Download PEQWS_Mod01_Prob04_v03 - Courses

Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Practice Examination Questions With Solutions
Module 1 – Problem 4
Filename: PEQWS_Mod01_Prob04.doc
Note: Units in problem are enclosed in square brackets.
Time Allowed: 20 minutes.
Problem Statement:
The circuit that follows depicts two batteries connected together. Battery A is
charging battery B, i.e., A is delivering positive power to B. Battery A has a
voltage of 16[V] across its terminals, with the polarity indicated. The power
delivered to B is a function of time, pDEL, where
pDEL  48e10t [W]; for t  0, and for t in [s].
a) Find the current iCHAR, with the polarity as indicated in the figure.
b) Find the energy delivered to battery B during the first 100[ms] after t = 0.
c) Assuming that the batteries are left connected for an effectively infinite amount
of time after t = 0, what will be the total energy delivered to battery B?
d) What percentage of the total energy is delivered in the first 100[ms]?
iCHAR
+
Battery
A
+
Battery
B
Page 1.4.1
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Problem Solution:
The problem statement was
The circuit that follows depicts two batteries connected together. Battery A is
charging battery B, i.e., A is delivering positive power to B. Battery A has a
voltage of 16[V] across its terminals, with the polarity indicated. The power
delivered to B is a function of time, pDEL, where
pDEL  48e10t [W]; for t  0, and for t in [s].
a) Find the current iCHAR, with the polarity as indicated in the figure.
b) Find the energy delivered to battery B during the first 100[ms] after t = 0.
c) Assuming that the batteries are left connected for an effectively infinite amount
of time after t = 0, what will be the total energy delivered to battery B?
d) What percentage of the total energy is delivered in the first 100[ms]?
iCHAR
+
Battery
A
+
Battery
B
Solution:
a) The first step in this problem is to write an expression for the power in one of
the two batteries, with polarities correct. To help us do this, let’s go ahead and
label the voltage right in the diagram, so that it will look the same way we are used
to seeing it. This diagram follows.
Page 1.4.2
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
iCHAR
+
16[V]
Battery
A
+
16[V]
Battery
B
Let’s choose to use battery B. For battery B the voltage (16[V]) and the
current (iCHAR) are defined in the active sign convention. We are told that the
expression for the power delivered to battery B is
pDEL  48e10t [W]; for t  0, and for t in [s].
Remember that this is the same thing as saying that battery B absorbs this much
power. So, for our notation, we can rewrite this as
pabs,batteryB  48e10t [W]; for t  0, and for t in [s].
We also know that when we use the active convention, the expression for the
power absorbed is
pabs ,batteryB  vi  16[V]iCHAR .
Therefore, we can set these expressions equal to each other, and we get
pabs ,batteryB  48e10t [W]  16[V]iCHAR .
Solving, we get
iCHAR  3e10t [A]; for t  0, and for t in [s].
b) We want the energy delivered to battery B in the first 100[ms] after t = 0. The
most direct way to get this is to integrate the power delivered, which we know.
Thus, we can write that
wabs ,batteryB  
0.1[s]
0
wabs ,batteryB 
p(t )dt  
0.1[s]
0
10 t
48e
10
0.1[s]
|
0
48e 10t dt =
 4.8  e-1  1 [J].
Page 1.4.3
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
The solution is
wabs ,batteryB  3.03[J].
Remember that in such a problem, the solution is always a definite integral. One
of the keys is to get the correct limits in the definite integral.
c) This part is essentially the same problem as in part b), except that the upper
limit of integration is infinity. Thus, we can write that


0
0
wabs ,batteryB   p(t )dt   48e10t dt =
wabs ,batteryB
48e10t 

 4.8  e-  1 [J].
|
10 0
The solution is
wabs ,batteryB  4.8[J].
d) To get the percentage of the energy delivered in the first 100[ms], we need only
use the solution of part b), as a percentage of the solution of part c). We have
% wabs ,1st 100[ms] 
3.03
100%.
4.8
The solution is
% wabs ,1st 100[ms]  63.2%.
Note: Some students would contend that we have defined the reference polarity of
the current in this problem incorrectly. This is not the way we want to think about
these problems. By our definitions, one cannot define reference polarities
incorrectly. What we have done is define the reference polarity opposite from the
actual polarity. This is not incorrect; it simply results in a negative answer. We
need to become comfortable with this kind of result.
Problem taken from Quiz 1, Spring 1994, University of Houston, Department of Electrical and Computer Engineering, Cullen College of
Engineering.
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