Physics 234
Homework Chapter 31
Q8, Q11, P11, P32, P36, P50
Q8) The two equations that characterize the voltage across the generator and the
current in the circuit are,
   m sin  d t
and
i  I sin  d t   .
A possible phasor diagram for the RLC circuit is shown below for the case when VL >
VC.
VL
VL  VC
m

I
VR
VC
For this particular case, the phase constant, , is positive. For cases where VC > VL, the
phase constant would be negative.
a)
For cases when the load is primarily capacitive implies that VC > VL, the case
where the phase constant is negative. Cases (1) and (4) would apply in this case.
b)
The phasor diagram shown above is consistent with a situation where the current
lags the voltage across the generator (the alternating emf, m). The phasor diagram is also
the case corresponding to a positive phase constant, and would therefore apply to cases
(2) and (3).
Q11) The situation in Figure 31-25, shows the current slightly leading a voltage across
the generator. It therefore represents a negative phase constant, , and can be represented
by the phasor diagram, below.
VL
I
VC  VL

VR
m
VC
In this case, the voltage across the capacitor is slightly larger than the voltage across the
inductor.
a)
If one slightly increases the inductance of the circuit, the difference, VC – VL, will
become smaller (since VL = dL) and the phase constant will also become smaller, which
is apparent from examining the phasor diagram. Another way to look at this is,
VC  VL  I ( X C  X L )  I (
1
  d L ).
d C
As dL approaches 1/dC, VC – VL eventually becomes zero at resonance. Therefore, a
slight increase in inductance will shift the current and voltage, , curves closer to one
another. In other words, it will shift the current curve rightward.
The amplitude of the current in the circuit can be determined by the equation,
I
m
Z

m
R  (XC  X L)
2
2

m
1
R2  (
  d L) 2
d C
.
Therefore, a slight increase in L will decrease the difference, XC – XL, decreasing the
denominator in the equation above, and increasing the current amplitude, I.
b)
Increasing C slightly decreases XC, since XC = 1/dC. Similar to the situation in
part a), this will decrease the difference, VC – VL. The same reasoning as in part a) will
still apply so that the current curve will shift rightward and the current amplitude will
increase.
c)
Increasing d increases XL = d L and decreases XC = 1/dC. This brings XL and
XC even closer together than in parts a) and b) and decreases the difference, VC – VL,
even more so than in parts a and b. The effects are the same qualitatively so that the
current curve will shift rightward and the current amplitude will increase.
P11) (a) Since the frequency of oscillation f is related to the inductance L and
capacitance C by f  1 / 2  LC , the smaller value of C gives the larger value of f.
Consequently, f max  1/ 2 LCmin , f min  1/ 2 LCmax , and
Cmax
365 pF
f max


 6.0.
f min
Cmin
10 pF
(b) An additional capacitance C is chosen so the ratio of the frequencies is
r
160
. MHz
 2.96.
0.54 MHz
Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds
to that of the tuning capacitor. If C is in picofarads (pF), then
C  365 pF
C  10 pF
 2.96.
The solution for C is,


C  365 pF  2.96 2 C  10 pF   C (1  2.96 2 )  2.96 2 10 pF  365 pF
87.616 pF  365 pF
 35.7 pF ~ 36 pF .
(1  2.96 2 )
C 
(c) We solve f  1 / 2  LC for L. For the minimum frequency, C = 365 pF + 36 pF =
401 pF and f = 0.54 MHz. Thus
L
1
 2 
2
Cf
2

 2 
2
 40110
1
12
F  0.54 106 Hz 
2
 2.2 104 H.
P32) (a) The circuit consists of one generator across one inductor; therefore, m = VL.
The current amplitude is
I
m
XL

m
25.0 V

 5.22 103 A .
 d L (377 rad/s)(12.7 H)
(b) When the current is at a maximum, its derivative is zero. Thus, Eq. 30-35 gives L = 0
at that instant. Stated another way, since (t) and i(t) have a 90° phase difference, then
(t) must be zero when i(t) = I. The fact that  = 90° = /2 rad is used in part (c).
(c) Consider Eq. 31-28 with    m / 2 . In order to satisfy this equation, we require
sin(dt) = –1/2. This occurs when the value of dt is either 210º or at 330 º. Now we
note that the problem states that  is increasing in magnitude, which (since it is already
negative) means that it is becoming more negative. The sin function is becoming more
negative at 210º (with increasing angle) and this is, therefore, the correct value of dt.
Solving for i, we have,




i  I sin  d t     5.22 mAsin 210   90  5.22mAsin 120   4.52 mA.
P36) (a) The circuit has a resistor and a capacitor (but no inductor). Since the
capacitive reactance decreases with frequency, then the asymptotic value of Z must be the
resistance: R = 500 .
(b) We describe three methods here (each using information from different points on the
graph):
method 1: At d = 50 rad/s, we have Z  700 , which gives C = (d Z2 - R2 )1 = 41 F.
method 2: At d = 50 rad/s, we have XC  500 , which gives C = (d XC)1 = 40 F.
method 3: At d = 250 rad/s, we have XC  100 , which gives C = (d XC)1 = 40 F.
P50)
(a) The phasor diagram is shown below.
VL


VR
(b) We have VR = VL, which implies
I R = I XL → R = d L
which yields f = d/2 = R/2L = 318 Hz.
(c)  = tan1(VL /VR) = +45.
(d) d = R/L = 2.00×103 rad/s.
(e) I = (6 V)/ R2 + XL2 = 3/(40 2)  53.0 mA.