Download PHY 104 Exam #2

Jaquin
Spring 2011
PHY 104 Exam #2
Answer all the questions that appear below.
1) A 5.00 pF parallel plate capacitor with circular plates is to be used in a circuit in which it will be
subjected to potentials up to 200 V. The electric field between the plates is no to exceed 50,000
N/C.
A) What should the separation of the plates be?
q
Use the relationship between Voltage and Electric Fields for

parallel plate capacitors. The work done by the E field to move
a test charge q across the plates is W  qE  d  qV. Therefore,
V
200 V
 .004 m . The plates should be separated by just 4 mm.
V  E  d and d  
E 50,000 N
C
B) What charge will the plates hold at maximum potential?
Q
Use the definition of capacitance C  Max and solve for QMax. The charge on the capacitor at
VMax
maximum potential is QMax  C  VMax  5  1012 F  200 V  1  109 C  1 nC .
2) What is the net capacitance of the following capacitor network?
8.5 F
2.4 F
4.6 F
3.1 F
First collapse the two capacitors in parallel using CEQ  C1  C2  8.5 F  3.1 F  11.6 F
2.4 F
4.6 F
11.6
F
Jaquin
Spring 2011
Next collapse the series of three capacitors using
1
1
1
1
1
1
1
. The
 




CEQ C1 C2 C3 2.4 F 4.6 F 11.6 F
value of the CEQ works out to be 1.39 F.
3) A copper wire at 20C has a diameter of 2.54 mm , a length of 25 cm and carries 5.00 A of current.
How many electrons pass any given point in the wire over a 4 second period?
There is much more information given in this problem that what is needed. WWe only need to know the
current.
I = 5A
By definition I 
q
t
. Therefore the charge that passes past some point in the wire after 4 seconds is
given by q  I  t  5 A  4 s  5
C
1e


 4 s  20 C  20 C  
 1.25  1020 e .
-19

s
1.60  10 C 
Thus 1.2510 20 electrons pass any given point in the wire over a 4 second period.
4) A lead cylinder with length L and radius r has a resistance of R.
The cylinder is dropped from a great height onto a solid surface
and compresses its length by 15%. Assuming that the deformed
lead is still has a cylindrical shape; by what percentage did the
resistance change?
L
L'
The picture to the right illustrates the shape of the cylinder before and
after dropping. The volume of the cylinder, V=L∙A, remains
L A L A
unchanged. Therefore, L∙A= L'∙A'. Thus A 

 1.18 A .
.85L
L
The resistance of the cylinder across its ends is given by R 
L
A
A=πr2
before the fall and R 
A'=πr'2
 L
A
after the
fall. Examining the resistance after the fall R', we see that
L  .85L   .85  L
L
R 


 0.720
 0.720 R

A 1.18 A  1.18  A
A
Thus the resistance of the cylinder decreases by 28% or is 72% of its original resistance.
5) 26 gauge copper magnet wire has a diameter of 0.4 mm. What length of this wire is needed to have
a resistance of 25 Ohms at room temperature (20C)?
A = πr2
Use the relation for resistance R 
L
A
and solve for L.
Jaquin
Spring 2011
L
A R

  0.0002 m  25   187 m
2

1.68  10-8  - m
The copper wire would need to be 187 m long to have a resistance of 25 Ohms.
6) What would be the resistance of this copper wire if it were cooled to liquid nitrogen temperatures of
-200C?
We can use the temperature dependent relation for resistance, R  R0  1    T  , where  is the
temperature coefficient of resistivity in units of C-1. For copper the temperature coefficient of
resistivity is 0.0039 C-1. So the new resistance will be
R  R0  1    T 

 25   1  0.0039

  220 C 
 25   1  0.0039 C-1   200 C - 20 C

C
-1


 25   0.142
 3.55 
The resistance of the copper wire drops to 3.55 Ohms at -200 Celsius.
7) With a 1,500 M resistor across its terminals, the terminal voltage of a certain battery is 2.50 V.
With only a 5.00  resistor across its terminals, the terminal voltage is 1.75 V. What is the internal
emf and the internal resistance of this battery?
a) This is a bit of a trick question in that the 1,500 M (1x109 ) resistor acts essentially like an
open circuit (i.e. infinite resistance). Thus the internal emf of the battery is just 2.50 V since
there is no appreciable internal resistance loss due to
1
the astronomically high 1,500 M  resistor.
Battery
b) The internal resistance can be solved if you know the
terminal emf (1.75 v) and the current through the
r
battery using E - Ir = Terminal Voltage, where E is the
I R=5
I
internal emf of the battery and r is the internal
E=2.5 v
resistance of the battery. The potential between point s
1 & 2 (a.k.a. the terminal voltage) in the circuit as
shown in the diagram is 1.75 V. Therefore, from
Ohm’s Law, there is I = V/R = (1.75 v)/ (5 ) = 0.35 A
flowing through the resistor and also flowing through the battery. Thus we can solve for the
internal resistance
E  Ir  VTerminal
r
E  V 2.5 v - 1.75 v

 2.14 
I
0.35 A
Thus the internal resistance of the battery is 2.14 Ohms.
2
Jaquin
Spring 2011
8) A toaster using a NichromeTM heating element operates on 120 V. When it is switched on it will
draw a steady current of 1.4 A. How much energy is consumed by the Toaster over a toasting cycle
of 1.5 minutes?
Power P  V  I  120 v  1.4 A  168 Watts
Work(or Energy)
Power 
Time
J
 Energy E  P  t  168 Watts  90 seconds  168  90 s  15,120 Joules
s
The energy is consumed by the Toaster over a toasting cycle of 1.5 minutes is 15,120 Joules.
9) Calculate the equivalent resistance of the following resistance network.
20 Ω
20 Ω
20 Ω
20 Ω
20 Ω
20 Ω
20 Ω
First, re-draw the circuit so as to resemble the circuit at the left. Then
combine the two resistors in parallel. Their equivalent resistance is 10
.
Next, combine the two pairs of series resistors. The two on top will
have an equivalent resistance of 30  and the two on the bottom will
have an equivalent resistance of 40 .
Next, combine the two new parallel resistors. Their equivalent
resistance will be 17.14 
Finally, combine the three series resistors into a single equivalent
resistance of 57.14 
Jaquin
Spring 2011
10) Calculate the current flowing through the 4 Ω resistor given that the current through the 1 Ω resistor
is 3 A as shown, the current through the 2 Ω resistor is -3 A as shown, and the battery supplies 90 C
of charge every 20 seconds as indicated.
5Ω
3Ω
1Ω
V
2Ω
4Ω
This problem is much simpler than it first appears. The key is to simply solving this is to interpret the
statement “the battery supplies 90 C of charge every
20 seconds as indicated”. This means that the
90 C
 4.5 A flowing
battery has a current of
1.5 A
1.5 A
20 s
3.0 A
6.0 A
through it in the upward direction indicated in the
figure. Thus the currents at the junction just below
3Ω
1Ω
the 2  resistor are as shown below in red. The
V
2Ω
4Ω
3.0 A
4.5 A
unknown current must be 7.5 A according to
I=?
Kirchhoff’s Junction Rule. This is also the current
4.5 A
through the 4  resistor.
11) Determine the current through each of the resistors in the circuit shown below and identify the
resistor that dissipates the most power.
R1
I1
V2
R3
V1
I2
R4
V1
V2
R1
R2
R3
R4
R3
This is a full blown Kirchhoff’s Loop Law problem.
appear below.
The loop equations for the current directions I have drawn above
10 V
6V
10 Ω
20 Ω
2Ω
8Ω
Jaquin
Spring 2011
V2  I1R1  I1  I 2 R2  0
V1  I1  I 2 R1  I 2 R3  R4   0
Redistributing terms…
I1 R1  R2   I 2 R2  V2
I1R2  I 2 R2  R3  R4   V2
Substituting in values and dropping units …
I1 R1  R2   I 2 R2  V2
I1R2  I 2 R2  R3  R4   V2
I obtained solutions of the two equations for I1 and I2 of …
I1  0.04 A
I 2  0.36 A
The current through each resistor is then…
Resistor
R1=10 
R2=20 
R3=2 
R4=8 
The R2 resistor draws the most power.
Power (I2R)
0.016 W
3.20 W
0.26 W
1.04 W
Current
0.04 A to Right
0.40 A to Right
0.36 A to Left
0.36 A Down
12) An uncharged 5 µF capacitor is connected in series with a 50 Ω resistor and a 12.0 V battery.
R
V
C
Switch
A) What is the time constant of the RC circuit?


The time constant is given by   RC  5  106 F  50   2.5  104 second  0.25 ms .
B) Sketch the graphs of current in the circuit and the charge on the capacitor as functions of time.
Jaquin
Spring 2011
I0
Q0
-t/RC
I(t) = I0e
I
Q
Q(t) = Q01-e-t/RC
t
t
C) What will be the charge on the capacitor 0.5 ms after the charging begins?
t


The governing equation is Q(t )  Q0  1  e RC  . The maximum charge, which would be obtained


only after a long time (greater than 10 time constants), is calculated from
Q0  C  V  5  106 F  12 v  60 C .
Thus at t = 0.5 ms, the charge on the capacitor is
0.5 ms


Q(0.5 ms )  60 C  1  e 0.25 ms   60 C  1  e  2  60 C  1  0.135  51.9 C


Thus the capacitor has 51.6 C of charge or 86% of its final charge after 0.5 ms.


D) What will be the potential across the resistor at 0.5 ms after charging begins?
Use Ohm’s Law, V=IR, to calculate the potential across the resistor. However, first the current must
t
be found using the governing equation for RC circuits I (t )  I 0  e RC . The constant I0 is the
maximum current that is experienced immediately before the capacitor has time to charge.
V 12 v
I0  
 0.24 A
R 50 
Thus at t = 0.5 ms, the current through the resistor is
I (0.5 ms )  0.24 A  e
0.5 ms
0.25 ms
 0.24 A  e 2  0.24 A  0.135  0.0325 A
Thus the potential across the resistor is 32.5 mA or 13.5% of its initial maximum value after 0.5 ms.