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:
30 100: AC
circuit with R,L,C.
11 December 2012
R
2 n
I0
Physics 122, Fall 2012
1
Problem 29
29-45
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
t t
(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.
rpm
(c) Calculate the counter torque: the torque necessary to
produce the motor’s rated p
p
performance at 1000 rpm.
p
Part (c) is not in the book; it’s added for clarity, and other
reasons y
you’ll see momentarily.
y
11 December 2012
Physics 122, Fall 2012
2
Problem 29
29-45
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
Physics 122, Fall 2012
3
Problem 29
29-45
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
0 2  
0 2 
1
2
R
750
0 2 

V  2  
250 V  190 V.
1
r  R 1000
11 December 2012
Physics 122, Fall 2012
4
Problem 29
29-45
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
hil it’s
it’ producing
d i currentt I is
i
   B  NIAB
Note that this requires exertion of constant mechanical
power, as you may have guessed from conservation of
energy:
dW   d   dt
dW
    NIAB   NIB  I  I  Pelec
Pmech 
dt
11 December 2012
Physics 122, Fall 2012
5
Problem 28
28-43
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 formula
the
f
l iin partt ((a)) reduces
d
to
t that
th t ffor a circular
i l lloop.
11 December 2012
Physics 122, Fall 2012
6
Problem 28
28-43
43 (continued)
First, and as usual, we split
the problem up into tractable
parts – the line segments –
and
d identify
id tif the
th method
th d off
calculation – the Biot-Savart
field law.
 Each segment has length
2 x0  2 R sin  n  ,
n = 7 shown.
h
2 n
R
y0 R
and its midpoint lies
x0
y0  R cos  n 
I0
x0
from the center.
center
11 December 2012
Physics 122, Fall 2012
7
Problem 28
28-43
43 (continued)
 The segments all produce
B in the same direction:
out of the page.
 The
h contribution to B from
each segment can be
calculated separately.
separately (We
did a similar calculation in
class on 25 October 2012.))
Follow recipe:
g segment,
g
y
 Point x along
through center of polygon
(bisecting the segment).
11 December 2012
y
dB
y0
r  r

Id
 x0
Physics 122, Fall 2012
x
x0
x
8
Problem 28
28-43
43 (continued)
 Identify an appropriate
infinitesimal current
element:
y
dB
Id   Idx xˆ ,
lying at x .
 Get r – r’ for this element:
y0

2
r  r   x 2  y02
r
 r   xˆ cos   yˆ sin 
  xˆ
11 December 2012
x
x 2  y02
 yˆ
r  r
Id
 x0
x
x0
x
y0
x 2  y02
Physics 122, Fall 2012
9
Problem 28
28-43
43 (continued)
 Now we have all the ingredients for the B-S law, and we
simply integrate:
x0

0 Id   r  r  0 I
dx 
B1 

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  
x 2  y02
11 December 2012

x
dx 
2
 y02



32
1
cos  d  
2

cos  sin 2 
dx 

32
y0
2
2
x   y0
y0 2 x dx 
Physics 122, Fall 2012

10
Problem 28
28-43
43 (continued)
 As x    x0  x0 ,



x0
x0



  arccos
  arccos
 2
 2
2 
2
x

y
x

y
0 
0
 0
 0
 
 
    
         
2 n
2 n
2 n 2 n
 So
 2  n
0 Iy0
B1 
zˆ
4

sin 3  y0 d
 2  n
0 I

ˆ

z   cos  
4 y0
11 December 2012




y03
sin 2 
2  n
2  n
0 I



ˆ
sin  
z
2 y0
n
Physics 122, Fall 2012
11
Problem 28
28-43
43 (continued)
 Thus the total B from all n segments is
0 In



ˆ
sin   .
Bz
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
p
((see lecture notes for 30 October 2012).
)
11 December 2012
Physics 122, Fall 2012
12
Problem 30
30-100
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
i t    for
f the
th currentt lleaving
i the
th source. ]
R
V
11 December 2012
C
Physics 122, Fall 2012
L
13
Problem 30
30-100
100 (continued)
Never mind the trial solution,
because we’ll use complex
exponentials instead of trig
f
functions.
ti
 Impedance first: parallel
LC has
R
V
1
1
1   2 LC

 iC 
ZLC i L
i L
so
Z  R  ZLC  R 
11 December 2012
C
L
i L
1   2 LC
Physics 122, Fall 2012
14
Problem 30
30-100
100 (continued)
 Write Z in magnitudephase form:
R
Z  Z0 ei
2

 L

2

Z0   R  

2


1

LC





  arctan
t  IIm Z R
Re Z 
12
V
C
L


L


 arctan 

2


 R 1   LC 

 
11 December 2012
Physics 122, Fall 2012
15
Problem 30
30-100
100 (continued)
 Then, since V  V0 e 
(so Re V  V0 sin t ),
i t  2 
R
V V0 it   2 
I 
e
Z Z0
is the total current drawn
from the voltage source,
and is the current flowing
through the resistor. Real
part:
V0
Re I 
sin t    .
Z0
11 December 2012
V
Physics 122, Fall 2012
C
L
16
Problem 30
30-100
100 (continued)
 The voltage across the parallel LC is
VLC
V0 it   2  i L
 IZLC 
e
Z0
1   2 LC
V0 it  
L
e

2
1   LC Z0
 Thus the currents through
g C and L are
VLC
 2 LC V0 it   2 
IC 
e
 iCVLC 
2
XC
1   LC Z0
VLC VLC
V0 it   2 
1


IL 
e
XL
i L 1   2 LC Z0
11 December 2012
Physics 122, Fall 2012
17
Problem 30
30-100
100 (continued)
 Real parts of the currents:
V0
Re I R 
sin t    .
Z0
Re IC  
 2 LC V0
2
sin t   
1   LC Z0
V0
1
Re I L 
sin t   
1   2 LC Z0
with Z0 and  as given above.
11 December 2012
Physics 122, Fall 2012
18