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1) Express the Voltage Sources as a phasor.
2) Find the Impedance values for the given elements.
a. Let ๐ = ๐๐๐๐ ๐๐๐
b. Let ๐ = ๐๐๐๐๐๐ ๐๐ณ๐ณ๐ณ๐ณ/๐๐
c. Let ๐ = ๐๐๐๐๐๐๐๐ ๐๐ณ๐ณ๐ณ๐ณ/๐๐
132.63 mH
193.42 ยตF
80.5 ฮฉ
๐๐๐๐.๐๐๐๐๐๐๐๐๐๐(๐๐๐๐๐๐๐๐ โ ๐๐๐๐๐๐) V + -
**Make sure to convert the Amplitude to the RMS value.
21.21โ2
= 15 ๐ โด ๐๐๐๐โกโ๐๐๐๐๐๐ ๐ฝ๐ฝ
๐ = 2๐๐ = (2๐)(60) = 377 ๐๐๐/๐
๐๐ฟ = ๐๐๐ฟ = ๐(377)(132.63๐ฅ10โ3) = ๐๐๐๐๐๐ ฮฉ
๐๐ถ = โ๐1๐๐ถ
= โ๐1
(235)(193.42๐ฅ10โ6)
๐๐ถ = โ๐๐๐๐๐๐ ฮฉ
๐๐ = ๐ = ๐๐๐๐.๐๐ ฮฉ
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.
๐๐ = 10 + 1
โ๐12+
121
+1
(โ๐5 + ๐20)โ1
= 10 + [๐0.083 + 0.048โ ๐0.067]โ1
๐๐ = 10 +1
0.048 + ๐0.016= 10 +
10.051โก18.43๐
= 10 + 19.61โกโ18.43๐
๐๐ = 10 + 18.6โ ๐6.2 = 28.6โ ๐6.2 = ๐๐๐๐.๐๐๐๐โกโ๐๐๐๐.๐๐๐๐๐๐ ฮฉ
5) Find (Is ), (I1 ), (I2), and (I3). Use Current Divider Rule when finding (I1), (I2), and (I3).
๐21 = ๐๐๐๐ ฮฉ
๐๐ถ2 = โ๐ 1(377)(530.5๐ฅ10โ6) = โ๐๐๐๐ ฮฉ
๐๐ฟ = ๐(377)(53.05๐ฅ10โ3) = ๐๐๐๐๐๐ ฮฉ
๐ฐ๐ฐ๐๐
10 ฮฉ
๐๐๐๐.๐๐๐๐๐๐๐๐(๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐) V ๐๐๐๐๐๐ ๐๐๐๐ ๐๐๐๐.๐๐๐๐ ๐๐๐๐ ๐๐๐๐ ฮฉ
๐๐๐๐๐๐.๐๐ ๐๐๐๐
+
- ๐ฝ๐ฝ๐ณ๐ณ ๐ฐ๐ฐ๐๐ ๐ฐ๐ฐ๐๐
๐ฐ๐ฐ๐๐
+ -
๐ฐ๐ฐ๐๐
10 ฮฉ
๐๐๐๐ ฮฉ
+
- ๐ฝ๐ฝ๐ณ๐ณ ๐ฐ๐ฐ๐๐ ๐ฐ๐ฐ๐๐
๐ฐ๐ฐ๐๐
+ -
๐ฝ๐ฝ๐๐ = ๐๐๐๐โก๐๐๐๐๐๐ V โ๐๐๐๐๐๐ ฮฉ ๐๐๐๐๐๐ ฮฉ
โ๐๐๐๐ ฮฉ
๐๐ =19.8โ2
โก30๐ = ๐๐๐๐โก๐๐๐๐๐๐ ๐ฝ๐ฝ
๐10 = ๐๐๐๐ ฮฉ
๐๐ถ1 = โ๐ 1(377)(221๐ฅ10โ6) = โ๐๐๐๐๐๐ ฮฉ
๐ผ๐ =๐๐ ๐๐
=14โก30๐
29.26โกโ12.23๐= ๐๐.๐๐๐๐๐๐โก๐๐๐๐.๐๐๐๐๐๐ ๐จ๐จ
๐๐๐ = 1
โ๐12+
121
+1
(โ๐5 + ๐20)โ1
๐๐๐ = 19.61โกโ18.43๐ ฮฉ
๐ผ1 =๐๐๐๐๐ถ1
(๐ผ๐ ) =19.61โกโ18.43๐
12โกโ90๐(0.478โก42.23๐) = ๐๐.๐๐๐๐๐๐โก๐๐๐๐๐๐.๐๐๐๐ ๐จ๐จ
๐ผ2 =๐๐๐๐21
(๐ผ๐ ) =19.61โกโ18.43๐
21โก0๐(0.478โก42.23๐) = ๐๐.๐๐๐๐๐๐โก๐๐๐๐.๐๐๐๐ ๐จ๐จ
๐ผ3 =๐๐๐๐๐ถ2+๐ฟ
(๐ผ๐ ) =19.61โกโ18.43๐
15โก90๐(0.478โก42.23๐) = ๐๐.๐๐๐๐๐๐โกโ๐๐๐๐.๐๐๐๐ ๐จ๐จ
๐๐ฟ = (๐ผ3)(๐๐ฟ) = (0.625โกโ66.2๐)(20โก90๐) = 12.5 โก23.8๐ ๐ฝ
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 โturn offโ the sources by
replacing both current and voltage sources as SHORTS.d. ____ When solving the Thevenin Impedance, we must โturn offโ the sources by
replacing the current source a SHORT and the voltage sources as an OPEN.e. ____ When solving the Thevenin Impedance, we must โturn offโ 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.
๐๐๐ณ๐ณ๐๐๐ณ๐ณ๐ณ๐ณ
๐๐โก๐๐๐๐๐๐ ๐จ๐จ
๐๐๐๐ ฮฉ ๐๐๐๐๐๐ ฮฉ
๐๐๐๐ ฮฉ -๐๐๐๐ ฮฉ
๐๐.๐๐ ฮฉ
๐๐๐๐.๐๐ ฮฉ
T F F
F
T
T
First step is to remove the load and find ๐๐๐ถ๐ถ๐๐. Remember the current source is replaced with an OPEN.
๐๐๐ถ๐ถ๐๐
๐๐๐๐ ฮฉ ๐๐๐๐๐๐ ฮฉ
๐๐๐๐ ฮฉ -๐๐๐๐ ฮฉ
+
-
๐๐๐ป =(16โก0๐)(8โกโ90๐)
16 โ ๐8=
(16โก0๐)(8โกโ90๐)17.89โกโ26.57๐
๐๐๐ป = 7.15โกโ63.43๐ = ๐๐.๐๐ โ ๐๐๐๐.๐๐ ฮฉ
6) Find the voltage across the Inductor (VL).
Now โturn onโ the source and find ๐ฝ๐ฝ๐ถ๐ถ๐๐ at the open terminals.
๐๐๐ = (16)||(โ๐8) = 7.15โกโ63.43๐ ฮฉ
๐๐๐ป = (๐ผ๐ )๐๐๐ = (8โก30๐)(7.15โกโ63.43๐) = ๐๐๐๐.๐๐โกโ๐๐๐๐.๐๐๐๐๐๐ ๐ฝ๐ฝ
9) Is Maximum Power being delivered to the Load and why?
๐๐๐ถ๐ถ๐๐ = ๐๐๐ณ๐ณโ
3.2โ ๐6.4 = 3.2 + ๐6.4
7.15โกโ63.43๐ = 7.15โก63.43๐
10) The acronym used for Capacitors in AC circuits is โICEโ. Explain what is meant with thisacronym.
Current LEADS Voltage for a capacitive circuit. If we are just analyzing a Capacitorโs current andvoltage relationship we will see the current LEADS the voltage by 900.
๐ฝ๐ฝ๐ถ๐ถ๐๐
๐๐๐๐ ฮฉ ๐๐๐๐๐๐ ฮฉ
๐๐๐๐ ฮฉ -๐๐๐๐ ฮฉ
+
-
๐๐โก๐๐๐๐๐๐ ๐จ๐จ
๐๐๐ป is the voltage across the 16 ฮฉ and the โj8 ฮฉ. We can use Nodal Analysis or combine the 16 ฮฉ and โj8 ฮฉ impedances in parallel and find the voltage across them.
๐๐.๐๐ ฮฉ
๐๐๐๐.๐๐โกโ๐๐๐๐.๐๐๐๐๐๐ ๐ฝ๐ฝ +-
โ๐๐๐๐.๐๐ ฮฉ
๐๐.๐๐ ฮฉ
๐๐๐๐.๐๐ ฮฉ
๐ช๐ช +
- ๐ฝ๐ฝ๐ช๐ช
๐ฐ๐ฐ๐ช๐ช ๐ฐ๐ฐ๐ช๐ช ๐ฝ๐ฝ๐ช๐ช
๐ฝ๐ฝ๐ฝ๐ฝ
This implies the Current Angle (๐ฝ๐ฝ๐๐) equals to ๐ฝ๐ฝ๐ฝ๐ฝ + ๐๐๐๐๐๐.
๐๐๐๐
YES!
11) The acronym used for Inductors in AC circuits is โELIโ. Explain what is meant with this acronym.
Voltage LEADS Current for an inductive circuit. If we are just analyzing an Inductorโs 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
๐ฝ๐ฝ๐๐
๐ฝ๐ฝ๐น๐น
**Time in milliseconds
๐ฝ๐ฝ๐๐ + - ๐๐๐ณ๐ณ๐๐๐ณ๐ณ๐ณ๐ณ
๐๐ ฮฉ
+
-
๐ฝ๐ฝ๐ณ๐ณ ๐ฝ๐ฝ๐น๐น +-
๐ณ๐ณ +
- ๐ฝ๐ฝ๐ณ๐ณ
๐ฐ๐ฐ๐ณ๐ณ ๐ฐ๐ฐ๐ณ๐ณ
๐ฝ๐ฝ๐ณ๐ณ
๐ฝ๐ฝ๐๐
This implies the Voltage Angle (๐ฝ๐ฝ๐ฝ๐ฝ) equals to ๐ฝ๐ฝ๐๐ + ๐๐๐๐๐๐.
๐๐๐๐
12) For the Graph on the other page answer the following questions.
a. What is the period (T) of the waveforms?
๐ = ๐ ms
b. What is the frequency of the waveforms (๐)?
๐ =1๐
=1
5๐ฅ10โ3= ๐๐๐๐๐๐ ๐๐๐
c. What is the time difference (โ๐๐) between ๐ฝ๐ฝ๐๐ ๐๐๐ ๐ฝ๐ฝ๐น๐น?
The โ๐๐ between the peak of ๐ฝ๐ฝ๐๐ and the peak of ๐ฝ๐ฝ๐น๐น is about 0.5 ms.โ๐ก = ๐๐.๐๐ ๐๐๐๐
d. What is the phase difference (โ๐ฝ๐ฝ) between ๐ฝ๐ฝ๐๐ ๐๐๐ ๐ฝ๐ฝ๐น๐น?
โ๐ =โ๐ก๐
(360๐) =0.5๐ฅ10โ3
5๐ฅ10โ3(360๐) = ๐๐๐๐๐๐
e. Write the function for ๐ฝ๐ฝ๐๐(๐๐).
๐๐ (๐ก) = ๐๐ ๐ฌ๐ข๐ง[๐๐๐ (๐๐๐๐๐๐)๐๐] ๐ฝ๐ฝ
f. Express ๐ฝ๐ฝ๐๐(๐๐) in phasor form.
๐๐ =6โ2
โก0๐ = ๐๐.๐๐๐๐โก๐๐๐๐ ๐ฝ๐ฝ
g. Write the function for ๐ฝ๐ฝ๐น๐น(๐๐). (Use ๐๐ as the reference angle)
๐๐ (๐ก) = ๐๐ ๐ฌ๐ข๐ง[๐๐๐ (๐๐๐๐๐๐)๐๐ โ ๐๐๐๐๐๐] ๐ฝ๐ฝ
h. Express ๐ฝ๐ฝ๐น๐น(๐๐) in phasor form.
๐๐ =3โ2
โกโ36๐ = ๐๐.๐๐๐๐โกโ๐๐๐๐๐๐ ๐ฝ๐ฝ
13) Since we know VS a๐๐ VR, find the voltage across the Load (VL). Refer to the circuit atthe bottom of the other page. (Hint: use KVL)
โ๐๐ + ๐๐ฟ + ๐๐ = 0
๐๐ฟ = ๐๐ โ ๐๐ = 4.24โก0๐ โ 2.12โกโ36๐ = 4.24โ (1.72โ ๐1.25) = 2.52 + ๐1.25 ๐
๐๐ฟ = ๐๐.๐๐๐๐โก๐๐๐๐.๐๐๐๐๐๐ ๐ฝ๐ฝ
14) Find ZLoad; Is the Load Capacitive or Inductive?
๐ผ๐ = ๐ผ๐ =๐๐ ๐
=2.12โกโ36๐
2= 1.06โกโ36๐ ๐ด
๐๐ฟ๐๐๐ =๐๐ฟ๐ผ๐
=2.81โก26.38๐
1.06โกโ36๐= 2.65โก62.38๐ = ๐๐.๐๐๐๐ + ๐๐๐๐.๐๐๐๐ ฮฉ
Since the imaginary component of ๐๐ฟ๐๐๐ is positive then the Load is Inductive.
15) Draw the Load equivalent circuit below.
๐ฝ๐ฝ๐๐ + -
๐๐ ฮฉ
+
-
๐ฝ๐ฝ๐ณ๐ณ ๐ฝ๐ฝ๐น๐น +-
๐๐.๐๐๐๐ ฮฉ
๐๐๐๐.๐๐๐๐ ฮฉ
๐ฝ๐ฝ๐๐ + -
๐๐ ฮฉ
+
-
๐ฝ๐ฝ๐ณ๐ณ ๐ฝ๐ฝ๐น๐น +-
๐ฐ๐ฐ๐๐
๐ฐ๐ฐ๐น๐น
16) The following circuit is operating at a frequency of ๐ = 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 (๐ฐ๐ฐ๐๐).
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 (๐ช๐ช) isdetermined by choosing a Capacitance Reactive Power (๐ธ๐๐) equal to the InductanceReactive Power (๐ธ๐ณ๐ณ).
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?
๐๐๐๐๐๐ ๐พ๐พ
+
-
๐๐๐๐๐๐โก๐๐๐๐๐๐ ๐ฝ๐ฝ
๐ฐ๐ฐ๐๐
๐๐๐๐๐๐ ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ
๐ = 1602 + 4002 = 430.81 ๐๐๐ด
๐ = tanโ1 400160
= 68.2๐
๐๐๐๐๐๐ ๐พ๐พ
๐๐๐๐๐๐ ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ
๐๐๐๐.๐๐๐๐
๐ = ๐๐๐ ๐ผ๐ โ
๐ผ๐ โ
=430.81โก68.2๐
220โก40๐= 1.96โก28.2๐ ๐ด
๐ผ๐ = ๐๐.๐๐๐๐โกโ๐๐๐๐.๐๐๐๐ ๐ฝ๐ฝ
|๐๐ฟ| = |๐๐| = 400 ๐๐๐ด๐
๐๐ =๐๐2
๐๐ => ๐๐ =
(220)2
400= 121 ฮฉ
๐๐ =1๐๐ถ
=> ๐ถ =1
(500)(121)= ๐๐๐๐.๐๐๐๐ ๐๐๐๐
d. Find the source current (๐ฐ๐ฐ๐๐) when a Capacitor is connected in parallel with the Load.Assume the Capacitance Reactive Power is โ๐๐๐๐๐๐ ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ.
๐๐๐๐๐๐ ๐พ๐พ
+
-
๐๐๐๐๐๐โก๐๐๐๐๐๐ ๐ฝ๐ฝ
๐ฐ๐ฐ๐๐
๐๐๐๐๐๐ ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ
โ๐๐๐๐๐๐ ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ
๐ฐ๐ฐ๐๐ ๐ฐ๐ฐ๐๐
๐ผ1 = 1.96โกโ28.2๐ ๐ด
(๐ผ2)โ =๐
๐๐๐ =400โกโ900
220โก40๐= 1.82โกโ130๐ ๐ด
๐ผ2 = 1.82โก130๐ ๐ด
๐ผ๐ = ๐ผ1 + ๐ผ2
๐ผ๐ = (1.96โกโ28.2๐) + (1.82โก130๐)
๐ผ๐ = ๐๐.๐๐๐๐๐๐โก๐๐๐๐.๐๐๐๐๐๐ ๐ฝ๐ฝ
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.