two points.
Reference
ground
The metal chassis that houses the assembly or
a large conductive area on a printed circuit
board is used as a common or reference point;
also called common.
Kirchhoff’s A law stating that (1) the sum of the voltage
voltage law drops around a closed loop equals the source
voltage in that loop or (2) the algebraic sum of
all of the voltages (drops and source) is zero.
Open
A circuit condition in which the current path is
broken.
Short
A circuit condition in which there is zero or an
abnormally low resistance between two points;
usually an inadvertent condition.
Voltage divider A circuit consisting of series resistors across
which one or more output voltages are taken.
3
07-Dec-11
Quiz
1. In a series circuit with more than one resistor, the
current is
Quiz
2. In a series circuit with more than one resistor, the
voltage is
a. larger in larger resistors
a. larger across larger resistors
b. smaller in larger resistors
b. smaller across larger resistors
c. always the same in all resistors
c. always the same across all resistors
d. there is not enough information to say
d. there is not enough information to say
Quiz
3. If three equal resistors are in series, 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
Quiz
4. A series circuit cannot have
a. more than two resistors
b. more than one voltage source
c. more than one path
d. all of the above
d. there is not enough information to say
Quiz
5. In a closed loop, the algebraic sum of all voltages (both
sources and drops)
a. is zero
Quiz
6. The current in the 10 kΩ resistor is
a. 0.5 mA
b. 2.0 mA
b. is equal to the smallest voltage in the loop
c. 2.4 mA
c. is equal to the largest voltage in the loop
d. 10 mA
VS +
24 V
R1
10 kΩ
R2
2.0 kΩ
d. depends on the source voltage
4
07-Dec-11
Quiz
Quiz
8. The smallest output voltage available from the voltage
divider is
7. The output voltage from the voltage divider is
a. 2.0 V
a. 0 V
b. 4.0 V
VS +
24 V
c. 12 V
R1
10 kΩ
R2
2.0 kΩ
d. 20 V
b. 1.5 V
VOUT
R1
10 k Ω
VS +
15 V
c. 5.0 V
R2
10 kΩ
d. 7.5 V
Quiz
9. The total power dissipated in a series 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
VOUT
Quiz
10. The meaning of the voltage VAB is the voltage at
a. Point A with respect to ground
b. Point B with respect to ground
c. The average voltage between points A and B.
d. The voltage difference between points A and B.
d. sum of the power in all resistors
Quiz
Summary
Resistors in parallel
Answers:
1. c
6. b
2. a
7. b
3. c
8. a
4. c
9. d
5. a
10. d
Resistors that are connected to the same two points are said to
be in parallel.
A
R1
R2
R3
R4
B
5
07-Dec-11
Summary
Summary
Parallel circuits
Parallel circuit rule for voltage
Because all components are connected across the same voltage
source, the voltage across each is the same.
A parallel circuit is identified by the fact that it has
more than one current path (branch) connected to a common
voltage source.
VS
For example, the source voltage is 5.0 V. What will a volt- meter read if
it is placed across each of the resistors?
+5.0 V
+
+
R1
R2
R3
+5.0 V
+
VS
+5.0 V
Summary
Parallel circuit rule for resistance
R1
680 Ω
VS
+
R2
1.5 kΩ
R3
2.2 kΩ
R2
1.5 kΩ
R3
2.2 kΩ
R1
R2
The resistance of two parallel resistors can be found by
either:
RT =
1
1
1
+
R1 R2
or
RT =
R1 R2
R1 + R2
What is the total resistance if R1 = 27 kΩ and R2 = 56
kΩ?
18.2 kΩ
Summary
Parallel circuit
R1
680 Ω
Special case for resistance
of two parallel resistors
For example, the resistors in a parallel circuit are 680 Ω, 1.5 kΩ, and
2.2 kΩ. What is the total resistance?
386 Ω
+
+5.0 V
+
Summary
The total resistance of resistors in parallel is
the reciprocal of the sum of the reciprocals of the
individual resistors.
VS
+5.0 V
+
R4
Summary
R1
680 Ω
R2
1.5 kΩ
Tabulating current, resistance, voltage and power is a useful way to
summarize parameters in a parallel circuit.
Continuing with the previous example, complete the parameters
listed in the Table.
R3
2.2 kΩ
Kirchhoff’s current law
Kirchhoff’s current law (KCL) is generally stated as:
The sum of the currents entering a node is equal to the sum of the
currents leaving the node.
Notice in the previous example that the current from the source is
equal to the sum of the branch currents.
I1= 7.4 mA
R1= 0.68 kΩ
V1= 5.0 V
P1= 36.8 mW
I1= 7.4 mA
R1= 0.68 kΩ
V1= 5.0 V
P1= 36.8 mW
I2= 3.3 mA
R2= 1.50 kΩ
V2= 5.0 V
P2= 16.7 mW
I2= 3.3 mA
R2= 1.50 kΩ
V2= 5.0 V
P2= 16.7 mW
I3= 2.3 mA
R3= 2.20 kΩ
V3= 5.0 V
P3= 11.4 mW
I3= 2.3 mA
R3= 2.20 kΩ
V3= 5.0 V
P3= 11.4 mW
IT= 13.0 mA
RT= 386 Ω
VS= 5.0 V
PT= 64.8 mW
IT= 13.0 mA
RT= 386 Ω
VS= 5.0 V
PT= 64.8 mW
6
07-Dec-11
Summary
Current divider
Current divider
When current enters a node (junction) it divides into currents with
values that are inversely proportional to the resistance values.
The most widely used formula for the current divider is the tworesistor equation. For resistors R1 and R2,
 R2 
I1 = 
 I T and
 R1 + R2 
 R1 
I2 = 
 IT
 R1 + R2 
Notice the subscripts. The resistor in the numerator is not the
same as the one for which current is found.
Summary
Power in parallel circuits
Power in each resistor can be calculated with any of the standard
power formulas. Most of the time, the voltage is
known, so the equation
Summary
P=
V 2 is most convenient.
R
As in the series case, the total power is the sum of the
powers dissipated in each resistor.
What is the total power if 10 V is applied to the parallel combination
of R1 = 270 Ω and R2 = 150 Ω?
1.04 W
Key Terms
Assume that R1is a 2.2 kΩ resistor that is in parallel
with R2, which is 4.7 kΩ. If the total current into the
resistors is 8.0 mA, what is the current in each
resistor?
 R2 
 4.7 kΩ 
5.45 mA
I1 = 
 IT = 
 8.0 mA =
 6.9 kΩ 
 R1 + R2 
 R1 
 2.2 kΩ 
2.55 mA
I2 = 
 IT = 
 8.0 mA =
 6.9 kΩ 
 R1 + R2 
Notice that the larger resistor has the smaller current.
Summary
Assume there are 8 resistive wires that form a rear window defroster
for an automobile.
(a) If the defroster dissipates 90 W when connected to a 12.6 V source,
what power is dissipated by each resistive wire?
(b) What is the total resistance of the defroster?
(a) Each of the 8 wires will dissipate 1/8 of the total
power or 90 W
= 11.25 W
8 wires
2
(12.6 V ) = 1.76 Ω
V2
(b) The total resistance is R =
=
P
90 W
What is the resistance of each wire? 1.76 Ω x 8 = 14.1 Ω
Quiz
Parallel The relationship in electric circuits in which two or
more current paths are connected between two separate
points (nodes).
Branch One current path in a parallel circuit.
Kirchhoff’s A law stating the total current into a node equals the
current law total current out of the node.
Node A point or junction in a circuit at which two or more
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
components are connected.
Current divider A parallel circuit in which the currents divide inversely
proportional to the parallel branch resistances.
7
07-Dec-11
Quiz
2. The number of nodes in a parallel circuit is
Quiz
3. The total resistance of the parallel resistors is
a. one
a. 2.52 kΩ
b. two
b. 3.35 kΩ
c. three
c. 5.1 kΩ
d. any number
d. 25.1 kΩ
Quiz
4. If three equal resistors are in parallel, the total resistance is
R1
10 kΩ
R2
10 kΩ
Quiz
5. In any circuit the total current entering a node is
a. one third the value of one resistor.
a. less than the total current leaving the node.
b. the same as one resistor.
b. equal to the total current leaving the node.
c. three times the value of one resistor.
c. greater than the total current leaving the node.
d. the product of the three resistors
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
Quiz
7. The total current leaving the source is
 R1 
 IT
 RT 
a. 1.0 mA
R 
b. I1 =  2  I T
 RT 
c. 6.0 mA
a. I1 = 
R3
5.1 kΩ
b. 1.2 mA
VS +
12 V
R1
10 kΩ
R2
2.0 kΩ
d. 7.2 mA

c. I1 =  R2

 IT
 R1 + R2 
 R 
d. I1 =  1  I T
 R1 + R2 
8
07-Dec-11
Quiz
Quiz
9. The voltage across R2 is
8. The current in R1 is
a. 0 V
a. 6.7 mA
b. 0.67 V
b. 13.3 mA
c. 20 mA
I = 20 mA
R1
100 Ω
R2
200 Ω
c. 1.33 V
I = 20 mA
R1
100 Ω
R2
200 Ω
d. 4.0 V
d. 26.7 mA
Quiz
Quiz
10. The total power dissipated in a parallel circuit is equal to the
Answers:
a. power in the largest resistor.
1. d
6. c
b. power in the smallest resistor.
2. b
7. d
c. average of the power in all resistors.
3. a
8. b
d. sum of the power in all resistors.
4. a
9. c
5. b
10. d
Summary
Summary
Identifying series-parallel relationships
Combination circuits
Most practical circuits have
combinations of series and parallel
components.
Most practical circuits have various combinations of
series and parallel components. You can frequently
simplify analysis by combining series and parallel
components.
An important analysis method is to form an equivalent
circuit. An equivalent circuit is one that has
characteristics that are electrically the same as
another circuit but is generally simpler.
Components that are connected in
series will share a common path.
Components that are connected in
parallel will be connected across
the same two nodes.
1
2
9
07-Dec-11
Summary
Summary
Equivalent circuits
Equivalent circuits
For example:
Another example:
is equivalent to
R1
1.0 kΩ
R1
is equivalent to
2.0 kΩ
R2
1.0 kΩ
R1
R2
R1,2
500 Ω
1.0 kΩ 1.0 kΩ
There are no electrical measurements that can
distinguish the boxes.
There are no electrical measurements that can
distinguish the boxes.
Summary
Summary
is equivalent to
R1
1.0 kΩ
R3
R 1,2
R3
R2
4.7 kΩ
3.7 kΩ
4.7 kΩ
2.7 kΩ
is equivalent to
Kirchhoff’s voltage law and Kirchhoff’s current law
can be applied to any circuit, including combination
circuits.
For example,
applying KVL, the
path shown will
have a sum of 0 V.
2.07 kΩ
Start/Finish
Summary
I
I
−
26.5 mA
+
A
I
VS
5.0 V
+
+
8.0 mA
−
−
R1
270 Ω
100 Ω
R3
330 Ω
R6
100 Ω
R4
100 Ω
100 Ω
Start/Finish
Combination circuits
R5
100 Ω
VS +
10 V
R1
270 Ω
R2
330 Ω
R3
470 Ω
Tabulating current, resistance, voltage and power is a
useful way to summarize parameters. Solve for the
unknown quantities in the circuit shown.
R2
470 Ω
R4
18.5 mA
R11
270 Ω
Summary
Kirchoff’s current law can also be applied to the same
circuit. What are the readings for node A?
+
VS
5.0 V
So will
this path!
R3
330 Ω
R6
There are no electrical
measurements that can
distinguish between the
three boxes.
R 1,2,3
R2
470 Ω
R5
100 Ω
I1= 21.6 mA
I2= 12.7 mA
I3= 8.9 mA
IT= 21.6 mA
R1= 270 Ω
R2= 330 Ω
R3= 470 Ω
RT= 464 Ω
V1= 5.82 V P1= 126 mW
V2= 4.18 V P2= 53.1 mW
V3= 4.18 V P3= 37.2 mW
VS= 10 V PT= 216 mW
10
07-Dec-11
Summary
Loaded voltage divider
R2
330 Ω
R3
470 Ω
Notice that the current in R1 is
equal to the sum of the branch currents in R2 and R3.
The sum of the voltages around the outside loop is zero.
R1= 270 Ω
R2= 330 Ω
R3= 470 Ω
RT= 464 Ω
V1= 5.82 V P1= 126 mW
V2= 4.18 V P2= 53.1 mW
V3= 4.18 V P3= 37.2 mW
VS= 10 V PT= 216 mW
VS =
+15 V
R1
330 Ω
Stiff voltage divider
A
R3
2.2 kΩ
Form an equivalent series circuit by combining R2 and
R3; then apply the voltage-divider formula to the
equivalent circuit: R2,3 = R2 R3 = 470 Ω 2.2 kΩ = 387 Ω
 R2,3 
387 Ω


V3 = V2,3 = 
V =
15 V = 8.10 V
 R + R  S  330 Ω + 387 Ω 
2,3 
 1
Summary
Loading effect of
a voltmeter
VS +
10 V
R3
Summary
R2
470 Ω
What is the voltage
across R3?
A
R2
A voltage-divider with a resistive load is a combinational
circuit and the voltage divider is said to be loaded. The
loading reduces the total resistance from node A to ground.
Summary
Loaded voltage divider
R1
+
The voltage-divider equation
was developed for a series
circuit. Recall that the output
voltage is given by
R 
V2 =  2  VS
 RT 
R1
VS
A stiff voltage-divider is one in
R2
RL
which the loaded voltage nearly
the same as the no-load voltage.
To accomplish this, the load
current must be small compared
to the bleeder current (or RL is large compared to the divider resistors).
If R1 = R2 = 1.0 kΩ, what value of RL will make the divider a
stiff voltage divider? What fraction of the unloaded voltage is
the loaded voltage?
RL > 10 R2; RL should be 10 kΩ or greater. For a 10 kΩ load,
 R2 || RL 
0.91 kΩ


This is 95% of the
VL = 
 VS = 
 VS = ( 0.476 ) VS
 1.0 kΩ + 0.91 kΩ 
unloaded voltage.
 R1 + R2 || RL 
Summary
R1
470 kΩ
+
Wheatstone bridge
4.04
10 VV
R2
Assume VS = 10 V, but the
+
4.04 V
470 kΩ
meter reads only 4.04 V
when it is across either R1
or R2.
Can you explain what is happening?
All measurements affect the quantity being measured. A
voltmeter has internal resistance, which can change the
resistance of the circuit under test. In this case, a 1 MΩ
internal resistance of the meter accounts for the readings.
The Wheatstone bridge consists
R3
R1
of a dc voltage source and four VS
Output
resistive arms forming two
voltage dividers. The output is
R2
R4
taken between the dividers.
Frequently, one of the bridge
resistors is adjustable.
When the bridge is balanced, the output voltage is zero,
and the products of resistances in the opposite diagonal
arms are equal.
-
I1= 21.6 mA
I2= 12.7 mA
I3= 8.9 mA
IT= 21.6 mA
Summary
+
Kirchhoff’s laws can be applied
as a check on the answer.
R1
270 Ω
VS +
10 V
11
07-Dec-11
Summary
Summary
Wheatstone bridge
VS
12 V
+
R1
470 Ω
R3
330 Ω
Output
-
Example: What is the
value of R2 if the bridge
is balanced? 384 Ω
Thevenin’s theorem
R2
R4
270 Ω
Thevenin’s theorem states that any two-terminal,
resistive circuit can be replaced with a simple
equivalent circuit when viewed from two output
terminals. The equivalent circuit is:
RTH
VTH
Summary
Summary
Thevenin’s theorem
Thevenin’s theorem
VTH is defined as the open circuit voltage between the two
output terminals of a circuit.
RTH is defined as the total resistance appearing between
the two output terminals when all sources have been
replaced by their internal resistances.
What is the Thevenin voltage for the circuit? 8.76 V
RTH
What is the Thevenin resistance for the circuit? 7.30 kΩ
Output terminals
R1
VS
12 V
R2
27 kΩ
VTH
Summary
Thevenin’s theorem
Thevenin’s theorem is useful for solving the Wheatstone
bridge. One way to Thevenize the bridge is to create two
Thevenin circuits − from A to ground and from B to ground.
The resistance between point
R1
R2
A and ground is R1||R3 and the VS
+
RL
A
B
resistance from B to ground is
R2||R4. The voltage on each
R3
R4
side of the bridge is found
using the voltage divider rule.
Remember, the
load resistor
has no affect on
the Thevenin
parameters.
10 kΩ
RL
68 kΩ
Summary
Thevenin’s theorem
For the bridge shown, R1||R3 = 165 Ω and
R2||R4 = 179 Ω. The voltage from A to ground
(with no load) is 7.5 V and from B to ground
(with no load) is 6.87 V .
VS
+15 V
+
R1
330 Ω
A
R3
330 Ω
RL
150 Ω
R2
390 Ω
B
R4
330 Ω
The Thevenin circuits for each of the
bridge are shown on the following slide.
12
07-Dec-11
Summary
Summary
Thevenin’s theorem
Maximum power transfer
VTH
7.5 V
165 Ω
The maximum power is transferred from a source to a
load when the load resistance is equal to the internal
source resistance.
B RTH'
RTH A RL
150 Ω
VTH'
6.87 V
179 Ω
RS
VS +
RL
Putting the load on the Thevenin circuits and
applying the superposition theorem allows you to
calculate the load current. The load current is: 1.27 mA
Summary
Summary
Maximum power transfer
Superposition theorem
Summary
+
-
VS1
12 V
6.8 kkΩ
kΩ
Ω
6.8
VS2S2
-++1.56 mA
18 V
RR222
6.8 kkΩ
Ω
6.8
Ω
II22
-
Set up a table of
pertinent information
and solve for each
quantity listed:
VS1
12 V
6.8 kΩ
I2
+
-
VS2
18 V
R2
6.8 kΩ
Troubleshooting
RR333
RR111
2.7 kkΩ
Ω
2.7
Ω
2.7 kΩ
Summary
++
What does the ammeter
read for I2?
What does the
ammeter read for
I2? (See next slide
for the method and
the answer).
-
V 2 ( 5.0 V )
=
= 0.5 W
RL
50 Ω
2
PL =
R3
R1
RL
50 Ω
+
VS +
10 V
-
RS
50 Ω
The superposition theorem is a way to determine currents
and voltages in a linear circuit that has multiple sources by
taking one source at a time and algebraically summing the
results.
+
What is the power delivered to the matching load?
The voltage to the
load is 5.0 V. The
power delivered is
The maximum power transfer theorem assumes the
source voltage and resistance are fixed.
Source 1:
RT(S1)= 6.10 kΩ I1= 1.97 mA I2= 0.98 mA
Source 2:
RT(S2)= 8.73 kΩ I3= 2.06 mA I2= 0.58 mA
Both sources
I2= 1.56 mA
The total current is the algebraic sum.
The effective troubleshooter must think logically about
circuit operation.
Understand normal circuit operation and
find out the symptoms of the failure.
Decide on a logical set of steps to find the
fault.
Following the steps in the plan, make
measurements to isolate the problem.
Modify the plan if necessary.
13
07-Dec-11
Troubleshooting
Selected Key Terms
R1
330 Ω
A
R2
470 Ω
R3
The output of the voltage2.2 kΩ
divider is 6.0 V. Describe how
you would use analysis and
planning in finding the fault.
From an earlier calculation, V3 should equal 8.10 V. A low
voltage is most likely caused by a low source voltage or
incorrect resistors (possibly R1 and R2 reversed). If the circuit is
new, incorrect components are possible.
Decide on a logical set of steps to locate the fault. You could
decide to 1) check the source voltage, 2) disconnect the load and
check the output voltage, and if it is correct, 3) check the load
resistance. If R3 is correct, check other resistors.
Loading The effect on a circuit when an element that
draws current from the circuit is connected
across the output terminals.
Load current The output current supplied to a load.
Bleeder The current left after the load current is
current subtracted from the total current into the circuit.
Wheatstone A 4-legged type of bridge circuit with which an
bridge unknown resistance can be accurately measured
using the balanced state. Deviations in resistance
can be measured using the unbalanced state.
Selected Key Terms
Thevenin’s A circuit theorem that provides for reducing
theorem any two-terminal resistive circuit to a single
equivalent voltage source in series with an
equivalent resistance.
Maximum power The condition, when the load resistance
transfer equals the source resistance, under which
maximum power is transferred to the load.
Superposition A method for analyzing circuits with two or
more sources by examining the effects of each
source by itself and then combining the
effects.
Quiz
1. Two circuits that are equivalent have the same
a. number of components
b. response to an electrical stimulus
c. internal power dissipation
d. all of the above
Quiz
2. If a series equivalent circuit is drawn for a complex
circuit, the equivalent circuit can be analyzed with
Quiz
3. For the circuit shown,
a. R1 is in series with R2
a. the voltage divider theorem
b. R1 is in parallel with R2
b. Kirchhoff’s voltage law
c. R2 is in series with R3
c. both of the above
d. R2 is in parallel with R3
R1
VS
R2
R3
-
VS =
+15 V
+
Summary
d. none of the above
14
07-Dec-11
Quiz
Quiz
4. For the circuit shown,
5. A signal generator has an output voltage of 2.0 V with no
load. When a 600 Ω load is connected to it, the output
drops to 1.0 V. The Thevenin resistance of the generator is
R4
a. R1 is in series with R2
VS
+
c. R2 is in parallel with R3
R2
R1
b. R4 is in parallel with R1
-
d. none of the above
a. 300 Ω
R3
b. 600 Ω
c. 900 Ω
d. 1200 Ω.
Quiz
Quiz
6. For the circuit shown, Kirchhoff's voltage law
7. The effect of changing a measured quantity due to
connecting an instrument to a circuit is called
a. applies only to the outside loop
b. applies only to the A junction.
a. loading
c. can be applied to any closed path.
b. clipping
d. does not apply.
VS +
10 V
A
R1
270 Ω
c. distortion
d. loss of precision
R2
330 Ω
R3
470 Ω
Quiz
Quiz
8. An unbalanced Wheatstone bridge has the voltages
shown. The voltage across R4 is
a. 4.0 V
+
-
d. 7.0 V
R1
7.0 V
R2
R3
+ RL 1.0 V
b. 5.0 V
c. 6.0 V
R4
d. 7.0 V
VS
12 V
R3
R1
+ RL -
-
c. 6.0 V
VS
12 V
+
a. 4.0 V
b. 5.0 V
9. Assume R2 is adjusted until the Wheatstone bridge is
balanced. At this point, the voltage across R4 is measured
and found to be 5.0 V. The voltage across R1 will be
R2
R4
5.0 V
15
07-Dec-11
Quiz
10. Maximum power is transferred from a fixed source
when
Quiz
Answers:
1. b
6. c
a. the load resistor is ½ the source resistance
2. c
7. a
b. the load resistor is equal to the source resistance
3. d
8. a
c. the load resistor is twice the source resistance
4. d
9. d
d. none of the above
5. b
10. b
16