UNIVERSITY OF MASSACHUSETTS DARTMOUTH
COLLEGE OF ENGINEERING
EGR 101 INTRODUCTION TO ENGINEERING I
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This assignment is due Thursday, October 23, 2013.
Use engineering paper.
Show all of your steps.
Present your work neatly and clearly.
Box your final answers.
Problem 1
Consider the following circuit that consists of series and parallel capacitors. Each
capacitor has capacitance equal to 0.047  F .
 Find the total capacitance CT in  F . Let C represent a 0.047 µF capacitor.
2C
C/2
C/2
C
Total Capacitance is 2C in series with C.
1
1
1
3

 
CT 2C C 2C
2C 20.047 F 

3
3
CT  0.031F
CT 
Problem 2
Consider the following circuit in which R1  R2  5M  and each capacitor has
capacitance equal to 12 pF.
Let C represent a 12 pF capacitor.
C/2
2C
1
2
1
5
 

C eq C 2C 2C
C eq 
2C 212 pF 

 4.8 pF
5
5
Req  R1  R2  5M  5M
Req  10M
 Draw a simple circuit with one total resistor and one total capacitor, and label
their values with proper units.
 Find the time constant in ms.
R1
10MΩ
C1
4.8pF
  RC  5  10 6 4.8  10 9 
  24ms
Problem 3
Consider a capacitor with a plate area equal to 900cm 2 . The plate spacing is 0.01 mm.
The design requires a capacitance of 923.9 nF. In order to select the dielectric material,
calculate the following:
 The material permittivity  in F/m, and
 The material relative permittivity  r .
 What material do you select from Table 12.1 on p. 368 based on your relative
permittivity?
 By what factor should you increase or decrease the capacitor spacing if the
capacitor is 923.9 pF instead of 923.9 nF?
C
A

d
Cd
r 
0 A
 0 r A
d
 1m 
923.9  10 12 F 0.01mm

1000mm 

r 
2

2  1m
12  F 
8.85  10  900cm  

m
 100cm 
9.239  10 12
r 
 11.6
796.5  10 15
   r  0  11.68.85  10 12   102.66  10 12 
F

m
From Table 12.1 on page 368 of the text, Tantalum Oxide has a relative permittivity
of 11.6.
C
A
d
2
F

2  1m
102.66  10  900cm  

A
m
100cm 


d

 10  10 3 m 
12
C
923.9  10 F 
d  10mm
12
The spacing between the plates needs to be increased from 0.01 mm to 10 mm. This
is 1000 times the separation distance in the original capacitor. This makes sense,
since the capacitance changed by a factor of 1000.
Problem 4
Consider the following circuit in which the SPDT switch is set to discharge the capacitor
and has been closed for a long time. The capacitor is 10 F and R2  1k  .
 Use Ohm’s law to find R3 in order to limit the initial discharge current to 5 mA.
 Report in ms the time constant of the circuit in the discharge mode.

In the discharge mode, the voltage across the capacitor is given by
vC  t   VS e

t

where   RC .
t
V 
i t    S e RC
R
VS
VS

 5mA
R R2  R3
10V
 5mA
R23
R23  2k
R3  R23  R2  2k  1k
R3  1k
  RC  ( R2  R3 )C  2k10F 
  20ms
 Find an expression for the current as a function of time, i  t  .
iC
dvC
dt
vC t   VS e

t

dvC t 
V
 S
dt

 t 
e 




 VS t
dvc
iC
 C  
e
dt
 RC




t
V S 
e
R
V
i0    S
R
it   
 Plot i  t  in the interval t  0 using Excel with a time interval equal to 5 103 s
and fill in the first 25 rows.
a. Submit a copy of the graph.
b. Submit a print of the table that consists of two columns: the left column
containing time and the right column containing current values.
t
i(t)
sec
Amperes
0
-0.005
0.005
-0.003894004
0.01
-0.003032653
0.015
-0.002361833
0.02
-0.001839397
0.025
-0.001432524
0.03
-0.001115651
0.035
-0.00086887
0.04
-0.000676676
0.045
-0.000526996
0.05
-0.000410425
0.055
-0.000319639
0.06
-0.000248935
0.065
-0.000193871
0.07
-0.000150987
0.075
-0.000117589
0.08
-9.15782E-05
0.085
-7.13212E-05
0.09
-5.5545E-05
0.095
-4.32585E-05
0.1
-3.36897E-05
0.105
-2.62376E-05
0.11
-2.04339E-05
0.115
-1.59139E-05
0.12
-1.23938E-05
0.125
-9.65227E-06
0.13
-7.5172E-06