1) Express the Voltage Sources as a phasor.

2) Find the Impedance values for the given elements.

a. Let 𝒇 = 𝟏𝟏𝟎𝟎 𝟑𝟑𝒛

b. Let 𝝎 = 𝟏𝟏𝟑𝟑𝟓𝟓 𝒓𝑳𝑳𝑳𝑳/𝟐𝟐

c. Let 𝝎 = 𝟏𝟏𝟓𝟓𝟎𝟎𝟎𝟎 𝒓𝑳𝑳𝑳𝑳/𝟐𝟐

132.63 mH

193.42 µF

80.5 Ω

𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏𝟐𝟐𝒊𝒊𝟐𝟐(𝟏𝟏𝟏𝟏𝟎𝟎𝒕𝒕 − 𝟏𝟏𝟑𝟑𝒐𝒐) V + -

**Make sure to convert the Amplitude to the RMS value.

21.21√2

= 15 𝑉 ∴ 𝟏𝟏𝟓𝟓∡−𝟏𝟏𝟑𝟑𝒐𝒐 𝑽𝑽

𝜔 = 2𝜋𝑓 = (2𝜋)(60) = 377 𝑟𝑎𝑑/𝑠

𝑍𝐿 = 𝑗𝜔𝐿 = 𝑗(377)(132.63𝑥10−3) = 𝒋𝒋𝟓𝟓𝟎𝟎 Ω

𝑍𝐶 = −𝑗1𝜔𝐶

= −𝑗1

(235)(193.42𝑥10−6)

𝑍𝐶 = −𝒋𝒋𝟏𝟏𝟏𝟏 Ω

𝑍𝑅 = 𝑅 = 𝟖𝟖𝟎𝟎.𝟓𝟓 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.

𝑍𝑇 = 10 + 1

−𝑗12+

121

+1

(−𝑗5 + 𝑗20)−1

= 10 + [𝑗0.083 + 0.048− 𝑗0.067]−1

𝑍𝑇 = 10 +1

0.048 + 𝑗0.016= 10 +

10.051∡18.43𝑜

= 10 + 19.61∡−18.43𝑜

𝑍𝑇 = 10 + 18.6− 𝑗6.2 = 28.6− 𝑗6.2 = 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏∡−𝟏𝟏𝟏𝟏.𝟏𝟏𝟑𝟑𝒐𝒐 Ω

5) Find (Is ), (I1 ), (I2), and (I3). Use Current Divider Rule when finding (I1), (I2), and (I3).

𝑍21 = 𝟏𝟏𝟏𝟏 Ω

𝑍𝐶2 = −𝑗 1(377)(530.5𝑥10−6) = −𝒋𝒋𝟓𝟓 Ω

𝑍𝐿 = 𝑗(377)(53.05𝑥10−3) = 𝒋𝒋𝟏𝟏𝟎𝟎 Ω

𝑰𝑰𝟐𝟐

10 Ω

𝟏𝟏𝟏𝟏.𝟖𝟖𝟐𝟐𝒊𝒊𝟐𝟐(𝟑𝟑𝟑𝟑𝟑𝟑𝒕𝒕 + 𝟑𝟑𝟎𝟎𝒐𝒐) V 𝟏𝟏𝟏𝟏𝟏𝟏 𝝁𝝁𝝁𝝁 𝟓𝟓𝟑𝟑.𝟎𝟎𝟓𝟓 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 Ω

𝟓𝟓𝟑𝟑𝟎𝟎.𝟓𝟓 𝝁𝝁𝝁𝝁

+

- 𝑽𝑽𝑳𝑳 𝑰𝑰𝟏𝟏 𝑰𝑰𝟏𝟏

𝑰𝑰𝟑𝟑

+ -

𝑰𝑰𝟐𝟐

10 Ω

𝟏𝟏𝟏𝟏 Ω

+

- 𝑽𝑽𝑳𝑳 𝑰𝑰𝟏𝟏 𝑰𝑰𝟏𝟏

𝑰𝑰𝟑𝟑

+ -

𝑽𝑽𝟐𝟐 = 𝟏𝟏𝟏𝟏∡𝟑𝟑𝟎𝟎𝒐𝒐 V −𝒋𝒋𝟏𝟏𝟏𝟏 Ω 𝒋𝒋𝟏𝟏𝟎𝟎 Ω

−𝒋𝒋𝟓𝟓 Ω

𝑉𝑠 =19.8√2

∡30𝑜 = 𝟏𝟏𝟏𝟏∡𝟑𝟑𝟎𝟎𝒐𝒐 𝑽𝑽

𝑍10 = 𝟏𝟏𝟎𝟎 Ω

𝑍𝐶1 = −𝑗 1(377)(221𝑥10−6) = −𝒋𝒋𝟏𝟏𝟏𝟏 Ω

𝐼𝑠 =𝑉𝑠𝑍𝑇

=14∡30𝑜

29.26∡−12.23𝑜= 𝟎𝟎.𝟏𝟏𝟑𝟑𝟖𝟖∡𝟏𝟏𝟏𝟏.𝟏𝟏𝟑𝟑𝒐𝒐 𝑨𝑨

𝑍𝑒𝑞 = 1

−𝑗12+

121

+1

(−𝑗5 + 𝑗20)−1

𝑍𝑒𝑞 = 19.61∡−18.43𝑜 Ω

𝐼1 =𝑍𝑒𝑞𝑍𝐶1

(𝐼𝑠) =19.61∡−18.43𝑜

12∡−90𝑜(0.478∡42.23𝑜) = 𝟎𝟎.𝟑𝟑𝟖𝟖𝟏𝟏∡𝟏𝟏𝟏𝟏𝟑𝟑.𝟖𝟖𝒐𝒐 𝑨𝑨

𝐼2 =𝑍𝑒𝑞𝑍21

(𝐼𝑠) =19.61∡−18.43𝑜

21∡0𝑜(0.478∡42.23𝑜) = 𝟎𝟎.𝟏𝟏𝟏𝟏𝟏𝟏∡𝟏𝟏𝟑𝟑.𝟖𝟖𝒐𝒐 𝑨𝑨

𝐼3 =𝑍𝑒𝑞𝑍𝐶2+𝐿

(𝐼𝑠) =19.61∡−18.43𝑜

15∡90𝑜(0.478∡42.23𝑜) = 𝟎𝟎.𝟏𝟏𝟏𝟏𝟓𝟓∡−𝟏𝟏𝟏𝟏.𝟏𝟏𝒐𝒐 𝑨𝑨

𝑉𝐿 = (𝐼3)(𝑍𝐿) = (0.625∡−66.2𝑜)(20∡90𝑜) = 12.5 ∡23.8𝒐 𝑽

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 “turn off” the sources by

replacing both current and voltage sources as SHORTS.d. ____ When solving the Thevenin Impedance, we must “turn off” the sources by

replacing the current source a SHORT and the voltage sources as an OPEN.e. ____ When solving the Thevenin Impedance, we must “turn off” 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.

𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳

𝟖𝟖∡𝟑𝟑𝟎𝟎𝒐𝒐 𝑨𝑨

𝟏𝟏𝟏𝟏 Ω 𝒋𝒋𝟏𝟏𝟎𝟎 Ω

𝟏𝟏𝟏𝟏 Ω -𝒋𝒋𝟖𝟖 Ω

𝟑𝟑.𝟏𝟏 Ω

𝒋𝒋𝟏𝟏.𝟏𝟏 Ω

T F F

F

T

T

First step is to remove the load and find 𝒁𝒁𝑶𝑶𝟑𝟑. Remember the current source is replaced with an OPEN.

𝒁𝒁𝑶𝑶𝟑𝟑

𝟏𝟏𝟏𝟏 Ω 𝒋𝒋𝟏𝟏𝟎𝟎 Ω

𝟏𝟏𝟏𝟏 Ω -𝒋𝒋𝟖𝟖 Ω

+

-

𝑍𝑇𝐻 =(16∡0𝑜)(8∡−90𝑜)

16 − 𝑗8=

(16∡0𝑜)(8∡−90𝑜)17.89∡−26.57𝑜

𝑍𝑇𝐻 = 7.15∡−63.43𝑜 = 𝟑𝟑.𝟏𝟏 − 𝒋𝒋𝟏𝟏.𝟏𝟏 Ω

6) Find the voltage across the Inductor (VL).

Now “turn on” the source and find 𝑽𝑽𝑶𝑶𝟑𝟑 at the open terminals.

𝑍𝑒𝑞 = (16)||(−𝑗8) = 7.15∡−63.43𝑜 Ω

𝑉𝑇𝐻 = (𝐼𝑠)𝑍𝑒𝑞 = (8∡30𝑜)(7.15∡−63.43𝑜) = 𝟓𝟓𝟑𝟑.𝟏𝟏∡−𝟑𝟑𝟑𝟑.𝟏𝟏𝟑𝟑𝒐𝒐 𝑽𝑽

9) Is Maximum Power being delivered to the Load and why?

𝒁𝒁𝑶𝑶𝟑𝟑 = 𝒁𝒁𝑳𝑳∗

3.2− 𝑗6.4 = 3.2 + 𝑗6.4

7.15∡−63.43𝑜 = 7.15∡63.43𝑜

10) The acronym used for Capacitors in AC circuits is “ICE”. Explain what is meant with thisacronym.

Current LEADS Voltage for a capacitive circuit. If we are just analyzing a Capacitor’s current andvoltage relationship we will see the current LEADS the voltage by 900.

𝑽𝑽𝑶𝑶𝟑𝟑

𝟏𝟏𝟏𝟏 Ω 𝒋𝒋𝟏𝟏𝟎𝟎 Ω

𝟏𝟏𝟏𝟏 Ω -𝒋𝒋𝟖𝟖 Ω

+

-

𝟖𝟖∡𝟑𝟑𝟎𝟎𝒐𝒐 𝑨𝑨

𝑉𝑇𝐻 is the voltage across the 16 Ω and the –j8 Ω. We can use Nodal Analysis or combine the 16 Ω and –j8 Ω impedances in parallel and find the voltage across them.

𝟑𝟑.𝟏𝟏 Ω

𝟓𝟓𝟑𝟑.𝟏𝟏∡−𝟑𝟑𝟑𝟑.𝟏𝟏𝟑𝟑𝒐𝒐 𝑽𝑽 +-

−𝒋𝒋𝟏𝟏.𝟏𝟏 Ω

𝟑𝟑.𝟏𝟏 Ω

𝒋𝒋𝟏𝟏.𝟏𝟏 Ω

𝑪𝑪 +

- 𝑽𝑽𝑪𝑪

𝑰𝑰𝑪𝑪 𝑰𝑰𝑪𝑪 𝑽𝑽𝑪𝑪

𝜽𝜽𝑽𝑽

This implies the Current Angle (𝜽𝜽𝒊𝒊) equals to 𝜽𝜽𝑽𝑽 + 𝟏𝟏𝟎𝟎𝒐𝒐.

𝟎𝟎𝒐𝒐

YES!

11) The acronym used for Inductors in AC circuits is “ELI”. Explain what is meant with this acronym.

Voltage LEADS Current for an inductive circuit. If we are just analyzing an Inductor’s 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

𝑽𝑽𝟐𝟐

𝑽𝑽𝑹𝑹

**Time in milliseconds

𝑽𝑽𝟐𝟐 + - 𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳

𝟏𝟏 Ω

+

-

𝑽𝑽𝑳𝑳 𝑽𝑽𝑹𝑹 +-

𝑳𝑳 +

- 𝑽𝑽𝑳𝑳

𝑰𝑰𝑳𝑳 𝑰𝑰𝑳𝑳

𝑽𝑽𝑳𝑳

𝜽𝜽𝒊𝒊

This implies the Voltage Angle (𝜽𝜽𝑽𝑽) equals to 𝜽𝜽𝒊𝒊 + 𝟏𝟏𝟎𝟎𝒐𝒐.

𝟎𝟎𝒐𝒐

12) For the Graph on the other page answer the following questions.

a. What is the period (T) of the waveforms?

𝑇 = 𝟓 ms

b. What is the frequency of the waveforms (𝒇)?

𝑓 =1𝑇

=1

5𝑥10−3= 𝟏𝟏𝟎𝟎𝟎𝟎 𝟑𝟑𝒛

c. What is the time difference (∆𝒕𝒕) between 𝑽𝑽𝟐𝟐 𝑎𝑛𝑑 𝑽𝑽𝑹𝑹?

The ∆𝒕𝒕 between the peak of 𝑽𝑽𝟐𝟐 and the peak of 𝑽𝑽𝑹𝑹 is about 0.5 ms.∆𝑡 = 𝟎𝟎.𝟓𝟓 𝟑𝟑𝟐𝟐

d. What is the phase difference (∆𝜽𝜽) between 𝑽𝑽𝟐𝟐 𝑎𝑛𝑑 𝑽𝑽𝑹𝑹?

∆𝜃 =∆𝑡𝑇

(360𝑜) =0.5𝑥10−3

5𝑥10−3(360𝑜) = 𝟑𝟑𝟏𝟏𝒐𝒐

e. Write the function for 𝑽𝑽𝟐𝟐(𝒕𝒕).

𝑉𝑠(𝑡) = 𝟏𝟏 𝐬𝐢𝐧[𝟏𝟏𝝅(𝟏𝟏𝟎𝟎𝟎𝟎)𝒕𝒕] 𝑽𝑽

f. Express 𝑽𝑽𝟐𝟐(𝒕𝒕) in phasor form.

𝑉𝑠 =6√2

∡0𝑜 = 𝟏𝟏.𝟏𝟏𝟏𝟏∡𝟎𝟎𝒐𝒐 𝑽𝑽

g. Write the function for 𝑽𝑽𝑹𝑹(𝒕𝒕). (Use 𝑉𝑠 as the reference angle)

𝑉𝑅(𝑡) = 𝟑𝟑 𝐬𝐢𝐧[𝟏𝟏𝝅(𝟏𝟏𝟎𝟎𝟎𝟎)𝒕𝒕 − 𝟑𝟑𝟏𝟏𝒐𝒐] 𝑽𝑽

h. Express 𝑽𝑽𝑹𝑹(𝒕𝒕) in phasor form.

𝑉𝑅 =3√2

∡−36𝑜 = 𝟏𝟏.𝟏𝟏𝟏𝟏∡−𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽

13) Since we know VS a𝑛𝑑 VR, find the voltage across the Load (VL). Refer to the circuit atthe bottom of the other page. (Hint: use KVL)

−𝑉𝑠 + 𝑉𝐿 + 𝑉𝑅 = 0

𝑉𝐿 = 𝑉𝑠 − 𝑉𝑅 = 4.24∡0𝑜 − 2.12∡−36𝑜 = 4.24− (1.72− 𝑗1.25) = 2.52 + 𝑗1.25 𝑉

𝑉𝐿 = 𝟏𝟏.𝟖𝟖𝟏𝟏∡𝟏𝟏𝟏𝟏.𝟑𝟑𝟖𝟖𝒐𝒐 𝑽𝑽

14) Find ZLoad; Is the Load Capacitive or Inductive?

𝐼𝑆 = 𝐼𝑅 =𝑉𝑅𝑅

=2.12∡−36𝑜

2= 1.06∡−36𝑜 𝐴

𝑍𝐿𝑜𝑎𝑑 =𝑉𝐿𝐼𝑠

=2.81∡26.38𝑜

1.06∡−36𝑜= 2.65∡62.38𝑜 = 𝟏𝟏.𝟏𝟏𝟏𝟏 + 𝒋𝒋𝟏𝟏.𝟑𝟑𝟓𝟓 Ω

Since the imaginary component of 𝑍𝐿𝑜𝑎𝑑 is positive then the Load is Inductive.

15) Draw the Load equivalent circuit below.

𝑽𝑽𝟐𝟐 + -

𝟏𝟏 Ω

+

-

𝑽𝑽𝑳𝑳 𝑽𝑽𝑹𝑹 +-

𝟏𝟏.𝟏𝟏𝟏𝟏 Ω

𝒋𝒋𝟏𝟏.𝟑𝟑𝟓𝟓 Ω

𝑽𝑽𝟐𝟐 + -

𝟏𝟏 Ω

+

-

𝑽𝑽𝑳𝑳 𝑽𝑽𝑹𝑹 +-

𝑰𝑰𝟐𝟐

𝑰𝑰𝑹𝑹

16) The following circuit is operating at a frequency of 𝝎 = 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 (𝑰𝑰𝒔𝒔).

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 (𝑪𝑪) isdetermined by choosing a Capacitance Reactive Power (𝑸𝒄𝒄) equal to the InductanceReactive Power (𝑸𝑳𝑳).

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?

𝟏𝟏𝟏𝟏𝟑𝟑 𝑾𝑾

+

-

𝟐𝟐𝟐𝟐𝟑𝟑∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽

𝑰𝑰𝒔𝒔

𝟑𝟑𝟑𝟑𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽

𝑆 = 1602 + 4002 = 430.81 𝑉𝑉𝐴

𝜃 = tan−1 400160

= 68.2𝑜

𝟏𝟏𝟏𝟏𝟑𝟑 𝑾𝑾

𝟑𝟑𝟑𝟑𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽

𝟏𝟏𝟔𝟔.𝟐𝟐𝒐𝒐

𝑆 = 𝑉𝑉𝑠 𝐼𝑠 ∗

𝐼𝑠 ∗

=430.81∡68.2𝑜

220∡40𝑜= 1.96∡28.2𝑜 𝐴

𝐼𝑠 = 𝟏𝟏.𝟐𝟐𝟏𝟏∡−𝟐𝟐𝟔𝟔.𝟐𝟐𝒐𝒐 𝑽𝑽

|𝑄𝐿| = |𝑄𝑐| = 400 𝑉𝑉𝐴𝑅

𝑄𝑐 =𝑉𝑉2

𝑋𝑐 => 𝑋𝑐 =

(220)2

400= 121 Ω

𝑋𝑐 =1𝜔𝐶

=> 𝐶 =1

(500)(121)= 𝟏𝟏𝟏𝟏.𝟐𝟐𝟑𝟑 𝝁𝝁𝝁𝝁

d. Find the source current (𝑰𝑰𝒔𝒔) when a Capacitor is connected in parallel with the Load.Assume the Capacitance Reactive Power is −𝟑𝟑𝟑𝟑𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽.

𝟏𝟏𝟏𝟏𝟑𝟑 𝑾𝑾

+

-

𝟐𝟐𝟐𝟐𝟑𝟑∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽

𝑰𝑰𝒔𝒔

𝟑𝟑𝟑𝟑𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽

−𝟑𝟑𝟑𝟑𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽

𝑰𝑰𝟏𝟏 𝑰𝑰𝟐𝟐

𝐼1 = 1.96∡−28.2𝑜 𝐴

(𝐼2)∗ =𝑆

𝑉𝑉𝑠=400∡−900

220∡40𝑜= 1.82∡−130𝑜 𝐴

𝐼2 = 1.82∡130𝑜 𝐴

𝐼𝑠 = 𝐼1 + 𝐼2

𝐼𝑠 = (1.96∡−28.2𝑜) + (1.82∡130𝑜)

𝐼𝑠 = 𝟑𝟑.𝟑𝟑𝟐𝟐𝟔𝟔∡𝟑𝟑𝟑𝟑.𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽

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.