Problem 31.48

Heimadæmi 9
Due: 11:00pm on Thursday, March 17, 2016
You will receive no credit for items you complete after the assignment is due. Grading Policy
Problem 31.48
Figure shows a low­pass filter; the output voltage is taken across
the capacitor in an L­R­C series circuit.
Part A
Derive an expression for Vout /Vs, the ratio of the output and source voltage amplitudes, as a function of the angular
frequency ω of the source.
ANSWER:
V
out
Vs
V
out
Vs
V
out
Vs
V
out
Vs
1
=
2
ω C √R +(ω L−
1
2
)
ω C
1
=
2
ω C √R −(ω L−
1
2
)
ω C
1
=
2
ω C √R −(ω L+
1
2
)
ω C
1
=
2
ω C √R +(ω L+
1
2
)
ω C
Correct
Part B
Find an expression for Vout /Vs when ω is large.
ANSWER:
1
1
2
(L C) ω
1
(L C) ω
(L C) ω2
Correct
Part C
Find an expression for Vout /Vs in the limit of small frequency.
ANSWER:
1
1
2
(L C) ω
1
(L C) ω
(L C) ω2
Correct
Problem 31.55
−−−−−−−−−−−−−−−−−−
The impedance of an L­R­C parallel circuit can be written as 1/Z
2
2
= √ (1/R ) + [ωC − 1/ωL]
.
Part A
Show that at the resonance angular frequency ω0
current through the ac source is a minimum.
ANSWER:
−−
−
= 1/√LC
, the impedance Z is a maximum and therefore the
Using −−−−−−−−−−−−−−−−−−
2
2
1/Z = √ (1/R ) + [ωC − 1/ωL]
we see that the impedance is a maximum when the square
Z
root is a minimum, and that occurs when ωC − 1/ωL
resonance angular frequency ω0
−−
−
= 1/√LC
. Since I
. But that occurs when ω
= 0
=
−−
−
= 1/√LC
which is the
V /Z, the current is then a minimum when Z is a
maximum.
Using −−−−−−−−−−−−−−−−−−
2
2
1/Z = √ (1/R ) + [ωC − 1/ωL]
we see that the impedance is a maximum when the square
Z
root is a maximum, and that occurs when ωC − 1/ωL
resonance angular frequency ω0
−−
−
= 1/√LC
. Since I
. But that occurs when ω
= 0
= V /Z
−−
−
= 1/√LC
which is the
, the current is then a minimum when Z is a
maximum.
Using −−−−−−−−−−−−−−−−−−
2
2
1/Z = √ (1/R ) + [ωC − 1/ωL]
we see that the impedance is a maximum when the square
Z
root is a minimum, and that occurs when ωC − 1/ωL
resonance angular frequency ω0
−−
−
= 1/√LC
. Since I
. But that occurs when ω
= 0
= Z/V
−−
−
= 1/√LC
which is the
, the current is then a minimum when Z is a
maximum.
Using −−−−−−−−−−−−−−−−−−
2
2
1/Z = √ (1/R ) + [ωC − 1/ωL]
we see that the impedance is a maximum when the square
Z
root is a maximum, and that occurs when ωC − 1/ωL
resonance angular frequency ω0
−−
−
= 1/√LC
. Since I
. But that occurs when ω
= 0
= Z/V
−−
−
= 1/√LC
which is the
, the current is then a minimum when Z is a
maximum.
Correct
Part B
A 100­Ω resistor, a 0.100­μF capacitor, and a 0.300­H inductor are connected in parallel to a voltage source with
amplitude 240 V.What is the resonance angular frequency?
ANSWER:
ω0
= 5.77×103 rad/s Correct
Part C
For the circuit that was described in part B find the maximum current through the source at the resonance frequency.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
I
= 2.40 A
Correct
Part D
For the circuit that was described in part B find the maximum current in the resistor at resonance.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
IR
= 2.40 A
Correct
Part E
For the circuit that was described in part B find the maximum current in the inductor at resonance.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
IL
= 0.139 A
Correct
Part F
For the circuit that was described in part B find the maximum current in the branch containing the capacitor at
resonance.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
IC
= 0.139 A
Correct
Problem 31.51
In an L­R­C series circuit the magnitude of the phase angle is 54.2 ∘ with the source voltage lagging the current. The
reactance of the capacitor is 360 Ω, and the resistor resistance is 174 Ω. The average power delivered by the source is
133 W .
Part A
Find the reactance of the inductor.
ANSWER:
XL
= 119 Ω Correct
Part B
Find the rms current.
ANSWER:
Irms
= 0.874 A Correct
Part C
Find the rms voltage of the source.
ANSWER:
Vrms
= 260 V Correct
Exercise 31.3
The voltage across the terminals of an ac power supply varies with time according to V
amplitude is V0 = 43.0 V .
= V0 cos(ωt)
. The voltage
Part A
What is the root­mean­square potential difference Vrms ?
ANSWER:
Vrms
= 30.4 V Correct
Part B
What is the average potential difference Vav between the two terminals of the power supply?
ANSWER:
Vav
= 0 V Correct
Exercise 31.1
You have a special lightbulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.30 A ,
even for an instant.
Part A
What is the largest root­mean­square current you can run through this bulb?
ANSWER:
Irms
= 0.919 A Correct
Exercise 31.7
The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 179 V and frequency 56.0 Hz applied
across the capacitor is to produce a current amplitude of 0.854 A through the capacitor.
Part A
What capacitance C is required?
ANSWER:
C
= 1.36×10−5 F Correct
Score Summary:
Your score on this assignment is 99.1%.
You received 5.94 out of a possible total of 6 points.

Problem 31.48

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