Transcript PWA_Mod04_Prob04_v06

Filename: PWA_Mod04_Prob04.ppt
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
R1=
20[W]
+
-
Find the Thévenin Equivalent at terminals A and B.
R 2=
25[W]
Problems With Assistance
v =
6[V]
+
Module 4 – Problem
4
S1
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
Go
straight to
the First
Step
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
Go
straight to
the
Problem
Statement
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Overview of this Problem
In this problem, we will use the following
concepts:
• Equivalent Circuits
• Thévenin’s Theorem
• Equivalent Resistance with Dependent
Sources
Go
straight to
the First
Step
Go
straight to
the
Problem
Statement
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Textbook Coverage
The material for this problem is covered in your textbook in
the following sections:
• Circuits by Carlson: Sections #.#
• Electric Circuits 6th Ed. by Nilsson and Riedel: Sections
#.#
• Basic Engineering Circuit Analysis 6th Ed. by Irwin and
Wu: Section #.#
• Fundamentals of Electric Circuits by Alexander and
Sadiku: Sections #.#
• Introduction to Electric Circuits 2nd Ed. by Dorf: Sections
#-#
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Coverage in this Module
The material for this problem is covered in
this module in the following presentation:
• DPKC_Mod04_Part04
Next slide
Dave Shattuck
University of Houston
Problem Statement
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
How should
we start this
problem?
What is the
first step?
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution – First Step
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
-
a)
Define the opencircuit voltage.
b)
Label the
terminals of
resistor R4 and
remove it.
c)
Define the shortcircuit current.
d)
Combine resistors
R2 and R3 in
parallel.
e)
Combine resistors
R1 and R2 in
series.
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
How should we start
this problem?
What is the first
step?
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
Dave Shattuck
University of Houston
Your choice for First Step –
Define the open-circuit voltage
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
This is a good choice
for the first step.
We could find the
open-circuit voltage,
and that will be the
Thévenin voltage. We
could also have solved
for the short-circuit
current first. Either is
a reasonable first
choice. Let’s define
the open-circuit
voltage, and then
solve for it.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step –
Label the terminals of resistor R4 and remove it
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
This is not a good
choice for the first step.
The resistor R4 is in
parallel with the two
terminals at which we
are finding the
equivalent. However,
we are not finding the
equivalent seen by R4,
so this resistor will be a
part of the equivalent.
Thus, it needs to be left
in place. Please go
back and try again.
Dave Shattuck
University of Houston
Your choice for First Step –
Define the short-circuit current
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
This is a good choice
for the first step. Since
we are finding the
Thevenin equivalent,
we will eventually
have to find the opencircuit voltage, or solve
for it after finding the
equivalent resistance.
However, this is a very
reasonable first choice.
So, let’s label the shortcircuit current and
solve for it.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step was –
Combine resistors R2 and R3 in parallel
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
This is a good first step,
but is not the choice that
we would like to follow
here, simply due to the
approach we are going to
take with this problem.
Note that the resistors R2
and R3 are in parallel,
and that combining them
would not change the
voltage vX that we need
for the vS2 dependent
source. This was a fine
choice, but we would
like for you to go back
and try again.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for First Step was –
Combine resistors R1 and R2 in series
This is not a good
choice.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
-
Please go back and try
again.
+
R3=
15[W]
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
The problem here is
that resistors R1 and R2
are not in series, since
they do not have the
same current going
through them.
R4=
33[W]
iX
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Defining the Open-Circuit
Voltage and Solving for It
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
+
-
vS2=
10[W]iX
A +
vD
R5=
22[W]
vX
vOC iS1=
2[S]vX
- B
R4=
33[W]
We have defined the opencircuit voltage, and
called it vOC. In
preparation for solving
for this voltage, we
have also chosen a
reference node for the
node-voltage method,
and defined the other
node voltage, vD. Note
that there are only three
essential nodes, so we
will have two
equations, plus two
more for the dependent
source variables. Thus
we will need to write
four equations.
Let’s go to the next slide
and write these
equations.
iX
Next slide
vOC
v v
v v
 2[S]vX  OC D  OC D  0,
33[W]
15[W]
25[W]
© Brooks/Cole Publishing Co.
vD  10[W]iX vD  vOC vD  vOC vD  6[V]  vD vD  6[V]  vD




 0,
22[W]
15[W]
25[W]
20[W]
20[W]
v
iX  OC , and
33[W]
R1=
vX  vD  vOC .
20[W]
Dave Shattuck
University of Houston
+
-
vS1=
6[V]
+
R3=
15[W]
+
+
-
Writing the NodeVoltage Equations
R 2=
25[W]
vS2=
10[W]iX
vX
A -
Let’s go to the next slide
and simplify these
equations.
+
vD
R5=
22[W]
We have written the four
equations needed here.
vOC iS1=
2[S]vX
- B
R4=
33[W]
iX
Find the Thévenin
Equivalent at terminals A
and B.
Next slide
vOC
vOC  vD vOC  vD
 2[S]  vD  vOC  

 0, and
33[W]
15[W]
25[W]
v
vD  10[W] OC
33[W] vD  vOC vD  vOC


 0.
22[W]
15[W]
25[W]
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Simplifying the NodeVoltage Equations
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
-
Let’s go to the next slide
and solve these
equations.
+
R3=
15[W]
+
We canceled the last two
terms of the second
equation, and plugged
the last two equations
into the first two.
vS2=
10[W]iX
A +
vD
R5=
22[W]
vX
vOC iS1=
2[S]vX
- B
R4=
33[W]
iX
Find the Thévenin
Equivalent at terminals A
and B.
Next slide
Dave Shattuck
University of Houston
Solving the Node-Voltage Equations
© Brooks/Cole Publishing Co.
2.137[S]vOC  2.107[S]vD  0, and
0.120[S]vOC  0.152[S]vD  0.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
When we take these two
equations, and solve,
we find that
R 2=
25[W]
vS1=
6[V]
+
-
Thus, the open-circuit
voltage is zero.
+
R3=
15[W]
+
vOC = 0.
vS2=
10[W]iX
If you had chosen to find the
short-circuit current,
you would have found
that iSC is also zero.
A +
vD
R5=
22[W]
vX
vOC iS1=
2[S]vX
- B
R4=
33[W]
iX
Now, we have no choice but
to find the equivalent
resistance, REQ.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Defining the Short-Circuit
Current and Solving for It
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
vX
R3=
15[W]
A -
+
+
-
vS2=
10[W]iX
iSC
iS1=
2[S]vX
vE
R5=
22[W]
R4=
33[W]
iX
B
We have defined the shortcircuit current, and
called it iSC. In
preparation for solving
for this current, we
have also chosen a
reference node for the
node-voltage method,
and defined the other
node voltage, vE. Note
that there are only two
essential nodes, so we
will have one
equations, plus two
more for the dependent
source variables, and
once for iSC. Thus we
will need to write four
equations.
Let’s go to the next slide
and write these
equations.
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
R1=
20[W]
+
-
vE  10[W]iX
v
v
v  6[V]  vE vE  6[V]  vE
 E  E  E

 0,
22[W]
15[W] 25[W]
20[W]
20[W]
v
v
0
iSC  E  E 
 2[S]v X  0,
15[W] 25[W] 33[W]
0
iX 
, and
33[W]
v X  vE .
Writing the NodeVoltage Equations
R 2=
25[W]
vS1=
6[V]
+
We have written the four
equations needed here.
vX
R3=
15[W]
A -
Let’s go to the next slide
and simplify these
equations.
+
+
-
vS2=
10[W]iX
iSC
iS1=
2[S]vX
vE
R5=
22[W]
R4=
33[W]
iX
B
Find the Thévenin
Equivalent at terminals A
and B.
Next slide
Dave Shattuck
University of Houston
vE  0
vE
vE


 0, and
22[W] 15[W] 25[W]
vE
vE
iSC 

 2[S]vE  0.
15[W] 25[W]
© Brooks/Cole Publishing Co.
Simplifying the NodeVoltage Equations
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
-
Let’s go to the next slide
and solve these
equations.
+
R3=
15[W]
+
We canceled the last two
terms of the first
equation, and plugged
the last two equations
into the first two.
vS2=
10[W]iX
A +
vD
R5=
22[W]
vX
vOC iS1=
2[S]vX
- B
R4=
33[W]
iX
Find the Thévenin
Equivalent at terminals A
and B.
Next slide
Dave Shattuck
University of Houston
Solving the Node-Voltage Equations
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
When we take these two
equations, and solve,
we find that
R 2=
25[W]
vS1=
6[V]
+
-
iSC = 0.
Thus, the short-circuit
current is zero.
+
R3=
15[W]
+
vE  0
v
v
 E  E  0, and
22[W] 15[W] 25[W]
v
v
iSC  E  E  2[S]vE  0.
15[W] 25[W]
vS2=
10[W]iX
If you had chosen to find the
open-circuit voltage,
you would have found
that vOC is also zero.
A +
vD
R5=
22[W]
vX
vOC iS1=
2[S]vX
- B
R4=
33[W]
iX
Now, we have no choice but
to find the equivalent
resistance, REQ.
Dave Shattuck
University of Houston
Finding the Equivalent Resistance
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
-
1)
Combine resistors R2
and R3 in parallel.
vX
2)
Apply a test source.
A -
3)
Set independent
sources equal to zero.
4)
Set dependent sources
equal to zero.
5)
Set all sources equal to
zero.
vS1=
6[V]
+
R3=
15[W]
+
We wish to find the
equivalent resistance,
REQ. What should be
our first step?
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
Dave Shattuck
University of Houston
You Chose: Combine resistors R2 and R3 in parallel
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
You said that the first
thing to combine the
parallel resistors.
These resistors are in
parallel, and they
can be combined.
Thus, this is a
reasonable step, but
taking this as the
first step does not
emphasize the very
important step that
we must always
apply when finding
the equivalent
resistance. So,
please go back and
try again.
You Chose: Apply a test source
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
You said that the first thing
would be to apply a
test source. This is
not correct.
Applying a test source will
be important here, but
for this to work, we
need to get rid of the
independent sources.
This is very important.
Many students neglect
this step, and get the
wrong answer.
Therefore we
emphasize it by
always performing it
first. Let’s go back
and try again.
Dave Shattuck
University of Houston
You Chose: Set independent sources equal to zero
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
You said that the first
thing to do was to set
the independent
sources equal to zero.
This is the best first
step. Always do this
first, and you will not
forget to do it. Let’s
set the independent
source, vS1, equal to
zero.
Dave Shattuck
University of Houston
You Chose: Set dependent sources equal to zero
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
You said that the first
thing to do was to set
the dependent sources
equal to zero. This is
incorrect. We do not
set dependent sources
equal to zero. We do
set the independent
sources equal to zero.
Let’s do that. Go
back and try again.
You Chose: Set all sources equal to zero
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
iX
You said that the first
thing to do was to set
all of the sources
equal to zero. This is
incorrect. We do not
set dependent sources
equal to zero. We do
set the independent
sources equal to zero.
Let’s do that. Go
back and try again.
Setting Independent Sources Equal to Zero
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
R 2=
25[W]
+
R3=
15[W]
+
-
vX
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
R4=
33[W]
We have set the
independent source, vS1,
equal to zero. Before
going further, let’s
simplify this circuit. The
resistor R1 is in parallel
with a short circuit, and
will have no effect. It can
be removed. The resistors
R2 and R3 are in parallel,
as many of you have
already noted. Let’s go
ahead and combine them
in parallel, with a resistor
R6. Note that the voltage
vX is across this resistor.
We make these
simplifications in the next
slide.
iX
B
Next slide
What is the Next Step?
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
We have simplified the circuit. What is the next step to take?
1)
Combine resistors R6 and R5 in series.
2)
Apply a test source.
3)
Simplify by setting iS1 equal to iX.
Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
R4=
33[W]
iX
B
You Chose: Combine resistors R6 and R5 in series
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You said that the first thing would be to combine resistors R6 and R5 in series. This is not
a good thing to do. The resistors R6 and R5 are in series. However, if we did this,
the voltage vX would no longer be present, and the iS1 dependent current source
depends on vX.
Let’s go back and try again.
Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
R4=
33[W]
iX
B
You Chose: Apply a test source
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You said that the first thing would be to apply a test source. This is the best choice for
next step.
Let’s apply a test source, between terminals A and B.
Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
R4=
33[W]
iX
B
You Chose: Simplify by setting iS1 equal to iX
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
You said that the first thing would be to simplify by setting iS1 equal to iX. This is not a
good thing to do. The current iX is not equal to iS1, as we will see when we begin
writing equations.
Let’s go back and try again.
Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
R4=
33[W]
iX
B
Applying a Test Source
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
We have applied a test source, between terminals A and B. We chose a voltage source,
because we thought that this would make the solution a little easier. A current
source would have been nearly as good. We chose to give it a value, 1[V], just to
make it easier to solve this with a typical calculator. The key is to find the ratio of
vT/iT, since we have used the active sign convention to define iT. This ratio will give
us REQ. Let’s solve.
Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
+
vS2=
10[W]iX
vT=
1[V]
-
R 5=
22[W]
iT
iS1=
2[S]vX
R 4=
33[W]
iX
B
Next slide
Writing Equations for iT
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Let’s solve. We apply KCL to the top essential node, and get
1[V]  10[W]iX
1[V]

 2[S]v X  iT  0,
9.4[W]  22[W] 33[W]
1[V]
iX 
, and
33[W]
 1[V]  10[W]iX 
vX   
 9.4[W].
 9.4[W]  22[W] 
Find the Thévenin Equivalent at terminals A
and B.
R6=
9.4[W]
+
+
-
A
-
vX
+
vS2=
10[W]iX
vT=
1[V]
-
R 5=
22[W]
iT
iS1=
2[S]vX
R 4=
33[W]
iX
B
Next slide
Solving for iT
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
To solve these equations for iT, we plug the last two equations into the first, and get
 

 1[V] 
 1[V]  
1[V]  10[W] 
  1[V]  10[W] 



33[
W
]
33[
W
]
1[V]



  9.4[W]   i .
 2[S]   
  9.4[W]  22[W] 
 T
9.4[W]  22[W]
33[W]
 



 

Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
+
vS2=
10[W]iX
vT=
1[V]
-
R 5=
22[W]
iT
iS1=
2[S]vX
R 4=
33[W]
iX
B
Next slide
Finding iT
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
We plug in values and get
  1[V]  0.303[V] 

1[V]  0.303[V] 1[V]

 2[S]  
9.4[
W
]
  iT , or

31.4[W]
33[W]
31.4[W]



22.2[mA]  30.3[mA]  2[S]   22.2[mA]  9.4[ W]   iT , or
470[mA]  iT .
Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
+
vS2=
10[W]iX
vT=
1[V]
-
R 5=
22[W]
iT
iS1=
2[S]vX
R 4=
33[W]
iX
B
Next slide
Finding REQ
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Now we can solve for REQ and get
REQ
vT
1[V]


 2.13[W].
iT 470[mA]
Note that this value could be
negative. For example,
if we had defined vX with
the opposite polarity, this
answer would have been
negative, with the value
–2.74[W]. You could try
it yourself.
Find the Thévenin Equivalent at terminals A and B.
R6=
9.4[W]
+
+
-
A
-
vX
+
vS2=
10[W]iX
vT=
1[V]
-
R 5=
22[W]
iT
iS1=
2[S]vX
R 4=
33[W]
iX
B
Next slide
The Thevenin Equivalent
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The Norton equivalent is a zero valued voltage source, in series
with the equivalent resistance. This is, of course, just the
resistance. Thus, we have the circuit drawn below.
REQ  2.13[W].
Find the Thévenin Equivalent at terminals A and B.
R1=
20[W]
+
-
R 2=
25[W]
vS1=
6[V]
A
+
R3=
15[W]
+
-
vTH  0.
vX
REQ=
2.13[W]
A -
vS2=
10[W]iX
iS1=
2[S]vX
R5=
22[W]
B
R4=
33[W]
B
Go back to
Problem
Statement
iX
Go to Comments Slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
What Happened Here?
• It seemed like a strange problem, with the open-circuit voltage
and short-circuit current ending up as zero. Actually, if either is
zero, the other will be zero. When there are no independent
sources in the circuit, this will always happen.
• Here in this problem, there was an independent source.
However, if you look carefully you will note that that part of the
circuit was shorted out, and had no effect on the rest of the
circuit. This was clear from the node-voltage equations that we
wrote to solve for iSC and for vOC. Note that in both cases the
terms that included the independent source cancelled out. This
is why the independent source seemed to not be there.
Go back to
Overview
slide.