Prof. Anchordoqui
Problems set # 11
Physics 169
May 5, 2015
1. A semicircular conductor of radius R = 0.250 m is rotated about the axis AC at a constant
rate of 120 rev/min (Fig. 1). A uniform magnetic field in all of the lower half of the figure is
directed out of the plane of rotation and has a magnitude of 1.30 T. (i) Calculate the maximum
value of the emf induced in the conductor. (ii) What is the value of the average induced emf for
each complete rotation? (iii) How would the answers to (i) and (ii) change if B were allowed to
extend a distance R above the axis of rotation? Sketch the emf versus time (iv) when the field is
as drawn in Fig. 1 and (v) when the field is extended as described in (iii).
Solution (i) εmax = BAω = BπR2 ω/2 = 1.30 T · π2 · (0.25 m)2 · 4.00π rad/s = 1.60 V. (ii)
R 2π
R
BAω 2π
1
ε
dθ
=
ε̄ = 2π
2π
0
0 sin θ dθ = 0. (iii) The maximum and average ε would remain unchanged. (iv) See Fig. 2. (v) See Fig. 2.
2. An AC power supply produces a maximum voltage of V0 = 100 V. This power supply is
connected to a 24.0 − Ω resistor, and the current and resistor voltage are measured with an ideal
AC ammeter and an ideal AC voltmeter, as shown in Fig. 3 What does each meter read? Recall
that an ideal ammeter has zero resistance and an ideal voltmeter has infinite resistance.
Solution The meters measure the rms values of potential difference and current. These are
V0
Vrms = √
= 70.7 V and Irms = Vrms
R = 2.95 A.
2
3. (i) For the series RLC connection of Fig. 4, draw a phasor diagram for the voltages. The
amplitudes of the voltage drop across all the circuit elements involved should be represented with
phasors. (ii) An RLC circuit consists of a 150-Ω resistor, a 21-µF capacitor and a 460-mH inductor,
connected in series with a 120-V, 60-Hz power supply. What is the phase angle between the current
and the applied voltage? (iii) Which reaches its maximum earlier, the current or the voltage?
Solution (i) For the series connection, the instantaneous voltage across the system is equal to the
sum of voltage across each element. The phase angle between the voltage (across the system) and
C
the current (through the system) is: φ = arctan XL −X
. In Fig. 5 the phasor diagram for a series
R
RLC circuit is shown for both the inductive case XL > XC and the capacitive case XL < XC . On
~0 leads I~0 by a phase φ. On the other
the one hand, in the inductive case, V0,L > V0,c , we see that V
~
~0 by a phase φ. (ii) From the
hand, in the capacitance case, V0,C > V0,L , we have that I0 leads V
definition, the inductive reactance of the inductor is XL = ωL = 2π · 60 Hz · 0.46 H = 173 Ω. From
1
1
the definition, the capacitive reactance of the capacitor is XC = ωC
= 2π·60 Hz·21×10
−6 F = 126 Ω.
The phase angle between the voltage (across the system) and the current (through the system) is:
Ω−126 Ω
C
φ = arctan XL −X
= arctan 174 150
= 17.4◦ (iii) The voltage leads the current.
R
Ω
4. Figure 6 shows a parallel RLC circuit. The instantaneous voltage (and rms voltage) across
each of the three circuit elements is the same, and each is in phase with the current through the
resistor. The currents in the capacitor and the inductor lead or lag behind the current in the reh
i1/2
1 2
.
sistor. (i) Show that the rms current delivered by the source is Irms = Vrms R12 + ωC − ωL
(ii) Show that the phase angle between the voltage and the current is tan ϕ = −R X1C − X1L .
(iii) Draw a phasor diagram for the currents. The amplitudes of the currents across all the circuit
elements involved should be represented with phasors.
Solution We are asked to find the impedance and the phase angle for this system of elements
connected in parallel. It will be easier to analyze the complex impedance. For elements connected
in parallel the equivalent complex impedance of the system is equal to the sum of inverse impedance
1
1
1
of each element Z1eq = Z1R + Z1L + Z1C = R1 + iωL
=
+ iωC = R1 + i ωC − ωL
, or Zeq = 1 +i ωC−
1
(
R
ωL )
q
1
1
−i(ωC− ωL )
0
V0 1 1 2
1 2
R
+ ωC − ωL
, and
. (i) Hence Irms = √I02 = √12 ZVeq
= √
Zeq = Vrms
1 2
1 2
R
2
( R ) +(ωC− ωL )
1
−(ωC− ωL
)
ImZeq
(ii) tan ϕ = ReZeq
. (iii) The phasor diagram for a parallel RLC circuit is shown in
=
1
R
Fig. 7, for both the inductive case XL > XC and the capacitive case XL < XC .
5. Draw to scale a phasor diagram showing Z, XL , XC , and ϕ for an AC series circuit for which
R = 300 Ω, C = 11 µF, L = 0.2 H, and f = 500/π Hz.
Solution Before drawing the diagram, we find the complex impedance of each element. The
complex impedance of a resistor is equal to its resistance, ZR = R = 300 Ω. It is therefore
a real number which we marked in the complex plane on the Re axis of Fig. 8. The complex
impedance of an inductor is an imaginary number dependent on the angular frequency of the
−1 · 0.2 H· = 200i Ω. This
current and the inductance of the inductor ZL = iωL = 2π · 500
π s
number is on the Im axis. Finally, the complex impedance of the capacitor is also an imaginary
number dependent on the angular frequency of the current and the capacitance of the capacitor
i 500 −1
i
= − 2π
· 11 × 10−6 F = −90.9i Ω. This number is also on the Im axis. Since the
ZC = − ωC
π s
elements are connected in series the equivalent complex impedance is equal to the sum of the complex impedance of each element: Z = ZR + ZL + ZC = 300 Ω + 200i Ω − 90.9i Ω = (300 + 109.1i) Ω.
The complex impedance of each element is indicated in Fig. 8. One tic corresponds to 100 Ω. The
diagram also illustrates how to find the equivalent complex impedance. Note that the phase of the
complex impedance of the resistor is 0◦ , the phase of the complex impedance of the capacitor is
−90◦ and the phase of the complex impedance of the inductor is 90◦ .
6. Draw to scale a phasor diagram showing the relationship between the current (common for
all elements) and the voltages in an AC series circuit for which R = 300 Ω, C = 11 µF, L = 0.2 H,
f = 500/π Hz, and I0 = 20 mA.
Solution The complex voltage across the resistor can be found by multiplying the complex current (through the resistor) by the complex impedance of the resistor VR (t) = I(t)ZR . Consistent
with the given values, the absolute value of this voltage is V0,R = I0 R = 20 mA · 300 Ω = 6 V.
Since the phase of the complex impedance of the resistor is zero, the phase of the voltage across
the resistor agrees with the phase of the current. The complex voltage across the inductor can be
found by multiplying the complex current (through the inductor) by the complex impedance of the
inductor VL (t) = I(t)ZL . Consistent with the given values, the absolute value of this voltage is
V0,L = I0 XL = 20 mA · 200 ω = 4 V. Since the phase of the complex impedance of the inductor
is 90◦ , the voltage across the inductor leads with the current by a 90◦ phase angle. Finally, the
complex voltage across the capacitor can be found by multiplying the complex current (through the
capacitor) by the complex impedance of the capacitor VC (t) = I(t)ZC . Consistent with the given
values, the absolute value of this voltage is V0,C = I0 · XC = 20 mA · 91 Ω = 1.8 V. Since the phase
of the complex impedance of the capacitor is −90◦ , the voltage across the capacitor lags behind the
current by 90◦ . If we want to find the complex voltage across the entire system, we can add the
complex voltages across all three elements (they are connected in series). I indicated this operation
on the phasor diagram. That voltage can also be found by multiplying the current by the equivalent
impedance of the system. Notice that the phase difference between the voltage across the system and
the current through the system is equal to the phase angle of the system (the phase of the equivalent
p
impedance). The absolute value of the voltage is V0 = I0 Z = 20 mA· (300 Ω)2 + (109 Ω)2 ≈ 6.4 V.
7. The resistor in the Fig. 10 represents the midrange speaker in a three-speaker system. Assume its resistance to be constant at 8 Ω. The source produces signals of uniform amplitude
∆Vin = 10 V al all frequencies. The inductor and capacitor are to function as a bandpass filter
with ∆Vout /∆Vin = 1/2 at 200 Hz and 4, 000 Hz. (i) Determine the required values of L and C.
(ii) Find the maximum value of the ratio ∆Vout /∆V in. (iii) Find the frequency f0 at which the
ratio has its maximum value. (iv) Find the phase shift between ∆Vin and ∆Vout at 200 Hz at f0 ,
and at 4, 000 Hz. (v) Find the average power transferred to the speaker at 200 Hz, at f0 , and at
4, 000 Hz. (vi) Treating the filter as a resonant circuit, find its quality factor.
Solution Complex current in the series RLC circuit is determined by the input complex voltage
Vin
Vin
= R+i ωL−
and the complex impedance of the circuit I = Vin /Z = R+iωL+R+
. Complex
1
1
(
iωC
ωC )
voltage across the speaker is related to the complex current through the speaker, Vout = IR =
1
R2 −i(ωL− ωC
)R
R
V
=
vin . The number relating complex output voltage to complex
in
1
1 2
2
R+i(ωL− ωC )
R +(ωL− ωC )
input voltage is called the complex gain of the circuit. (i) From the relation between complex
R
voltages, the relation between the amplitudes is ∆Vout = |Vout | = q
∆Vin . The
1 2
2
R +(ωL− ωC
)
number relating the amplitudes of voltage is called the gain of the circuit. Gain G is a funcR
tion of frequency G = q
. It assumes a particular value at specific frequencies, de1 2
2
R +(ωL− ωC
)
termined by the difference between capacitive reactance and inductive reactance ω1 L − ω 11 C =
q
q
1
− G12 − 1R and ω2 L − ω 12 C =
− 1R. Solving simultaneously for the unknowns L =
G2
R
q
1
−1
G2
ω2 −ω1
=
R
q
1
−1
G2
2π(f2 −f1 )
=
√
8Ω 4−1
2π(4000 Hz−200 Hz
= 580 µH and C = −
ω1 −ω
q 2
1
Rω1 ω2
2 −1
G
4000 Hz−200 Hz√
2π·8Ω·200 Hz·4000 Hz· 4−1
=
f2 −f
q1
2πRf1 f2
1
−1
G2
=
= 54.6 µF. (ii) The maximum gain occurs when the sum of inductive
1
reactance and capacitive reactance is zero ωL ωC
= 0 from which G = √RR
= 1. (iii) The
2 +02
sum of inductive reactance and capacitive reactance is zero at the resonant frequency of this
ω0
circuit f0 = 2π√
= 2π√5.8×10−4 1H·5.46×10−5 F = 894 Hz. (iv) Complex gain also includes
LC
the phase relation. The phase difference α, between the output voltage and the input volt-
1
−(ωL− ωC
)
R
−4
2π·200 Hz·5.8×10
H−
ages is α = arctan
= − arctan
1
2πf L− 2πf
C
R
. Therefore at 200 Hz its value is α200
Hz
=
1
2π·200 Hz·5.46×10−5 H
−arctan
= 60◦ . (The output voltage leads the input voltage.)
8 Ω
At 894 Hz its value is α894 Hz = −arctan(0/8 Ω) = 0◦ . (The output voltage is in phase with the input
1
2π·4000 Hz·5.46×10−5 H−
−5
2π·4000 Hz·5.46×10
H
voltage.) At 4, 000 Hz its value is α4000 Hz = −arctan
=
8 Ω
◦
−60 . (The output voltage lags behind the input voltage.) (v) The average power delivered to
an element depends on the rms values of the voltage across the element and the current through
the element. Since the speaker acts like a resistor, Ohm’s law can be used to relate the rms current (through the speaker) to the voltage applied to the speaker. Knowing the gain of the circuit
the power can be directly related to the amplitude of the input voltage P = Irms Vrms cos ϕ =
(G∆Vin )2
V)2
V)2
∆V√
out ∆V
√out cos 0◦ =
. Therefore, P200 = (0.510
= 1.56 W, P894 = (110
= 6.25 W,
2R
2·8ω
2·8ω
R 2
2
2
V)
P4000 = (0.510
= 1.56 W. (vi) Parameters of the RLC circuit (resistance, capacitance and
2·8ω
−4 H
0L
inductance) affect the quality factor Q = ωR
= 2πfR0 L = 2π·894 Hz·5.8×10
= 0.41.
8 Ω
8. A person is working near the secondary coil of a transformer as shown in Fig. 11. The primary voltage is 120 V at 60 Hz. The capacitance Cs , which is the stray capacitance between the
hand and the secondary winding, is 20 pF. Assuming the person has a body resistance to ground
Rb = 50 kΩ, determine the rms voltage across the body.
Solution The secondary circuit consists of a “capacitor” with capacitance Cs = 20 µF, a “resistor” with resistance Rb = 50 kΩ and a source of electromotive force (the secondary coil of the
transformer). The source produces sinusoidal potential difference with rms value of 5000 V and
frequency of 60 Hz. At this frequency the impedance of the “capacitor” is ZC = 2πf1Cs and the
impedance of the “resistor” is equal to its resistance ZR = Rb . Since the “elements” are connected
in series, the equivalent resistance of the system (of the two elements) is Z = Rb2 + (2πf1Cs )2 . The
rms current in the “resistor” (and the circuit) is therefore Irms =
Vrms
Z
=
r
Vrms
Rb2 +
1
2πf Cs
Ohm’s law, the rms voltage across the body (resistor) is Vrms,b = Irms,b Rb = Rb r
Vrms
Rb2 +
Vrms
1
1+ 2πf C
r
s Rb
2
=
5 kV
r
1+ 2π·60
1
Hz·20 pF·50 Ω
2
2 .
From
1
2πf Cs
2
=
= 1.88 V.
9. A diode is a device that allows current to pass in only one direction (indicated by the arrowhead). Find, in terms of Vrms and R, the average power dissipated in the diode circuit shown in
Fig. 12.
Solution When the left side of the generator has a higher potential, the top diode conducts
(forward bias), and the bottom diode does not (reverse bias). The remaining circuit of three resistors acts like a single resistor. The resistors R are connected in series and the pair is connected
in parallel to the resistor 2R, see Fig. 13. Therefore the equivalent resistance of the system is
−1
1
1
+ =
Req
+
= R. Knowing the rms value of the voltage we could find the power in
2R
R+R
this circuit if the diode was not present. However, the presence of the diode reduces the power
to half its value (because only half of the time current would flow in the circuit). Therefore
change if B were allowed to extend a distance R above the
axis of rotation? Sketch the emf versus time (d) when the
field is as drawn in Figure P31.40 and (e) when the field is
extended as described in (c).
A
FG 1 R IJ
H2 K
a1.30 Tf 2 a0.250 mf b4.00
BA
1.60 V
BA
2
d
z
2
rad s
g
Figure P31.40
Chapter 31
227
Figure 1: Problem 1.
0
would
Figureremain
1
Figure 2: Solution of problem 1.
B
dt
FIG. P31.40
before, the power dissipated at this polarity is P − =
2
2
1 + 4 2= 11 Vrms .
P = P + + P − = Vrms Figure
BA cos t
a
2
1 Vrms
2 1.75 R
The overall (time) average power is
Tfe0.010 0 m j cos 2 a60.0 ft e8.00 mT m j cosa377 t f
a0.800
FIG. P31.40
2R
7
14 R
2
2
10. A transmission line that has a resistance per unit length of 4.5 × 10−4 Ω/m is to be used
to transmit 5 MW over 400 miles (6.44 × 105 m); see Fig. 14. The output voltage of the generator
is 4.5 kV. (i) What is the line loss if a transformer is used to step up the voltage to 500 kV?
t of the input power is lost to the line under these circumstances? (iii) What diffi3.02 (ii)
V What
sin 377
fraction
culties would be encountered in attempting to transmit the 5 MW at the generator voltage of 4.5 kV.
f a f
a j f aa f f e
j a f
Solution
order
to send
.0 tt (i)8In.00
cospower
0 m3.202cos
mT
m 2 out
377oft P
A 2sin60
377
= 5 MW at 500 kV potential difference, the current
P
in the grid (transmission lines) must be Irms,a = Vrms,a
= 10 A. The resistance of the two wires of
the transmission lines (connected is series) is R = 2Lλ = 580 Ω. Hence the loss of power in the lines
2 R = (10 A)2 580 Ω = 58 kW. (ii) A relative small fraction is
(dissipated in the lines) is Pa = Irms
2
P
58
kW
a
t P = 5 MW = 0.01 = 1%. (iii) If the power were send at 4.5 kV potential difference,
Wlostsin
in the377
lines
P
5 MW
the current in the grid (transmission lines) should be much larger Irms,c = Vrms,c
= 4.5
kV = 1.1 kA.
But even if nothing were connected at the end of the lines, with the potential difference of 4.5 kV,
kV
P current in the lines would be2 Irms,max = Vrms,c
the
= 4.5
R
58 Ω = 77 A. It would be impossible to send
t
24
.
1
mN
m
sin
377
5 MW of power into the lines.
a9.10 f a f
so
rrents
Figure 2
P + = 21 Vrms
R . When the polarity of the emf is reversed, the top diode is in reverse bias and
thethe
right.
t time the equivalent resistance of the system of rebottom diode is in forward bias. This
−1
1
sistors is different R− = R + R1 + R+2R
= 74 R; see Fig. 13. With the same arguments as
ure 2 at the right.
in
A cos
t
Bout
2
ure 1 at
227
Chapter 31
1
41. The rotating loop in an AC generator is a Figure
square
10.0 cm
on a side. It is rotated at 60.0 Hz in a uniform field of
0.800 T. Calculate (a) the flux through the loop as a
t emf induced in the loop, (c) the t
sin d function
0
of time, (b) the
aximum and average
nged.
R
C
2
B
2
2
R
a
f a f
Figure P33.3. What does each meter read? Note that an
e that at t 0 , the voltage
is and
zero
the current in the c
ideal across
ammeter the
has capacitor
zero resistance
thatwhile
an ideal
has infinite resistance.
a maximum. In fact, I Cvoltmeter
(t ) reaches
its maximum before VC (t ) by one quarter
(
/ 2 ). Thus, we say that
∆V
= 100 V
max
The current leads the voltage by /2 in a capacitive circuit
A
Section 33.3 Induct
8. An inductor is con
produces a 50.0-V
to keep the insta
80.0 mA?
R = 24.0 Ω
The RLC Series Circuit
V
der now the driven series RLC circuit shown in Figure 12.3.1.
Figure P33.3
Figure 3: Problem 2.
4. In the simple AC circuit shown in Figure 33.2, R " 70.0 #
and !v " !Vmax sin $t. (a) If !vR " 0.250 !Vmax for the
first time at t " 0.010 0 s, what is the angular frequency of
the source? (b) What is the next value of t for which
!vR " 0.250 !Vmax?
5. The current in the circuit shown in Figure 33.2 equals
60.0% of the peak current at t " 7.00 ms. What is the
smallest frequency of the source that gives this current?
6. Figure P33.6 shows three lamps connected to a 120-V AC
(rms) household supply voltage. Lamps 1 and 2 have
150-W bulbs; lamp 3 has a 100-W bulb. Find the rms
current and resistance of each bulb.
9. In a purely induct
!Vmax " 100 V. (a
50.0 Hz. Calculate
angular frequency
10. An inductor has a
the maximum cur
50.0-Hz source tha
11.
For the circu
$ " 65.0% rad/s, a
in the inductor at t
12. A 20.0-mH induct
outlet (!V rms "
energy stored in t
that this energy is z
13. Review problem. D
through an induc
outlet (!V rms " 12
Figure 4: Problem 3.
Figure 12.3.1 Driven series RLC Circuit
ing Kirchhoff’s loop rule, we obtain
V (t ) VR (t ) VL (t ) VC (t ) V (t ) IR L
dI
dt
Q
C
Figure 12.8.1 Phasor diagram for the series RLC circuit for (a) X
X and
leads to the
following
differential
equation:
Figure
5: Phasor diagram
for the series RLC
circuit for XL > XC (left) and XL < XC (right).
L
XL
(12
0
C
(b)
XC .
From Figure 12.8.1(a), we see that VL 0
VC 0 in the inductive case and V0 leads I 0 by a
phase . On the other hand, in the capacitive case shown in Figure 12.8.1(b), VC 0
VL 0
Parallel RLC Circuit
sider the parallel RLC circuit illustrated in Figure 12.6.1. The AC voltage sour
V0 sin t .
Figure 12.8.1 Phasor diagram for the series RLC circuit for (a) X L
XL
X C and (b)
XC .
From Figure 12.8.1(a), we see that VL 0
VC 0 in the inductive case and V0 leads I 0 by a
phase . On the other hand, in the capacitive case shown in Figure 12.8.1(b), VC 0
VL 0
and I 0 leads V0 by a phase .
4. When VL 0
VC 0 , or
frequency is
0
0 , the circuit is at resonance. The corresponding resonant
1/ LC , and the power delivered to the resistor is a maximum.
5. For parallel connection, draw a phasor diagram for the currents. The amplitudes of the
Figure 6: involved
Problem should
4.
currents across all the circuit elements
be represented with phasors. In
Figure 12.8.2 the phasor diagram for a parallel RLC circuit is shown for both the
inductive case X L X C and the capacitive case X L X C .
Figure 12.6.1 Parallel RLC circuit.
ke the series RLC circuit, the instantaneous voltages across all three circuit elem
, and C are the same, and each voltage is in phase with the current through
tor. However, the currents through each element will be different.
nalyzing this circuit, we make use of the results discussed in Sections 12.2 – 1
current in the resistor is
Figure 12.8.2 Phasor diagram for the parallel RLC circuit for (a) X L X C and (b)
FigureX7: Phasor
X C . diagram for the parallel RLC0 circuit for XL > XC (left) and XL < XC (right).
L
I R (t )
V (t )
R
V
sin t
R
I R 0 sin t
From Figure 12.8.2(a), we see that I L 0 I C 0 in the inductive case and V0 leads I 0 by a
phase . On the other hand, in the capacitive case shown in Figure 12.8.2(b), I C 0 I L 0
re I R 0
Im
V0 / R . and
TheI voltage
across the
inductor is
leads V by a phase .
0
0
Z
L
VL (t ) V (t ) V0ϕ sin t
h gives
L (t )
(12
Z
Z
C
Z
L
dI L
dt
Re
(12
12-25
R
V0
V0 also illustrates
V0 how5. to find the equivalent
sin The
'
cos Figure
sin t
t ' dtdiagram
t 8: Problem
I L 0 sin t
0 L
complex impedance.
NoteXthat
the
phase
of
the
complex
2
L
L
t
impedance of the resistor is 0, the phase of the complex
impedance of the capacitor is -90° and the phase of the complex
impedance of the inductor is 90°.
2
(12
elements) and the voltages in the circuit. Let's assume that
current has value of 20 mA and at a certain instant t it
represented by the phasor indicated in the following figure.
the figure one tic corresponds to 10 mA.
The complex volta
across the resistor can
found by multiplying
complex current (through
resistor) by the comp
Problem 33.49
impedance of the resistor
The resistor in the Figure P33.49 represents the midrange speakerVin (at ) = I (t ) ⋅ Z
R
R
Im
V
VL
V
Z
R
L
I
Z
ϕ
ϕ
Re
Z R
Z C
three-speaker system. Assume its resistanceV to be constant at 8 Ω. The source
C
represent an audio amplifier producing signals of uniform amplitude
∆Vin = 10V al all frequencies. The inductor and capacitor are to function as a
bandpass filter with ∆Vout /∆Vin = ½ at 200 Hz and 4,000 Hz. (a) Determine
the required values of L and C. (b) Find the maximum value of the ratio
∆Vout/∆Vin. (c) Find the frequency f0 at which the ratio has its maximum
value. (d) Find the phase shift between ∆Vin and ∆Vout at 200 Hz, at f0, and at
4000 Hz. (e) Find the average power transferred to the speaker at 200 Hz, at
f0, and at 4000 Hz. (f) Treating the filter as a resonant circuit, find its quality
factor.
Figure 9: Problem 6.
Consistent with the gi
values, the absolute value
this voltage is
Problem 33.23
L
C
5
I
∆Vin
∆Vout
R
A person is working mear the secondary
coil of a transformer as shown in Figur
33.25. The primary voltage is 120 V at 60 Hz. The capacitance Cs, which is th
stray capacitance between the hand and the secondary winding, is 20 pF. Assuming
the personComplex
has a body
ground
= 50 kΩ,bydetermine
the rms voltag
currentresistance
in Figure
the series
RLC
circuit
isbdetermined
the
10: to
Problem
7. R
across the input
body.
complex voltage and the complex impedance of the circuit
I=
Vin
Vin
Vin
=
=
1
Z R + iωL + 1
R + i ωL −
i ωC
ωC
Complex voltage across the speaker is related to the complex
current through the speaker.
5,000 V
8
Figure 11: Problem 8.
The secondary circuit consists of a “capacitor” with
capacitance Cs=20µF, a “resistor” with resistance Rb=50kΩ
diode
2R
at allows current to
diode
on (indicated by the
ows current to
erms
R
ndicatedof
by V
therms and R,
of Vrms andinR,the diodeR
sipated
diode
ed in the diode
P33.35.
35.
R
2R
R
R
R
diode
V
V
(because only half of theFigure
time12:current
flow in the circuit).
Problem would
9.
Therefore
2
1 Vrms
+
diode
2) P = ⋅
2 R
e of the
higher
2R
diode2R
pside
diodeof the
When the polarity
of the emf is
R
bias),a and higher
R
reversed,
the
top
diode
is
in
2R
does not
R
R
R
top
diode
reverse
bias
and
the
bottom
remaining
diode
isProblem
in forward
33.43bias. This
stors
acts diode
R
d bias),
and
stor. The time the equivalent
+
- resistance of
+
A transmission
line
that has a resistance
per unit length of 4.5⋅10-4 Ω/m is to
de
does
not
the
system
of
resistors
is
R
5
nected in
be used to
V transmit 5 MW over 400 miles (6.44⋅10 m). The
V output voltage of
different
the generator is 4.5 kV. (a) What is the line loss if a transformer is used to step
connected
he
remaining
3)
the voltage to 500 kV? (b) What fraction of the input power is lost to the
tor 2R. Therefore the up
equivalent
resistance
13: Solution
of problem 9.
−1 (c) What
under these Figure
circumstances?
difficulties would be encountered in
resistors acts −line
1 transmit1the 5 MW at 7the generator
attempting
to
voltage of 4.5 kV
R = R+
+
= R
resistor. The 500 kV R R + 2R + 4 −1
or arguments as before, the power dissipated at this
1 connectedWithinthe same
λ = 4.5⋅10 Ω/m
4.5
5 MW
= R polarity is kV
V
+rRis connected
2
1 Vrms
−
4) P = ⋅
. could
2we175
Rthe
resistor
Therefore
equivalent
resistance
s value of 2R.
the voltage
find
the
-4
L = 6.44⋅105 m
if the diode was not present. However, the
The overall (time) average power is
14:
de reduces the power to half its Figure
value
2 Problem 10.
2
Vrms
11 Vat
a) In order to
power of4P = 5MW
+ send
− out
rms500 kV potential
P= P +P =
⋅ 1+
= ⋅
difference, the current2in
R the grid7 (transmission
14 R lines) must be
−1
1
4
=R
I rms, a =
P
Vrms, a
= 10A