Physics 302/328 Homework
Chapter 8
‡ Problem 1: In the circuit below, find Z(w) at a frequency of 60 Hz, both magnitude
and phase. (Express in polar form.) {corrected}
In[13]:=
inductance = .01; capacitance = 10. * 10-6 ; f = 60.; w = 2 p f;
inductorImpedance = Â * w * inductance;
Print@"The inductive reactance is ", inductorImpedance, "W"D
1
capacitorImpedance = - Â
;
w * capacitance
Print@"The capacitive reactance is ", capacitorImpedance, "W"D
The inductive reactance is 0. + 3.76991 ÂW
The capacitive reactance is 0. - 265.258 ÂW
2
Chapter 8.07.nb
In[30]:=
impedance = inductorImpedance +
1
1.0
+
1
2. + capacitorImpedance
-1
;
Print@"The impedance in complex form is ", impedance, "W"D
Print@"The impedance in phasor form is ",
Abs@impedanceD, "W —", Arg@impedanceD ê Degree, "°"D
The impedance in complex form is 0.999957 + 3.76614 ÂW
The impedance in phasor form is 3.89663W —75.1303°
Solution:
Z = 3.90W ­∡75.13°
‡ Problem 2: Find the impedance, Z, shown in the circuit
below.
In[37]:=
impedance = 1. +
1
1+Â
+
-1
1
1.0 - Â
;
Print@"The impedance in complex form is ", impedance, "W"D
Print@"The impedance in phasor form is ",
Abs@impedanceD, "W —", Arg@impedanceD ê Degree, "°"D
The impedance in complex form is 2. + 0. ÂW
The impedance in phasor form is 2.W —0.°
Solution:
The last term (parallel L and C) is
1
0
= ¶, so the impedance is infinite.
‡ Problem 3: In the circuit below, determine the value of the capacitance such that
the current is in phase with the voltage at a frequency of 1000 Hz.
Chapter 8.07.nb
3
Problem 3: In the circuit below, determine the value of the capacitance such that
the current is in phase with the voltage at a frequency of 1000 Hz.
In[44]:=
w = 2 p * 1000.; inductance = .005; capacitance =
w2 * inductance
Print@"The capacitance needed is ", capacitance, "F"D
The capacitance needed is 5.06606 µ 10-6 F
‡
1
;
4
Chapter 8.07.nb
Solution:
For the voltage and current to be in phase, all imaginary parts of the impedance must
1
1
vanish. This means that w C = w L, or C = 2 ï C = 5.06 µF.
w L
‡ Problem 4: Find the frequency at which the circuit below is purely
resistive.
Solution:
For the impedance to be purely resistive, all imaginary parts of the impedance must vanish.
1
1
w
This means that w C = w L, or w =
. f = 2 p = 71.2 Hz
LC
In[46]:=
w=
1
H.005L H.001L
; f=
w
2p
;
Print@"The frequency at which this occurs is ", f, " Hz"D
The frequency at which this occurs is 71.1763 Hz
‡ Problem 5: Find the current V0 in the circuit below.
a) Find the impedance.
b) Find the current from I = V / Z (phasor form)
c) Use V = I Z at the capacitor to find V0
Problem 5: Find the current V0 in the circuit below.
a) Find the impedance.
Chapter 8.07.nb
b) Find the current from I = V / Z (phasor form)
c) Use V = I Z at the capacitor to find V0
Solution:
a) The impedance is 2 - j2 W = 2.83 W ­∡-45°.
In[73]:=
impedance = 2. - Â 12 + Â 10;
Print@"The voltage in phasor form is ",
Abs@voltageD, "V —", Arg@voltageD ê Degree, "°"D
The voltage in phasor form is 10.V —120.°
b) Then I = V / R = (10V ­∡ 120°) / (2.83 W ­∡-45°) = 3.536 A ­∡165°
In[78]:=
voltage = 10. Cos@120 DegreeD + Â * 10. Sin@120 DegreeD;
Print@"The voltage in phasor form is ",
Abs@voltageD, "V —", Arg@voltageD ê Degree, "°"D
current = voltage ê impedance; Print@"The current in phasor form is ",
Abs@currentD, "A —", Arg@currentD ê Degree, "°"D
The voltage in phasor form is 10.V —120.°
‡
The current in phasor form is 3.53553A —165.°
c) Then V0 =I R = (3.536 A ­∡165°) (12 W ­∡-90°) = 42.4 A ­∡75°
In[81]:=
capacitiveReactance = - Â 12;
OutputVoltage = current * capacitiveReactance;
Print@"V0 in phasor form is ",
Abs@OutputVoltageD, "V —", Arg@OutputVoltageD ê Degree, "°"D
V0 in phasor form is 42.4264V —75.°
Note: This is larger than the input voltage, but out of phase by 75°.
5