exam 2 review review packets...1)express the voltage sources as a phasor. 2)find the impedance...
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1) Express the Voltage Sources as a phasor.
2) Find the Impedance values for the given elements.
a. Let ð = ðððð ððð
b. Let ð = ðððððð ðð³ð³ð³ð³/ðð
c. Let ð = ðððððððð ðð³ð³ð³ð³/ðð
132.63 mH
193.42 µF
80.5 Ω
ðððð.ðððððððððð(ðððððððð â ðððððð) V + -
**Make sure to convert the Amplitude to the RMS value.
21.21â2
= 15 ð ⎠ððððâ¡âðððððð ðœðœ
ð = 2ðð = (2ð)(60) = 377 ððð/ð
ðð¿ = ððð¿ = ð(377)(132.63ð¥10â3) = ðððððð Ω
ðð¶ = âð1ðð¶
= âð1
(235)(193.42ð¥10â6)
ðð¶ = âðððððð Ω
ðð = ð = ðððð.ðð Ω
EE301 Exam 2 Review
Exam 2 Review
3) Redraw the circuit below. Express the source in phasor form and the elements as Impedances.
4) Find the total Impedance as seen by the Source.
ðð = 10 + 1
âð12+
121
+1
(âð5 + ð20)â1
= 10 + [ð0.083 + 0.048â ð0.067]â1
ðð = 10 +1
0.048 + ð0.016= 10 +
10.051â¡18.43ð
= 10 + 19.61â¡â18.43ð
ðð = 10 + 18.6â ð6.2 = 28.6â ð6.2 = ðððð.ððððâ¡âðððð.ðððððð Ω
5) Find (Is ), (I1 ), (I2), and (I3). Use Current Divider Rule when finding (I1), (I2), and (I3).
ð21 = ðððð Ω
ðð¶2 = âð 1(377)(530.5ð¥10â6) = âðððð Ω
ðð¿ = ð(377)(53.05ð¥10â3) = ðððððð Ω
ð°ð°ðð
10 Ω
ðððð.ðððððððð(ðððððððð + ðððððð) V ðððððð ðððð ðððð.ðððð ðððð ðððð Ω
ðððððð.ðð ðððð
+
- ðœðœð³ð³ ð°ð°ðð ð°ð°ðð
ð°ð°ðð
+ -
ð°ð°ðð
10 Ω
ðððð Ω
+
- ðœðœð³ð³ ð°ð°ðð ð°ð°ðð
ð°ð°ðð
+ -
ðœðœðð = ððððâ¡ðððððð V âðððððð Ω ðððððð Ω
âðððð Ω
ðð =19.8â2
â¡30ð = ððððâ¡ðððððð ðœðœ
ð10 = ðððð Ω
ðð¶1 = âð 1(377)(221ð¥10â6) = âðððððð Ω
ðŒð =ðð ðð
=14â¡30ð
29.26â¡â12.23ð= ðð.ððððððâ¡ðððð.ðððððð ðšðš
ððð = 1
âð12+
121
+1
(âð5 + ð20)â1
ððð = 19.61â¡â18.43ð Ω
ðŒ1 =ððððð¶1
(ðŒð ) =19.61â¡â18.43ð
12â¡â90ð(0.478â¡42.23ð) = ðð.ððððððâ¡ðððððð.ðððð ðšðš
ðŒ2 =ðððð21
(ðŒð ) =19.61â¡â18.43ð
21â¡0ð(0.478â¡42.23ð) = ðð.ððððððâ¡ðððð.ðððð ðšðš
ðŒ3 =ððððð¶2+ð¿
(ðŒð ) =19.61â¡â18.43ð
15â¡90ð(0.478â¡42.23ð) = ðð.ððððððâ¡âðððð.ðððð ðšðš
ðð¿ = (ðŒ3)(ðð¿) = (0.625â¡â66.2ð)(20â¡90ð) = 12.5 â¡23.8ð ðœ
7) When solving Thevenin Equivalent problems there are logical steps that hold true no matter ifthe circuit is DC or AC. Answer the following with True or False.
a. ____ If a Load is shown on the circuit, the first step is to remove the Load.b. ____ The Thevenin Impedance is always the impedance as seen by the Source.c. ____ When solving the Thevenin Impedance, we must âturn offâ the sources by
replacing both current and voltage sources as SHORTS.d. ____ When solving the Thevenin Impedance, we must âturn offâ the sources by
replacing the current source a SHORT and the voltage sources as an OPEN.e. ____ When solving the Thevenin Impedance, we must âturn offâ the sources by
replacing the current source an OPEN and the voltage sources as a SHORT.f. ____ When finding the Thevenin Voltage we use the open terminals were the Thevenin
perspective is.
8) Find the Thevenin Impedance and Voltage external to ZLoad for the circuit below. Draw theThevenin circuit with the Load attached.
ððð³ð³ððð³ð³ð³ð³
ððâ¡ðððððð ðšðš
ðððð Ω ðððððð Ω
ðððð Ω -ðððð Ω
ðð.ðð Ω
ðððð.ðð Ω
T F F
F
T
T
First step is to remove the load and find ððð¶ð¶ðð. Remember the current source is replaced with an OPEN.
ððð¶ð¶ðð
ðððð Ω ðððððð Ω
ðððð Ω -ðððð Ω
+
-
ððð» =(16â¡0ð)(8â¡â90ð)
16 â ð8=
(16â¡0ð)(8â¡â90ð)17.89â¡â26.57ð
ððð» = 7.15â¡â63.43ð = ðð.ðð â ðððð.ðð Ω
6) Find the voltage across the Inductor (VL).
Now âturn onâ the source and find ðœðœð¶ð¶ðð at the open terminals.
ððð = (16)||(âð8) = 7.15â¡â63.43ð Ω
ððð» = (ðŒð )ððð = (8â¡30ð)(7.15â¡â63.43ð) = ðððð.ððâ¡âðððð.ðððððð ðœðœ
9) Is Maximum Power being delivered to the Load and why?
ððð¶ð¶ðð = ððð³ð³â
3.2â ð6.4 = 3.2 + ð6.4
7.15â¡â63.43ð = 7.15â¡63.43ð
10) The acronym used for Capacitors in AC circuits is âICEâ. Explain what is meant with thisacronym.
Current LEADS Voltage for a capacitive circuit. If we are just analyzing a Capacitorâs current andvoltage relationship we will see the current LEADS the voltage by 900.
ðœðœð¶ð¶ðð
ðððð Ω ðððððð Ω
ðððð Ω -ðððð Ω
+
-
ððâ¡ðððððð ðšðš
ððð» is the voltage across the 16 Ω and the âj8 Ω. We can use Nodal Analysis or combine the 16 Ω and âj8 Ω impedances in parallel and find the voltage across them.
ðð.ðð Ω
ðððð.ððâ¡âðððð.ðððððð ðœðœ +-
âðððð.ðð Ω
ðð.ðð Ω
ðððð.ðð Ω
ðªðª +
- ðœðœðªðª
ð°ð°ðªðª ð°ð°ðªðª ðœðœðªðª
ðœðœðœðœ
This implies the Current Angle (ðœðœðð) equals to ðœðœðœðœ + ðððððð.
ðððð
YES!
11) The acronym used for Inductors in AC circuits is âELIâ. Explain what is meant with this acronym.
Voltage LEADS Current for an inductive circuit. If we are just analyzing an Inductorâs current andvoltage relationship we will see the voltage LEADS the current by 900.
The Graph and circuit below are for questions 12 through 15
6
3
0
-3
-6
3.75 2.5 1.25 5 6.25 7.5
V
t
ðœðœðð
ðœðœð¹ð¹
**Time in milliseconds
ðœðœðð + - ððð³ð³ððð³ð³ð³ð³
ðð Ω
+
-
ðœðœð³ð³ ðœðœð¹ð¹ +-
ð³ð³ +
- ðœðœð³ð³
ð°ð°ð³ð³ ð°ð°ð³ð³
ðœðœð³ð³
ðœðœðð
This implies the Voltage Angle (ðœðœðœðœ) equals to ðœðœðð + ðððððð.
ðððð
12) For the Graph on the other page answer the following questions.
a. What is the period (T) of the waveforms?
ð = ð ms
b. What is the frequency of the waveforms (ð)?
ð =1ð
=1
5ð¥10â3= ðððððð ððð
c. What is the time difference (âðð) between ðœðœðð ððð ðœðœð¹ð¹?
The âðð between the peak of ðœðœðð and the peak of ðœðœð¹ð¹ is about 0.5 ms.âð¡ = ðð.ðð ðððð
d. What is the phase difference (âðœðœ) between ðœðœðð ððð ðœðœð¹ð¹?
âð =âð¡ð
(360ð) =0.5ð¥10â3
5ð¥10â3(360ð) = ðððððð
e. Write the function for ðœðœðð(ðð).
ðð (ð¡) = ðð ð¬ð¢ð§[ððð (ðððððð)ðð] ðœðœ
f. Express ðœðœðð(ðð) in phasor form.
ðð =6â2
â¡0ð = ðð.ððððâ¡ðððð ðœðœ
g. Write the function for ðœðœð¹ð¹(ðð). (Use ðð as the reference angle)
ðð (ð¡) = ðð ð¬ð¢ð§[ððð (ðððððð)ðð â ðððððð] ðœðœ
h. Express ðœðœð¹ð¹(ðð) in phasor form.
ðð =3â2
â¡â36ð = ðð.ððððâ¡âðððððð ðœðœ
13) Since we know VS aðð VR, find the voltage across the Load (VL). Refer to the circuit atthe bottom of the other page. (Hint: use KVL)
âðð + ðð¿ + ðð = 0
ðð¿ = ðð â ðð = 4.24â¡0ð â 2.12â¡â36ð = 4.24â (1.72â ð1.25) = 2.52 + ð1.25 ð
ðð¿ = ðð.ððððâ¡ðððð.ðððððð ðœðœ
14) Find ZLoad; Is the Load Capacitive or Inductive?
ðŒð = ðŒð =ðð ð
=2.12â¡â36ð
2= 1.06â¡â36ð ðŽ
ðð¿ððð =ðð¿ðŒð
=2.81â¡26.38ð
1.06â¡â36ð= 2.65â¡62.38ð = ðð.ðððð + ðððð.ðððð Ω
Since the imaginary component of ðð¿ððð is positive then the Load is Inductive.
15) Draw the Load equivalent circuit below.
ðœðœðð + -
ðð Ω
+
-
ðœðœð³ð³ ðœðœð¹ð¹ +-
ðð.ðððð Ω
ðððð.ðððð Ω
ðœðœðð + -
ðð Ω
+
-
ðœðœð³ð³ ðœðœð¹ð¹ +-
ð°ð°ðð
ð°ð°ð¹ð¹
16) The following circuit is operating at a frequency of ð = 500 rad/s.
a. Draw the Power Triangle for the Load and label the Apparent, Real, and ReactivePower. Also include the Complex Power angle.
b. Find the source current (ð°ð°ðð).
c. Typically we like to reduce the source current. This can be accomplished by connectinga Capacitor in parallel to an Inductive Load. The Capacitor component value (ðªðª) isdetermined by choosing a Capacitance Reactive Power (ðžðð) equal to the InductanceReactive Power (ðžð³ð³).
Find the Capacitor component value so all of the Reactive Power at the Load iscancelled. In other words, what component value will correct the power factor tounity?
ðððððð ðŸðŸ
+
-
ððððððâ¡ðððððð ðœðœ
ð°ð°ðð
ðððððð ðœðœðœðœðœðœ
ð = 1602 + 4002 = 430.81 ðððŽ
ð = tanâ1 400160
= 68.2ð
ðððððð ðŸðŸ
ðððððð ðœðœðœðœðœðœ
ðððð.ðððð
ð = ððð ðŒð â
ðŒð â
=430.81â¡68.2ð
220â¡40ð= 1.96â¡28.2ð ðŽ
ðŒð = ðð.ððððâ¡âðððð.ðððð ðœðœ
|ðð¿| = |ðð| = 400 ðððŽð
ðð =ðð2
ðð => ðð =
(220)2
400= 121 Ω
ðð =1ðð¶
=> ð¶ =1
(500)(121)= ðððð.ðððð ðððð
d. Find the source current (ð°ð°ðð) when a Capacitor is connected in parallel with the Load.Assume the Capacitance Reactive Power is âðððððð ðœðœðœðœðœðœ.
ðððððð ðŸðŸ
+
-
ððððððâ¡ðððððð ðœðœ
ð°ð°ðð
ðððððð ðœðœðœðœðœðœ
âðððððð ðœðœðœðœðœðœ
ð°ð°ðð ð°ð°ðð
ðŒ1 = 1.96â¡â28.2ð ðŽ
(ðŒ2)â =ð
ððð =400â¡â900
220â¡40ð= 1.82â¡â130ð ðŽ
ðŒ2 = 1.82â¡130ð ðŽ
ðŒð = ðŒ1 + ðŒ2
ðŒð = (1.96â¡â28.2ð) + (1.82â¡130ð)
ðŒð = ðð.ððððððâ¡ðððð.ðððððð ðœðœ
17) Answer the following questions for the below Resonant Circuit
a. Find the value of L in millihenries if the resonant frequency is 1800 Hz.
b. Calculate XL and XC. How do they compare?
c. Find the magnitude of the current Irms at resonance.
d. Find the power dissipated by the circuit at resonance.
e. Calculate the Q of the circuit and the resulting bandwidth.
f. Calculate the cutoff frequencies, and calculate the power dissipated bythe circuit at these frequencies.
4
3.54 mA
3.54 mA 50.13
11.05
11.05162.9
1800 Hz + 81.5 Hz = 1881.5 Hz
1800 Hz - 81.5 Hz = 1718.5 Hz
50.13 25.06
18) Answer the following questions for the Low-Pass Filter
a. Determine fc.
b. Find Av = Vo/Vi at 0.1fc, and compare to the maximum value of 1 for the low-frequency range.
c. Find Av = Vo/Vi at 10fc, and compare to the minimum value of 0 for the low-frequency range.
d. Determine the frequency at which Av = 0.01 or Vo = (1/100)Vi.
19) Answer the following questions for the High-Pass Filter
a. Determine fc.
b. Find Av = Vo/Vi at 0.01fc, and compare to the minimum level of 0 for the low-frequency region.
c. Find Av = Vo/Vi at 100fc, and compare to the maximum level of 1 for the high-frequency region.
d. Determine the frequency at which Vo = (1/2)Vi.