Physics 122, Fall 2012
11 December 2012
Today in Physics 122: More examples in review
By class vote:
 Problem 29-45: DC
generator with
counter torque.
 Problem 28-43: B
f
from
““polygon”
l
”
current.
 Problem 30-100: AC
circuit with R,L,C.
11 December 2012
R
2 n
I0
Physics 122, Fall 2012
1
Problem 29-45
A DC generator is rated at 16 kW, 250 V and 64 A when it
rotates at 1000 rpm. The resistance of the armature windings
is 0.40 . Assume that the magnitude of the magnetic field is
constant
(a) Calculate the “no load” voltage at 1000 rpm – no load
meaning that no circuit is hooked up to the generator.
generator
(b) Calculate the full-load voltage (i.e. at 64 A) when the
generator is run at 750 rpm.
(c) Calculate the counter torque: the torque necessary to
produce the motor’s rated performance at 1000 rpm.
Part (c) is not in the book; it’s added for clarity, and other
reasons you’ll see momentarily.
11 December 2012
Physics 122, Fall 2012
2
Problem 29-45 (continued)
(a) The equivalent circuit of the
fully-loaded generator is at
right; the loop equation gives
us
  Ir  V  0
  V  Ir  280 V
With no load (no current
drawn), the output voltage
would be just   280 V.
r  0.4

V  250 V
I  64 A
R
The answer given for this problem in the back of the book,
2400 V, is wrong. Apparently it was worked out for a motor,
with a large (2200V) back EMF, instead of a generator.
Generators don’t have back EMF. They have counter torque.
11 December 2012
(c) University of Rochester
Physics 122, Fall 2012
3
1
Physics 122, Fall 2012
11 December 2012
Problem 29-45 (continued)
(b) If there are N windings in
the motor, and each has
area A, then the EMF is
d B
d

   NAB cos t 
dt
dt
  NAB sin
i  t  0 sin
i t ; 
r  0.4
V  250 V
I  64 A
R
0   NAB .
So
2
  
1 0 2

R
750
V 2   2 0 2 
250 V  190 V.

r  R 1000
1
0 2  
11 December 2012
Physics 122, Fall 2012
4
Problem 29-45 (continued)
(c) Again if there are N windings in the motor and each
winding has area A, then the maximum (counter-)torque I
must exert to keep the generator spinning at constant 
while it’s producing current I is
   B  NIAB
Note that this requires exertion of constant mechanical
power, as you may have guessed from conservation of
energy:
dW   d   dt
Pmech 
dW
    NIAB   NIB I  I  Pelec
dt
11 December 2012
Physics 122, Fall 2012
5
Problem 28-43
A wire is bent into the shape of a regular polygon with n
sides whose vertices are a distance R from the center. If the
wire carries a current I0,
(a) determine the magnetic field at the center;
(b) if n is allowed to become very large (n   ), show that
th fformula
the
l iin partt ((a)) reduces
d
tto th
thatt ffor a circular
i l lloop.
11 December 2012
(c) University of Rochester
Physics 122, Fall 2012
6
2
Physics 122, Fall 2012
11 December 2012
Problem 28-43 (continued)
First, and as usual, we split
the problem up into tractable
parts – the line segments –
and identify the method of
calculation – the Biot-Savart
field law.
 Each segment has length
n = 7 shown.
2 n
R
2 x0  2 R sin  n  ,
y0 R
and its midpoint lies
x0
y0  R cos  n 
I0
x0
from the center.
11 December 2012
Physics 122, Fall 2012
7
Problem 28-43 (continued)
 The segments all produce
B in the same direction:
out of the page.
 The contribution to B from
each segment can be
calculated separately
separately. (We
did a similar calculation in
class on 25 October 2012.)
Follow recipe:
 Point x along segment, y
through center of polygon
(bisecting the segment).
11 December 2012
y
dB
y0
r  r

Id
x
 x0
x0
Physics 122, Fall 2012
x
8
Problem 28-43 (continued)
 Identify an appropriate
infinitesimal current
element:
y
dB
Id   Idxxˆ ,
lying at x
x.
 Get r – r’ for this element:
y0

2
r  r   x 2  y02

r  r   xˆ cos   yˆ sin 
  xˆ
x
x 2  y02
11 December 2012
(c) University of Rochester
 yˆ
r  r
Id
 x0
x
x0
x
y0
x 2  y02
Physics 122, Fall 2012
9
3
Physics 122, Fall 2012
11 December 2012
Problem 28-43 (continued)
 Now we have all the ingredients for the B-S law, and we
simply integrate:
x
 Id  r
 r  0 I 0
dx
B1  0
xˆ    xˆ x   yˆ y0 

2
32
4
4
2
2
r  r
 x0 x   y 0



0 Iy0
zˆ
4
x0

 x0
 Substitute:
y0
sin  
2
x   y02
x
dx
2
 y02


32
cos  d  
11 December 2012

1
2
y0 2 x dx 
 x
2
 y02

32

cos  sin 2 
dx 
y0
Physics 122, Fall 2012
10
Problem 28-43 (continued)
 As x   x0  x0 ,



x0
x0
   arccos 
  arccos 
 x02  y02 
 x02  y02



 
 
    
         
2 n
2 n
2 n 2 n
 So




 2  n
sin 3  y0 d
3
2
 2  n y0 sin 
 Iy
B1  0 0 zˆ
4


0 I
 I
 
 2  n
zˆ   cos   2  n  zˆ 0 sin  
4 y0
2 y0
n
11 December 2012
Physics 122, Fall 2012
11
Problem 28-43 (continued)
 Thus the total B from all n segments is
B  zˆ
0 In
 
sin   .
2 y0
n
((b)) If n is large,
g we may
y use the small-angle
g approximation,
pp
sin  n    n ,
y0  R cos  n   R ,
whence
B  zˆ
0 I
,
2R
as expected (see lecture notes for 30 October 2012).
11 December 2012
(c) University of Rochester
Physics 122, Fall 2012
12
4
Physics 122, Fall 2012
11 December 2012
Problem 30-100
For the circuit shown below, V  V0 sin t. Calculate the
current in each element of the circuit, as well as the total
impedance. [Hint: try a trial solution of the form
I  I 0 sin t    for the current leaving the source. ]
R
V
C
11 December 2012
L
Physics 122, Fall 2012
13
Problem 30-100 (continued)
Never mind the trial solution,
because we’ll use complex
exponentials instead of trig
functions.
 Impedance first: parallel
LC has
R
V
C
1
1
1   2 LC

 iC 
ZLC i L
i L
so
Z  R  ZLC  R 
L
i L
1   2 LC
11 December 2012
Physics 122, Fall 2012
14
Problem 30-100 (continued)
 Write Z in magnitudephase form:
R
Z  Z0 ei
2

 L
 
Z0   R 2  
 
2

 1   LC  

  arctan  Im Z Re Z 
12
V
C
L


L


 arctan 

2
 R 1   LC  

 
11 December 2012
(c) University of Rochester
Physics 122, Fall 2012
15
5
Physics 122, Fall 2012
11 December 2012
Problem 30-100 (continued)
 Then, since V  V0 eit  2 
(so Re V  V0 sin  t ),
I
R
V V0 it   2 

e
Z Z0
is the total current drawn
from the voltage source,
and is the current flowing
through the resistor. Real
part:
V
Re I  0 sin t    .
Z0
11 December 2012
V
C
Physics 122, Fall 2012
L
16
Problem 30-100 (continued)
 The voltage across the parallel LC is
V0 it   2  i L
e
Z0
1   2 LC
 L V0 it  

e
1   2 LC Z0
VLC  IZLC 
 Thus the currents through C and L are
V
 2 LC V0 it   2 
IC  LC  iCVLC 
e
XC
1   2 LC Z0
V
V
V0 it   2 
1
I L  LC  LC 
e
XL
i L 1   2 LC Z0
11 December 2012
Physics 122, Fall 2012
17
Problem 30-100 (continued)
 Real parts of the currents:
V
Re I R  0 sin t    .
Z0
 2 LC V0
sin t   
1   2 LC Z0
V0
1
Re I L 
sin t   
1   2 LC Z0
Re IC  
with Z0 and  as given above.
11 December 2012
(c) University of Rochester
Physics 122, Fall 2012
18
6