# Final_MATH-Solutions

```Solutions to Final Exam for GP II,
MATH,
Spring 2012.
1. 10%
A network of resistors of resistance R1 and R2
extends to infinity towards the right (see
figure). Find the total resistance of the
network.
Hint:
What is the resistance of the network to the right of points c and d ?
Solution:
Let the total resistance be RT . Since the network is infinite, the resistance of the
network to the right of points c and d is also RT .

RT  R1 
2 R1R2   2 R1  R2  RT
RT R2
 R1 
RT  R2
RT  R2
RT2  2R1RT  2R1R2  0
RT  R1 
R12  2 R1R2
RT  R1 
R12  2 R1R2
since all R  0
2. ( 10% + 10% + 10% )
(a) What is Vab  Vb  Va when switch S is open?
(b) What is the current through switch S when it
is closed?
(c) What is the equivalent resistance when
switch S is closed?
Solution:
(a) Applying the voltage divider rule, we have
Va 
3
 36 V  12 V
 6  3 
Vb 
6
 36 V  24 V
 3  6 

Vab  24 V  12 V  12 V
(b)
For the upper loop:
6 I 1  3I  3I 2  0
For the lower loop:
3  I  I 1   3I  6  I  I 2   0
For the left leg:
6I 1  3  I1  I   36
For the left leg:
6I 1  3  I1  I   36
Simplifying gives
2I 1  I  I2  0
I1  4 I  2 I 2  0
3I 1  I  12
1st 2 eqs gives
3I 1  6 I  0

or
Current through S is
(c)
1
12  24
I 1   12   
A
3
7 7
I 1  2I  0
I 
12
A
7
I2  2
24 12 36


A
7
7
7
R
Equivalent resistance:
V
36
21



I1  I 2 24  36 5
7
7
3. ( 5% + 10% )
A long, straight, solid cylinder, oriented with its axis in the z-direction, carries a
current density J given by
 2 I   r 2 
 02 1     kˆ
J    a   a  

0

for r  a
for r  a
where a is the radius of the cylinder, r is the radial distance from the axis, and I0 is
a constant.
(a) Show that I0 is the total current.
(b) Find the magnetic field inside and outside the cylinder.
Solution:
(a)
Total current = 2
a

0
(b)
Ampere’s law

C
2I0
 a2
  r 2 
4I0  1 2 1 1 4 
1     r d r  2  a  2   a   I 0
a 2
a 4

  a  
B  d r  0 I .
To calculate the field at a pointof distance r from the axis, choose C to be a circle
of radius r , centered on the axis and passing through that point. Then
I
B  0 enclosed
2 r
For r &lt; a, the enclosed current is
I  2
r

0
2I0
 a2
4I
 20
a
B
  r 2 
1     r d r
  a  
2
2
1 2 1 1 4
r  r 
 2 r  a 2  4  r   I 0  a   2   a  


 0 I 0 I 0 r   r 

2   
2 r 2 a 2   a 
2



For r &gt; a, the enclosed current is I0 .
I
B 0 0
2 r
4. ( 15% + 10% )
Figure shows a small circuit within a larger one, both
lying on the surface of a table. The switch is closed
at t = 0 with the capacitor initially uncharged.
Assume that the small circuit has no appreciable
effect on the large one.
(a) What is the direction ( a to b or b to a ) of
the current in the resistor r
i. The instance after the switch is closed?
ii. One time constant after the switch is closed?
(b) Sketch a graph of the current in the small circuit as a function of time, calling
clockwise as positive.
Solution:
(a)
i. The instance after S is closed, I in the large circuit goes CCW to charge C.
This creates B pointing out of the paper. Magnetic flux  through the
small circuit thus jumps from 0 to a finite value, thus producing a
negative emf so that the current in r goes CW (from a to b).
ii. One time constant after S is closed, I in the large circuit decreases
exponentially towards 0 as C becomes more charged.  through small
circuit thus decreases with time, producing a positive emf so that the
current in r goes CCW (from b to a).
(b)
5. 20%
Vs in the figure is a signal source.
Derive an expression for the
amplitude ratio Vout / Vs as a
function of the angular
frequency  of the source. Find the limits for
0
and
Solution:
Vs 
Q
dI
L
 IR  0
C
dt
For a sinusoidal signal of frequency ,
I
Vs 
 i  LI  I R  0
iC
Voltage divider:
1
Vout 
iC
Vs
 1

R i
 L
 C

Amplitude ratio:
1
Vout

Vs

1
iC

2
 1

R i
  L   C R2    1   L 
 C

 C



1
 C R 
2
For
0,
Vout
 1.
Vs
For
,
Vout
 0.
Vs
  1   2 LC 
2
.
```