Problems With Assistance
Module 1 – Problem 1
Filename: PWA_Mod01_Prob01.ppt
Go
straight to
the First
Step
Go
straight to
the
Problem
Statement
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
In this problem, we will use the following
concepts:
 Power and energy
 Sign conventions
 Changing directions of energy flow
Overview of this Problem
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The material for this problem is covered in your
textbook in the following sections:
 Circuits by Carlson: Sections 1.1 & 1.2
 Electric Circuits 6th Ed. by Nilsson and Riedel:
Sections 1.5 & 1.6
 Basic Engineering Circuit Analysis 6th Ed. by
Irwin and Wu: Sections 1.2 & 1.3
 Fundamentals of Electric Circuits by Alexander
and Sadiku: Sections 1.3 through 1.6
 Introduction to Electric Circuits 2nd Ed. by Dorf:
Sections 1-2, 1-4 & 1-5
Textbook Coverage
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The material for this problem is covered in
this module in the following
presentations:
 DPKC_Mod01_Part01
 DPKC_Mod01_Part02
Coverage in this Module
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Problem Statement
A blender
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
Wall
Blender
at t = 0.
Socket
iB(t)
b)
Find the power absorbed by the blender
+
at t = 4.3[ms].
c)
Find the power absorbed by the blender
vS(t)
at t = 8.6[ms].
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
-
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Problem Statement (Preliminary Notes 1)
The coefficient of time in the argument of a sinusoid is
typically assumed to have units of [rad/s]. This is
The voltage in a wall socket has been usually
measured,
and
to be
equal to
not shown,
but found
is given here
for clarity.
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
Wall
Blender
at t = 0.
Socket
iB(t)
b)
Find the power absorbed by the blender
+
at t = 4.3[ms].
c)
Find the power absorbed by the blender
vS(t)
at t = 8.6[ms].
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
-
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Problem Statement (Preliminary Notes 2)
I have enclosed all units in square brackets. This is
not a convention, and you are not required to do so.
The voltage in a wall socket has been However,
measured,
and
found
to adds
be to
equal
it sets off
units,
and thus
clarityto
for
these problems.
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
Wall
Blender
at t = 0.
Socket
iB(t)
b)
Find the power absorbed by the blender
+
at t = 4.3[ms].
c)
Find the power absorbed by the blender
vS(t)
at t = 8.6[ms].
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
-
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Solution – First Step – Where to Start?
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
Wall
Blender
at t = 0.
Socket
iB(t)
b)
Find the power absorbed by the blender
+
at t = 4.3[ms].
c)
Find the power absorbed by the blender
vS(t)
at t = 8.6[ms].
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
How should we start this problem? What is the first step?
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Problem Solution – First Step
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
Wall
Blender
b)
Find the power absorbed by the blender Socket
iB(t)
at t = 4.3[ms].
+
c)
Find the power absorbed by the blender
at t = 8.6[ms].
vS(t)
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
How should we start this problem? What is the first step?
a)
Integrate the voltage over one period.
b)
Integrate the current over one period.
c)
Find an expression for the power with the right sign.
d)
Find the period of the sinusoids.
-
Dave Shattuck
University of Houston
Problem Solution – First Step -© Dept. of Elect. & Comp. Engr.
over one period.
Integrate the voltage
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
Wall
Blender
b)
Find the power absorbed by the blender Socket
iB(t)
at t = 4.3[ms].
+
c)
Find the power absorbed by the blender
at t = 8.6[ms].
vS(t)
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
This is not a good choice for a first step. While we have to
integrate power to get energy, integrating voltage does
not give us anything we can use in this problem.
Go back to Problem Solution – First Step.
-
Dave Shattuck
University of Houston
Problem Solution – First Step -© Dept. of Elect. & Comp. Engr.
over one period.
Integrate the current
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
Wall
Blender
b)
Find the power absorbed by the blender Socket
iB(t)
at t = 4.3[ms].
+
c)
Find the power absorbed by the blender
at t = 8.6[ms].
vS(t)
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
This is not a good choice for a first step. While we have to
integrate power to get energy, integrating current, while
it does give us total charge, does not give us anything
we can use in this problem.
Go back to Problem Solution – First Step.
-
Dave Shattuck
University of Houston
Problem Solution – First Step -© Dept. of Elect. & Comp. Engr.
sinusoid.
Find the period of the
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
Wall
Blender
b)
Find the power absorbed by the blender Socket
iB(t)
at t = 4.3[ms].
+
c)
Find the power absorbed by the blender
at t = 8.6[ms].
vS(t)
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
It would be useful to know the period of the sinusoid at some
point in this problem, but not at the beginning. In Part
d) we will use this, but it is not crucial at the beginning.
It would be better to start someplace else.
Go back to Problem Solution – First Step.
-
Dave Shattuck
University of Houston
Find an expression for
the power with the right sign.
Problem Solution – First Step --
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
Wall
Blender
b)
Find the power absorbed by the blender Socket
iB(t)
at t = 4.3[ms].
+
c)
Find the power absorbed by the blender
at t = 8.6[ms].
vS(t)
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
This is a good choice for a place to start. Once we have this
expression, we can start finding things of interest.
Now, what should the correct expression be?
-
Next slide
Dave Shattuck
University of Houston
Power – Using the Proper Sign
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
Wall
Blender
b)
Find the power absorbed by the blender Socket
iB(t)
at t = 4.3[ms].
+
c)
Find the power absorbed by the blender
at t = 8.6[ms].
vS(t)
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
We can write a power expression for either the blender or for
the wall socket. It makes sense to write the expression
for the blender, since all the questions relate to the
blender. We would write:
pabs ,bl (t )  vS iB
-
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
Next slide
Dave Shattuck
University of Houston
Solution for Part a)—Step 1
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
For Part a), the solution is obtained by substituting in t = 0 in
this expression, and evaluating. Go ahead and do this,
and write down your answer.
Wall
Socket
Blender
iB(t)
+
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
vS(t)
-
Next slide
Dave Shattuck
University of Houston
Solution for Part a) — Evaluating
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
For Part a), the solution is obtained by substituting in t = 0 in
this expression, and evaluating. Go ahead and do this,
and write down your answer.
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
For your answer, did you get:
Answer 1: pabs,bl (0[s])  750[W]?
Answer 2: pabs,bl (0[s])  747[W]?
Something else?
Wall
Socket
Blender
iB(t)
+
vS(t)
-
Your Solution for Part a) was
750[W]
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
For Part a), the solution is obtained by substituting in t = 0 in
this expression, and evaluating. Go ahead and do this,
and write down your answer.
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
If your answer was
pabs,bl (0[s])  750[W]
then you have made an error. You have neglected to
find the cos (-5o) when you evaluated, or made some
error getting it. Try again.
Wall
Socket
Blender
iB(t)
+
vS(t)
-
Your Solution for Part a) was
Something Else
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
For Part a), the solution is obtained by substituting in t = 0 in
this expression, and evaluating. Go ahead and do this,
and write down your answer.
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
If your answer was not
pabs,bl (0[s])  747[W]
then you have made an error. You could try again, or
you can look at the solution.
Wall
Socket
Blender
iB(t)
+
vS(t)
-
Your Solution for Part a) was
747[W]
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
a)
Find the power absorbed by the blender
at t = 0.
For Part a), the solution is obtained by substituting in t = 0 in
this expression, and evaluating.
If your answer was
pabs,bl (0[s])  747[W]
Wall
Socket
Blender
iB(t)
+
vS(t)
then you have the correct solution. What you did was
pabs ,bl (0)  163cos(377[rad/s]0) 4.60cos(377[rad/s]0  5 ) 
-
163cos(0) 4.60cos(0  5 ) 
1631 4.60  0.996   747[W].
Next part (b)
Dave Shattuck
University of Houston
Solution for Part b) — Step 1
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
b)
Find the power absorbed by the blender
at t = 4.3[ms].
For Part b), the solution is obtained by substituting in t =
4.3[ms] in this expression, and evaluating. Go ahead and
do this, and write down your answer.
Wall
Socket
Blender
iB(t)
+
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
vS(t)
-
Next slide
Dave Shattuck
University of Houston
Solution for Part b) — Evaluating
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
b)
Find the power absorbed by the blender
at t = 4.3[ms].
For Part b), the solution is obtained by substituting in t =
4.3[ms] in this expression, and evaluating. Go ahead and
do this, and write down your answer.
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
For your answer, did you get:
Answer 1: pabs,bl (4.3[ms])  1.39[W]?
Answer 2: pabs,bl (4.3[ms])  748[W]?
Something else?
Wall
Socket
Blender
iB(t)
+
vS(t)
-
Your Solution for Part b) was
748[W]
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
b)
Find the power absorbed by the blender
at t = 4.3[ms].
For Part b), the solution is obtained by substituting in t =
4.3[ms] in this expression, and evaluating. Go ahead and
do this, and write down your answer.
Wall
Socket
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
Blender
iB(t)
+
vS(t)
If your answer was
pabs,bl (4.3[ms])  748[W]
then you have made an error. You have neglected to convert radians to
degrees. Note that when you plug in 4.3[ms] for t, and multiply by 377, the
result will have units of [rad]. You need to convert to degrees, or use
radians as your units on your calculator. Note that you will also need to
convert 5 degrees to radians if you take this approach. Try again.
-
Your Solution for Part b) was
Something Else
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
b)
Find the power absorbed by the blender
at t = 4.3[ms].
For Part b), the solution is obtained by substituting in t =
4.3[ms] in this expression, and evaluating. Go ahead and
do this, and write down your answer.
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
If your answer was not
pabs,bl (4.3[ms])  1.39[W]
then you have made an error. You could try again, or
you can look at the solution.
Wall
Socket
Blender
iB(t)
+
vS(t)
-
Your Solution for Part b) was –
1.39[W]
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
b)
Find the power absorbed by the blender
For Part b), the solution is obtained by substituting in t = 4.3[ms]
at t = 4.3[ms].
in this expression, and evaluating.
If your answer was
pabs,bl (4.3[ms])  1.39[W]
then you have the correct solution. What you did was
pabs,bl (4.3[ms]) 







  377[rad/s] 4.3 103[s] 180 
  377[rad/s] 4.3  103[s] 180  



)   4.60cos  (
)  5
163cos  (


 [rad]
 [rad]







163cos(92.88 )4.60cos(92.88  5 ) 





163 0.0502  4.60  0.0370   1.39[W]
Next part (c)
Dave Shattuck
University of Houston
Solution for Part c) — Step 1
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
c)
Find the power absorbed by the blender
at t = 8.6[ms].
For Part c), the solution is obtained by substituting in t =
8.6[ms] in this expression, and evaluating. Go ahead and
do this, and write down your answer.
Wall
Socket
Blender
iB(t)
+
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
vS(t)
-
Next slide
Dave Shattuck
University of Houston
Solution for Part c) — Evaluating
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
c)
Find the power absorbed by the blender
at t = 8.6[ms].
For Part c), the solution is obtained by substituting in t =
8.6[ms] in this expression, and evaluating. Go ahead and
do this, and write down your answer.
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
For your answer, did you get:
Answer: pabs,bl (8.6[ms])  746[W]?
Something else?
Wall
Socket
Blender
iB(t)
+
vS(t)
-
Your Solution for Part c) was
Something Else
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
c)
Find the power absorbed by the blender
at t = 8.6[ms].
For Part b), the solution is obtained by substituting in t =
8.6[ms] in this expression, and evaluating. Go ahead and
do this, and write down your answer.
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W]
If your answer was not
pabs,bl (8.6[ms])  746[W]
then you have made an error. You could try again, or
you could look at the solution for Part b) which is very
similar, or you can look at the solution for Part c).
Wall
Socket
Blender
iB(t)
+
vS(t)
-
Your Solution for Part c) was
746[W]
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
c)
Find the power absorbed by the blender
For Part c), the solution is obtained by substituting in t = 8.6[ms]
at t = 8.6[ms].
in this expression, and evaluating.
If your answer was
pabs,bl (8.6[ms])  746[W]
then you have the correct solution. What you did was
pabs,bl (8.6[ms]) 







  377[rad/s] 8.6 103[s] 180 
  377[rad/s] 8.6  103[s] 180  



)   4.60cos  (
)  5
163cos  (


 [rad]
 [rad]







163cos(185.76 )4.60cos(185.76






5 ) 
163 0.9950  4.60  0.9999   746[W]
Next part (d)
Dave Shattuck
University of Houston
Solution for Part d) — Step 1
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
For Part d), the solution is obtained by integrating the power
over a single period. The power was
pabs ,bl (t )  vS iB
 163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 )[W].
Thus, we want to integrate this over one period. Any period
will be fine. Generally, we want to use the easiest
possible period. This is usually the period from t=0 to t
equal to one period. Let T be the period, and we write
T
T
0
0
Wall
Socket
Blender
iB(t)
+
vS(t)
-
wabs ,bl   pabs ,bl (t )dt   vS iB dt.
To do this, we need to find the period, T.
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Solution for Part d) —
Finding the Period
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
d)
Find the energy absorbed by the blender
one
cycle
of the
sinusoid.
We need toover
find the
period,
T. Note
that we
have the frequency
in terms of [rad/s]. This is called the angular frequency.
To get the period from this, we convert to frequency in
[cycles/s], or [Hertz], or [Hz]. Then we take the inverse.
Or, to help us remember, we can set it up as a series of
unit conversions. We can write
1[s]
2 [rad]
2 [s]


 T  16.67[ms].
377[rad] 1[cycle] 377[cycle]
Note that the [cycles] are usually left out of this result.
Symbolically, we can write
2

Wall
Socket
Blender
iB(t)
+
vS(t)
-
 T ; where  is the angular frequency.
Next slide
Solution for Part d) — The
Integral
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
So, for Part d), we need to integrate the power over a single
period. We have the integral
T


Blender
iB(t)
T
wabs,bl   pabs,bl (t )dt   vS iB dt
0
16.67[ms]
Wall
Socket
+
0


163cos(377[rad/s]t ) 4.60cos(377[rad/s]t  5 ) dt.
vS(t)
0
To do this integral, we have many choices. It is complicated
enough so that I am going to choose to use a computer
mathematics tool. There are many, but I am going to use
Mathcad, a Math Soft product.
-
Next slide
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Mathcad Evaluation
We have been given the following information.
We want to integrate these over a period. To make this easier, we convert
5[degrees] to radians. We use the fact that [radians]=180[degrees] to write the
following.
5 deg 
3.1416 rad
 0.08727 rad
180 deg
Next we define the variables for Mathcad, and then evaluate. Here, we have gone
ahead and checked our calculations for Parts a), b), and c), and then Part d).
vS ( t )
163 cos ( 377 t )
iB( t )
4.60 cos ( 377 t
0.08727)
p( t ) vS ( t )  iB( t )
answer for part a)
p( 0)  746.947
answer for part b)
p( 0.0043)  1.393
answer for part c)
p( 0.0086)  745.942
answer for part d)
2
This is shown as an example only. It involves
several kinds of notation that are special
for Mathcad. The point is that you should
find some tool to perform mathematical
computations. This could be a computer
program, an advanced calculator, or
fundamental calculus skills.
3 .14 16
3 77
p( t ) d t  6 .22 4
0
Next slide
Dave Shattuck
University of Houston
Solution for Part d) – Completed
© Dept. of Elect. & Comp. Engr.
The voltage in a wall socket has been measured, and found to be equal to
vS (t )  163cos(377[rad/s]t )[V].
A blender is connected to this wall socket, and the current in the blender is
measured and found to be
iB (t )  4.60cos(377[rad/s]t  5 )[A].
The diagram shows the voltage and current polarities in this connection.
d)
Find the energy absorbed by the blender
over one cycle of the sinusoid.
Thus, when we integrate the power over a single period, we get
T
T
0
16.67[ms]
0
wabs ,bl   pabs ,bl (t )dt   vS iB dt

 163cos(377[rad/s]t )4.60cos(377[rad/s]t  5 ) dt 
Wall
Socket
Blender
iB(t)
+
vS(t)
0
6.224[J].
-
The answer is wabs,bl = 6.224[J].
Some notes
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Note 1: Why was the sign of
the power changing?



In this problem, the blender is absorbing positive power
most of the time, but is actually delivering positive
power to the wall socket part of the time.
This is possible because of the inductive nature of the
motor in the blender. As we shall see later, energy can
be stored in an inductor, and delivered back to the
circuit later.
However, energy flow in inductors is not an important
concept right now. What is important right now is that
we need to be very careful to use the correct signs in
our power and energy expressions.
Next note
Go back to
Overview
slide.
Dave Shattuck
University of Houston
© Dept. of Elect. & Comp. Engr.
Note 2: Are all the integrals
this hard?
In this problem, the integral is complicated enough so that it
might not be included in the integral table in your textbook.
However, most mathematicians would suggest that it is not
that complicated, as integrals go.
 Whatever. The key here is that with present tools, even a
complicated integral can be solved relatively easily, so there is
no reason to fear them.
 In addition, later in this subject you may be introduced to
another technique for solving this kind of problem. The
technique is called complex power. Since these kinds of
problems are common with sinusoidal power sources, like wall
sockets, engineers have come up with this powerful approach
that makes these problems easier. This can be something to
look forward to.

Next note
Go back to
Overview
slide.
Dave Shattuck
University of Houston
Note 3: Why was the phase
of the current given in
degrees, and not radians?
© Dept. of Elect. & Comp. Engr.
In this problem, one of the major complications was that the
coefficient of time was given in [rad/s], and the phase of the sinusoid
was given in [degrees]. If the phase had been given in [rad] from the
beginning, the problem would have been easier.
 True enough. However, in later techniques in the circuits area, it is
convenient to use the phase in degrees, since it is a common unit for
angles. The technique is called phasor analysis. In any case, this is a
recurring problem in all of engineering.
 The only real long-term solution is to get in the habit of paying
attention to units all the time. In 1999, a very expensive space probe
that was headed to Mars was lost because of insufficient attention to
units. One must be alert to this problem. If your instructor is
emphasizing the importance of units in your course, this is at least
one reason why.

Go back to
Overview
slide.