Solution Set<br>
Week 9<br>
<p>
Chapter 22:
<p>
Discusssion Questions:
<p>
13.  A coil in a motor turning through a magnetic field will experience an
induced emf and an induced current that will send energy back to the
source, which is driving the motor. With no load, the motor produces almost
as much current as it draws and so costs very little to run. A free-turning
motor must have its speed limited by the back-current it generates -- with
no losses th ecurrent and it can no longer accelerate. With a load, the
motor does work and draws energy in excess of what it returns via the
back-current. In real life, a motor also produces a good deal of thermal
energy via friction and if it is to operaate continuously for any long
period of time it must be cooled. A jammed motor will not turn and not
generate a back-current. The driving current now undiminished by
back-current is too great. The wiring will heat up and the insulation will
fry and the motor will destroy itself.
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Problems:
<p>
32.  emf = 2 x pi x frequency x N x B x Area x sine(theta) and so,
<ul>
B = emf/[2 x pi x frequency x N x Area x sine(theta)]<br>
B = 20 V /[ 6.28 x 50 Hz x 150 x 0.016 m^2]; for the maximum V, we want
sine(theta) = 1<br>
B = 0.027 T
</ul>
The coil should spin at 2 x pi x 50 Hz = 310 Radians/sec
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54.
<ul>
emf = v x B x l = ( 2 x pi x frequency x R ) x 0.6 T x 0.30 m<br>
emf = 6.28 x 10 Hz x 0.10 m x 0.6 T x 0.30 m<br>
emf = 1.1 V
</ul>
</ul>
<p>
Chapter 23:
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Discussion Questions:
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3.  Reasons:
<ul>
<li>Because DC does not "peak",
for a given maximum voltage, the "delivered" voltage for DC is higher 
than for AC
<li>DC does not induce currents in nearby conductors 
<li> ...
</ul>
<p>
4.  The input to a transformer can be DC. All that you need is for the
amplitude of the current to change. The current does not have to change
direction
<p>
Problems:
<p>
1.  At 1/4 cycle, V(max) = + 120 V ===> V(t) = 120 V x sine(omega x time)
-- Note that at time = 1/4 cycle, omega x time = pi/2 since
omega = 2 x pi x frequency = 2 x pi / [period for variation] ===>
sine(omega x time) = sine(pi/2) = 1
<ul>
<li>1/4 cycle later ===> V = 120 V x sine(pi/2+pi/2) = 0 V
<li>1/8 cycle later ===> V = 120 V x sine(pi/2 + pi/2 + pi/4) = - 85 V
</ul>
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5.  maximum voltage = rms voltage x sqrt(2) = 100 V x 1.4 = 140 V
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7. For 120 V driving a 100-W light bulb, we have
<ul>
P = I x V ===> I = 100 W/120 V = 0.83 A 
</ul>
<p>
12.  X = 1/[omega x C] = 1/[2 x pi x 60 Hz x 8 x 10^(-5) F] = 33 ohms
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16.  
<ul>
<li>  There are two capacitors in parallel ===> C = C(1) + C(2) = 0.000005
F<br>
The capacitve reactance for the circuit is X = 1/(omega x C) = 530 ohms if
we assume that the current is at 60 Hz.<br>
===> I = V/X = 120 V/530 ohms = 0.23 A
<li> If we unhook either capacitor ===> C will decrease ===> X will
increase and so the current will go down
</ul>
<p>
23.  The Voltmeter reads the correct voltage, so the source is fine. The
problem must lie in the resistors. If all resistors are okay, then the
effective resistance is 
<ul>
R = R(1) + R(2) + R(3) =  65 ohms  ===> I = V/R = 1.8 A and we're okay
</ul>
To make the current larger ===> a resistor has to be bad, but which one?
<ul>
If either R(1) or R(2) is bad ===> R greater than R(3) ===> I less than
2.4 A which is okay.<br>
===> R(3) = 50 ohms must be bad
</ul>
<p>
26.  P = V^2/R ===> R = V^2/P = (120 V)^2 / 1,200 W = 12 ohms<br>
It draws a current, P = I^2 R ===> I = sqrt(P/R) = sqrt(1,200 W/12 ohms)
= 10 A
<p>
30.  The capacitve reactance is 1/[omega x C] = 29 ohms. The current
through the circuit is then V = I X ===> I = V/X = 120 V/29 ohms = 4.1
A