Home Work Solutions 11
11-1 Using the loop rule, derive the differential equation for an LC circuit:
d 2q 1
 q0
dt 2 C
Sol: The loop rule, for just two devices in the loop, reduces to the statement that the
magnitude of the voltage across one of them must equal the magnitude of the
voltage across the other. Consider that the capacitor has charge q and a voltage
(which we’ll consider positive in this discussion) V = q/C. Consider at this
moment that the current in the inductor at this moment is directed in such a way
that the capacitor charge is increasing (so i = +dq/dt). Equation 30-35 then
L
produces a positive result equal to the V across the capacitor: V = L(di/dt), and
we interpret the fact that di/dt > 0 in this discussion to mean that d(dq/dt)/dt =
d2q/dt2 < 0 represents a “deceleration” of the charge-buildup process on the
capacitor (since it is approaching its maximum value of charge). In this way we
can “check” the signs in Eq. 31-11 (which states q/C =  L d2q/dt2) to make sure
we have implemented the loop rule correctly.
11-2 A series circuit containing inductance L1 and capacitance C1 oscillates at angular
frequency ω. A second series circuit, containing inductance L2 and capacitance C2,
oscillates at the same angular frequency. In terms of ω, what is the angular frequency
of oscillation of a series circuit containing all four of these elements? Neglect
resistance. (Hint: Use the formulas for equivalent capacitance and equivalent
inductance; see Section 25-4 and Problem 47 in Chapter 30.)
Sol: For the first circuit  = (L1C1)–1/2, and for the second one  = (L2C2)–1/2. When
the two circuits are connected in series, the new frequency is
 

1

Leq Ceq
1
L1C1
1
 L1  L2  C1C2 /  C1  C2 
1
 C1  C2  /  C1  C2 
,
where we use  1  L1C1  L2 C2 .

1
 L1C1C2  L2C2C1  /  C1  C2 
11-3 An alternating source with a variable frequency, a capacitor with capacitance C,
and a resistor with resistance R are connected in series. The following figure gives the
impedance Z of the circuit versus the driving angular frequency ωd; the curve reaches
an asymptote of 500 Ω, and the horizontal scale is set by ωds = 300 rad/s. The figure
also gives the reactance XC for the capacitor versus ωd. What are (a) R and (b) C?
Sol: (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.
11-4 An alternating source with a variable frequency, an inductor with inductance L,
and a resistor with resistance R are connected in series. The following figure gives the
impedance Z of the circuit versus the driving angular frequency ωd, with the
horizontal axis scale set by ωds = 1600 rad/s. The figure also gives the reactance XL
for the inductor versus ωd. What are (a) R and (b) L?
Sol: (a) Since Z =
2
2
R + XL
and XL = d L, then as d  0 we find Z  R = 40 .
(b) L
= XL /d
= slope = 60 mH.
11-5 A typical light dimmer used to dim the stage lights in a theater consists of a variable inductor L
(whose inductance is adjustable between zero and Lmax) connected in series with a lightbulb B, as
shown in the following figure. The electrical supply is 120 V (rms) at 60.0 Hz; the light-bulb is
rated at 120 V, 1000 W. (a) What Lmax is required if the rate of energy dissipation in the lightbulb is
to be varied by a factor of 5 from its upper limit of 1000 W? Assume that the resistance of the
lightbulb is independent of its temperature. (b) Could one use a variable resistor (adjustable between
zero and Rmax) instead of an inductor? (c) If so, what Rmax is required? (d) Why isn't this done?
Sol:
(a) The power consumed by the light bulb is P = I2R/2. So we must let Pmax/Pmin =
(I/Imin)2 = 4, or

R 2   Lmax
  m / Zmin 
 Zmax 
 I 

 I   /Z   Z  
R
 min 
 m max 
 min 

2
2

2

2
2

  4.


We solve for Lmax:
Lmax 
3R



 / 1000 W  6.62  10
2 60.0 Hz 
3 120 V
2
2
H.
(b) Yes, one could use a variable resistor.
(c) Now we must let
2
 Rmax  Rbulb 

  4,
Rbulb


or
Rmax 
 4  1R
bulb
120 V   14.4 .
 1
1000 W
2
(d) This is not done because the resistors would consume, rather than temporarily store,
electromagnetic energy.
11-6 The following figure shows an “autotransformer.” It consists of a single coil (with an iron
core). Three taps Ti are provided. Between taps T1 and T2 there are 50 turns, and between taps T2 and
T3 there are 800 turns. Any two taps can be chosen as the primary terminals, and any two taps can be
chosen as the secondary terminals. For choices producing a step-up transformer, what are the (a)
smallest, (b) second smallest, and (c) largest values of the ratio Vs/Vp? For a step-down transformer,
what are the (d) smallest, (e) second smallest, and (f) largest values of Vs/Vp?
Sol:
For step-up transformer:
(a) The smallest value of the ratio Vs / Vp is achieved by using T2T3 as primary and
T1T3 as secondary coil: V13/V23 = (800 + 50)/800 = 1.07.
(b) The second smallest value of the ratio Vs / Vp is achieved by using T1T2 as primary
and T2T3 as secondary coil: V23/V13 = 800/200 = 4.00.
(c) The largest value of the ratio Vs / Vp is achieved by using T1T2 as primary and T1T3
as secondary coil: V13/V12 = (800 + 50)/50 = 17.0.
For the step-down transformer, we simply exchange the primary and secondary coils
in each of the three cases above.
(d) The smallest value of the ratio Vs / Vp is 1/17.0 = 0.0588.
(e) The second smallest value of the ratio Vs / Vp is 1/4.00 = 0.250.
(f) The largest value of the ratio Vs / Vp is 1/1.07 = 0.935.