Download PWA_Mod06_Prob01_v07

Filename: PWA_Mod06_Prob01.ppt
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below. The
voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
RX=100[W]
+
-
Problems With Assistance
C=
v
(t)
i (t)
Module 6 – Problem 1 v (t)150[mF]
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
X
X
S
C
-
+
LX=2.2[mH]
Go
straight to
the First
Step
vC (0)  100[mV].
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:
• Defining Equation for Inductors
• Defining Equations for Capacitors
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_Mod06_Part01
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Statement
The value for the current source in
the circuit shown is given below. The
voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
[mA]; for t  0.
vC (0)  100[mV].
Next slide
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solution – First Step – Where to Start?
The value for the current source in
the circuit shown is given below. The
voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
How should
we start this
problem?
What is the
first step?
Next slide
Problem Solution – First Step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
3000[s1 ]t
[mA]; for t  0. How should we start this problem? What is the first step?
vC (0)  100[mV].
a) Use Ohm’s Law to find the voltage across the resistor.
b)
Use source transformations to convert the current source
and resistor to a voltage source and resistor.
c)
Define the inductive voltage.
d)
Define the capacitive current.
e)
Define the inductive current.
f)
Define the capacitive voltage.
Dave Shattuck
University of Houston
Your choice for First Step –
© Brooks/Cole Publishing Co.
Use Ohm’s Law to find the voltage across the resistor
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
RX=100[W]
+
vX(t)
iS(t)
1
iS (t )  5e3000[s ]t [mA]; for t  0.
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vC (0)  100[mV].
This is not a good choice for the first step.
We can certainly find the voltage across the resistor. However, it will not help us much, since we
would also need to know the voltage across the current source to be able to find vX. To find the
voltage across the current source, we need to solve the rest of the circuit, which would make this
approach unnecessary. Finding the voltage across the resistor will not help much.
Go back and try again.
Your choice for First Step –
Use
© Brooks/Cole Publishing
Co. source transformations to convert the current source and
resistor to a voltage source and resistor
Dave Shattuck
University of Houston
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
1
iS (t )  5e3000[s ]t [mA]; for t  0.
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vC (0)  100[mV].
This is not a good choice.
The current source and resistor are in series, and not in parallel. Therefore, it would be a mistake
to try to use source transformations here. We might recognize that the resistor is going to have no
effect on anything here, except for the voltage across the current source, which we don’t need.
Please go back and try again.
Your choice for First Step –
Define the inductive voltage
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
1
iS (t )  5e3000[s ]t [mA]; for t  0.
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vC (0)  100[mV].
This is a good choice for the first step, and the one that we will choose here.
Our goal here will be to find the inductive voltage and capacitive voltage, and use them with
KVL to get vX. The capacitive voltage is already defined. Let’s go ahead and define the
inductive voltage.
Your choice for First Step –
Define the capacitive current
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
1
iS (t )  5e3000[s ]t [mA]; for t  0.
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vC (0)  100[mV].
This can be done, but it is not a good choice for the first step.
We are sophisticated enough about circuit analysis by this point to recognize that the current
through the capacitor is equal to the source current. We don’t need to define a new current
here.
Therefore, we recommend that you go back and try again.
Your choice for First Step –
Define the inductive current
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
1
iS (t )  5e3000[s ]t [mA]; for t  0.
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vC (0)  100[mV].
This can be done, but it is not a good choice for the first step.
We are sophisticated enough about circuit analysis by this point to recognize that the current
through the inductor is equal to the source current. We don’t need to define a new current
here.
Therefore, we recommend that you go back and try again.
Your choice for First Step –
Define the capacitive voltage
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
1
iS (t )  5e3000[s ]t [mA]; for t  0.
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vC (0)  100[mV].
This is not a good choice.
The capacitive voltage has already been defined. There is no need to define it again.
Please go back and try again.
Dave Shattuck
University of Houston
Defining the Inductive Voltage
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
RX=100[W]
+
vX(t)
iS(t)
1
iS (t )  5e3000[s ]t [mA]; for t  0.
vC (0)  100[mV].
LX=2.2[mH]
vL(t)
+
We have defined the inductive voltage. What should the second step be?
a)
Find the inductive voltage.
b)
Find the capacitive voltage.
c)
Find the resistive voltage.
d)
Find the voltage across the current source.
CX=
150[mF]
vC(t)
+
Your choice for Second Step –
Find the inductive voltage
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
RX=100[W]
+
vX(t)
iS(t)
[mA]; for t  0.
vC (0)  100[mV].
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
This is a good choice for the second step, and the one that we will choose here.
It would have been just as good a choice to start with the capacitive voltage. However, just by
arbitrary choice, we have chosen to find the inductive voltage first.
Let’s go ahead and find the inductive voltage.
Your choice for Second Step –
Find the capacitive voltage
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
RX=100[W]
+
vX(t)
iS(t)
[mA]; for t  0.
vC (0)  100[mV].
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
This is a good choice for the second step, but it is not the one that we will choose here.
It is a reasonable choice to start with the capacitive voltage. However, just by arbitrary choice,
we have chosen to find the inductive voltage first.
Let’s go ahead and find the inductive voltage.
Your choice for Second Step –
Find the resistive voltage
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
RX=100[W]
+
-
Loop A
vX(t)
iS(t)
Loop B
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
This is not a good choice for the second step.
If we were to find the resistive voltage, it would be because we were going to take KVL around
Loop A shown in this diagram. That would mean we would also need to find the voltage across
the current source. The only way to find the voltage across the current source is to find the
voltage across the resistor and vL and vC. If we found vL and vC, we could just use Loop B, and
not need the voltage across the resistor and the current source.
Please go back and try again.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Your choice for Second Step –
Find the voltage across the current source
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
RX=100[W]
+
-
Loop A
vX(t)
iS(t)
Loop B
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
vC (0)  100[mV].
This is not a good choice for the second step.
If we were to find the voltage across the current source, we would need to find the voltage across
the resistor and vL and vC. If we found vL and vC, we could just use Loop B, and not need the
voltage across the resistor and the current source.
Please go back and try again.
Dave Shattuck
University of Houston
Finding the Inductive Voltage – 1
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
To find the inductive voltage, we use the defining equation for the inductor. This equation is
given here, as a reminder. Be careful about the sign convention. Here, since vL and iS are in
the passive sign convention for the inductor, the equation has a positive sign. We have
diS (t )
vL (t )   LX
.
dt
Next step
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
We can substitute in, and get
Finding the Inductive Voltage – 2
RX=100[W]
+
CX=
150[mF]
vC(t)
vX(t)
iS(t)
-
-
+
LX=2.2[mH]
vL(t)
+
vL (t )   2.2 10
3


d  5 103  e3000t
dt
 [V].
Note that this will come out with units of [Volts], since we have used [Henries], [s], and [A] for
the other units. Now, we can perform the differentiation, and get the result in the next slide.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
Finding the Inductive Voltage – 3
RX=100[W]
+
CX=
150[mF]
vC(t)
vX(t)
iS(t)
-
-
+
LX=2.2[mH]
vL(t)
+
Performing the differentiation, we get
vL (t )  33e
3000[s1 ]t
[mV]; for t  0.
The next step is to find an expression for the capacitive voltage vC(t), for t > 0.
Dave Shattuck
University of Houston
Finding the Capacitive Voltage – 1
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
RX=100[W]
+
vX(t)
iS(t)
[mA]; for t  0.
vC (0)  100[mV].
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
To find the capacitive voltage, we use the defining equation for the capacitor. This equation is
given here, as a reminder. Be careful about the sign convention. Here, since vC and iS are in
the passive sign convention for the capacitor, the equation has a positive sign. We have
1
vC (t )  
CX

t
0
iS ( y )dy  vC (0).
Next step
Dave Shattuck
University of Houston
Finding the Capacitive Voltage – 2
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
We can substitute in, to get
RX=100[W]
+
vX(t)
iS(t)
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
1

vC (t )  
6
150

10


3
3000 y
5

10
e
dy

0.1

 [V ].
0 

t
Note that this will come out with units of [Volts], since we have used [Henries], [s], and [A] for
the other units. Now, we can perform the integration, and get the result in the next slide.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
Finding the Capacitive Voltage – 3
RX=100[W]
CX=
150[mF]
vC(t)
vX(t)
iS(t)
-
+
LX=2.2[mH]
vL(t)
+
[mA]; for t  0.
vC (0)  100[mV].
Performing the integration, we get
-
+
t


5 103
3000 y
vC (t )  
e
 0.1 [V], or

6 
0
  3000 150  10



vC (t )  0.0111 e 3000t  1  0.1 [V].
Now, we can write KVL.
Dave Shattuck
University of Houston
Writing KVL
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
RX=100[W]
+
vX(t)
iS(t)
Loop B
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
Writing KVL around Loop B, we get
vX  vC  vL  0, or
vX  vC  vL .
Now, we can get the
solution to part a).
Dave Shattuck
University of Houston
Solution to Part a)
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
RX=100[W]
+
vX(t)
iS(t)
Loop B
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
[mA]; for t  0.
vC (0)  100[mV].
Simplifying, we get the solution

v X (t )   0.0111 e

v X (t )  0.044e
3000 t
3000[s 1 ]t
 
 1  0.1  0.033e

 0.111 [V]; for t  0.
3000[s 1 ]t
 , or
Next, we do part b).
Dave Shattuck
University of Houston
Solution to Part b)
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
[mA]; for t  0.
vC (0)  100[mV].
RX=100[W]
+
vX(t)
iS(t)
Loop B
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
At this point, part b) involves simply plugging in the value for t, or
 3000[s1 ]104 [s]

v X (0.1[ms])   0.044e
 0.111 [V], or


v X (0.1[ms])   78.4  [mV].
Now for part c).
Dave Shattuck
University of Houston
Solution to Part c)
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
RX=100[W]
+
vX(t)
iS(t)
Loop B
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
[mA]; for t  0.
vC (0)  100[mV].
To get the energy stored in the inductor, we use the formula for this,
1
2
wL (0.1[ms])  LX  iS (0.1[ms])  , or
2
2
1
3
3
0.3
wL (0.1[ms])   2.2 10   5 10  e
[J]  15.1[nJ].
2


Now for part d).
Dave Shattuck
University of Houston
Solution to Part d)
© Brooks/Cole Publishing Co.
The value for the current source in
the circuit shown is given below.
The voltage across the capacitor at
t = 0 is also given.
a) Find vX(t) for t > 0.
b) Find vX(0.1[ms]).
c) Find the energy stored in the
inductor at t = 0.1[ms].
d) Find the energy stored in the
capacitor at t = 0.1[ms].
iS (t )  5e
3000[s1 ]t
RX=100[W]
+
vX(t)
iS(t)
Loop B
-
CX=
150[mF]
vC(t)
+
LX=2.2[mH]
vL(t)
+
[mA]; for t  0.
vC (0)  100[mV].
To get the energy stored in the capacitor, we use the formula for this,
1
2
wC (0.1[ms])  C X  vC (0.1[ms])  , or
2
Go to
2
1
wC (0.1[ms])  150 106  0.0111 e 0.3   0.111 [J]  792[nJ]. Comments
Slide
2


Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Do I have to remember how to
integrate?
• Yes, I am afraid that we need to remember our calculus pretty
well at this point. We are going to use it many different ways.
Reviewing calculus concepts is a good plan at several stages
during our training as engineers.
• Hopefully this will be fun. Calculus is a very powerful tool.
While it can be confusing at first (or at second, or at third), for
most people, with time, it becomes a easy to use tool to solve
key problems.
Go back to
Overview
slide.