EE301 Final Exam Review

Final Exam Review

1) Determine the total resistance (𝑽𝑽𝑻𝑻) seen by the source and the currents (𝑰𝑰𝒔𝒔), (𝑰𝑰𝟏𝟏), and (𝑰𝑰𝟐𝟐) in theDC circuit below.

2) Find the power delivered by the source (𝑷𝑷𝒔𝒔) and the power absorbed by all of the resistors(𝑷𝑷𝟏𝟏𝟑𝟑), (𝑷𝑷𝟏𝟏𝟐𝟐), (𝑷𝑷𝟏𝟏𝟐𝟐), and (𝑷𝑷𝟐𝟐). Does the power being delivered equal to the total power beingabsorbed?

3) Use nodal analysis to find the voltage at the node (𝑽𝑽). Now use Ohm’s Law to find the voltageacross the 12 Ω resistor. Do these voltages equal each other?

𝑰𝑰𝒔𝒔 +-

10 Ω

𝟑𝟑𝟑𝟑 𝑽𝑽 𝑰𝑰𝟏𝟏 𝑰𝑰𝟐𝟐

15 Ω

5 Ω 12 Ω

𝑉𝑉

𝐼1 =𝑅𝑒𝑞𝑅12

(𝐼𝑠) =7.512

(1.714) = 𝟏𝟏.𝟑𝟑𝟑𝟑𝟏𝟏 𝑽𝑽

𝐼2 =𝑅𝑒𝑞𝑅15+5

(𝐼𝑠) =7.5

15 + 5(1.714) = 𝟑𝟑.𝟏𝟏𝟑𝟑𝟑𝟑 𝑽𝑽

𝑅𝑇 = 10 + 1

12+

115 + 5

−1= 𝟏𝟏𝟑𝟑.𝟐𝟐 Ω

𝐼𝑠 =𝑉𝑉𝑠𝑅𝑇

=30

17.5= 𝟏𝟏.𝟑𝟑𝟏𝟏𝟑𝟑 𝑽𝑽

𝑅𝑒𝑞 = 1

12+

115 + 5

−1= 𝟑𝟑.𝟐𝟐 Ω

𝑃15 = (𝐼2)2(𝑅15) = (0.643)2(15) = 𝟏𝟏.𝟐𝟐𝟑𝟑 𝑾𝑾

𝑃5 = (𝐼2)2(𝑅5) = (0.643)2(5) = 𝟐𝟐.𝟑𝟑𝟑𝟑 𝑾𝑾

𝑃𝑠 = (𝐼𝑠)(𝑉𝑉𝑠) = (1.714)(30) = 𝟐𝟐𝟏𝟏.𝟑𝟑𝟐𝟐 𝑾𝑾

𝑃10 = (𝐼𝑠)2(𝑅10) = (1.714)2(10) = 𝟐𝟐𝟐𝟐.𝟑𝟑𝟔𝟔 𝑾𝑾

𝑃12 = (𝐼1)2(𝑅12) = (1.071)2(12) = 𝟏𝟏𝟑𝟑.𝟑𝟑𝟏𝟏 𝑾𝑾

𝑷𝑷𝒔𝒔 = 𝑷𝑷𝟏𝟏𝟑𝟑 + 𝑷𝑷𝟏𝟏𝟐𝟐 + 𝑷𝑷𝟏𝟏𝟐𝟐 + 𝑷𝑷𝟐𝟐

𝟐𝟐𝟏𝟏.𝟑𝟑𝟐𝟐 𝑾𝑾 = 29.38 + 13.76 + 6.20 + 2.07 = 𝟐𝟐𝟏𝟏.𝟑𝟑𝟏𝟏 𝑾𝑾

Yes! Power delivered equals total power absorbed.

𝑉𝑉 − 3010

+𝑉𝑉

12+

𝑉𝑉15 + 5

= 0

𝑉𝑉 1

10+

112

+1

20=

3010

𝑉𝑉[0.233] = 3

𝑉𝑉 =3

0.233= 12.876 𝑉𝑉

𝑉𝑉12 = (𝐼1)(𝑅12) = (1.071)(12) = 12.85 𝑉𝑉

Yes! This demonstrates there are many ways to solve a circuit.

4) Determine the total impedance (𝒁𝒁𝑻𝑻) seen by the source and the currents (𝑰𝑰𝒔𝒔), (𝑰𝑰𝟏𝟏), and (𝑰𝑰𝟐𝟐) inthe AC circuit below.

5) Find the complex power delivered by the source (𝑺𝑺𝑻𝑻 ) and the Real/Reactive power of theelements (𝑷𝑷𝟔𝟔), (𝑷𝑷𝟏𝟏𝟐𝟐), (𝑸−𝒋𝒋𝟏𝟏𝟑𝟑), (𝑷𝑷𝟏𝟏𝟑𝟑), (𝑷𝑷𝟏𝟏) and (𝑸𝒋𝒋𝟏𝟏𝟔𝟔). Does the source complex power equal tothe total Real and Reactive power of the elements?

𝑰𝑰𝒔𝒔

8 Ω

𝑰𝑰𝟏𝟏 𝑰𝑰𝟐𝟐

14 Ω

6 Ω 12 Ω

+ -

𝟐𝟐𝟐𝟐∡𝟏𝟏𝟐𝟐𝒐𝒐 𝑽𝑽 j18 Ω

-j10 Ω

𝑍𝑇 = 8 + 1

12 − 𝑗10+

114 + 6 + 𝑗18

−1= 8 +

115.62∡−39.81𝑜

+1

26.91∡41.99𝑜−1

𝐼𝑠 =𝑉𝑉𝑠𝑍𝑇

=25∡15𝑜

20.65∡−7.34𝑜= 𝟏𝟏.𝟐𝟐𝟏𝟏∡𝟐𝟐𝟐𝟐.𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽

𝑍𝑒𝑞 = 1

12 − 𝑗10+

114 + 6 + 𝑗18

−1= 12.76∡−11.94𝑜

𝐼1 =𝑍𝑒𝑞𝑍1

(𝐼𝑠) =12.76∡−11.94𝑜

12 − 𝑗10(1.21∡22.34𝑜) = 𝟑𝟑.𝟐𝟐𝟔𝟔𝟔𝟔∡𝟐𝟐𝟑𝟑.𝟐𝟐𝟏𝟏𝒐𝒐 𝑽𝑽

𝐼2 =𝑍𝑒𝑞𝑍2

(𝐼𝑠) =12.76∡−11.94𝑜

14 + 6 + 𝑗18(1.21∡22.34𝑜) = 𝟑𝟑.𝟐𝟐𝟑𝟑𝟑𝟑∡−𝟑𝟑𝟏𝟏.𝟐𝟐𝟐𝟐𝒐𝒐 𝑽𝑽

𝑍𝑇 = 8 + [0.064∡39.81𝑜 + 0.037∡−41.99𝑜]−1 = 8 + 10.0784∡11.94𝑜

= 𝟐𝟐𝟑𝟑.𝟏𝟏𝟐𝟐∡−𝟑𝟑.𝟑𝟑𝟑𝟑𝒐𝒐 Ω

𝑆𝑇 = 𝑉𝑉𝑠 𝐼𝑠 ∗

= (25∡15𝑜)(1.21∡−22.34𝑜)

𝑆𝑇 = 𝟑𝟑𝟑𝟑.𝟐𝟐𝟐𝟐∡−𝟑𝟑.𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽𝑽𝑽

𝑃8 = (𝐼𝑠)2(𝑅8) = (1.21)2(8) = 𝟏𝟏𝟏𝟏.𝟑𝟑𝟏𝟏 𝑾𝑾

𝑃12 = (𝐼1)2(𝑅12) = (0.988)2(12) = 𝟏𝟏𝟏𝟏.𝟑𝟑𝟏𝟏 𝑾𝑾

𝑄−𝑗10 = (𝐼1)2(𝑋𝑐) = (0.988)2(10) = −𝟐𝟐.𝟑𝟑𝟏𝟏 𝑽𝑽𝑽𝑽𝑽𝑽

𝑃14 = (𝐼2)2(𝑅14) = (0.574)2(14) = 4.61 W

𝑃6 = (𝐼2)2(𝑅6) = (0.574)2(6) = 1.98 W

𝑄𝑗18 = (𝐼2)2(𝑋𝐿) = (0.574)2(18) = 5.93 VAR

𝑺𝑺𝑻𝑻 = (𝑷𝑷𝟔𝟔 + 𝑷𝑷𝟏𝟏𝟐𝟐 + 𝑷𝑷𝟏𝟏𝟑𝟑 + 𝑷𝑷𝟏𝟏) + 𝒋𝒋(𝑸−𝒋𝒋𝟏𝟏𝟑𝟑 + 𝑸𝒋𝒋𝟏𝟏𝟔𝟔)

𝑆𝑇 = (11.71 + 11.71 + 4.61 + 1.98) + 𝑗(−9.76 + 5.93)

𝑆𝑇 = (30.01) + 𝑗(−3.83) = 𝟑𝟑𝟑𝟑.𝟐𝟐𝟐𝟐∡−𝟑𝟑.𝟐𝟐𝟑𝟑𝒐𝒐 𝑽𝑽𝑽𝑽

Remember that an asterisk (*) is the symbol for the complex conjugate.

𝐼2 = (𝐼)(𝐼∗) = (𝐼∡𝜃)(𝐼∡ − 𝜃) = 𝐼2∡𝜃 − 𝜃 = 𝐼2∡0𝑜 = 𝐼2

𝑃14 = (𝐼2)2(𝑅14) = (0.574)2(14) = 𝟑𝟑.𝟏𝟏𝟏𝟏 𝑾𝑾

𝑃6 = (𝐼2)2(𝑅6) = (0.574)2(6) = 𝟏𝟏.𝟐𝟐𝟔𝟔 𝑾𝑾

𝑄𝑗18 = (𝐼2)2(𝑋𝐿) = (0.574)2(10) = 𝟐𝟐.𝟐𝟐𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽

6) What is the difference between Apparent Power and Complex Power? Complex Power has a magnitude and angle while Apparent Power is just the magnitude of the Complex Power.

7) Find the power factor (𝝁𝝁𝒑𝒑) for the given Complex Power. Make sure to include if it is Leading or Lagging.

a. 𝑆 = 322∡30.25𝑜 𝑉𝑉𝐴 𝐹𝑝 = 𝟑𝟑.𝟔𝟔𝟏𝟏𝟑𝟑 𝑳𝑳𝑳𝑳𝒈𝒈𝒊𝒊𝒏𝒏𝒈

b. 𝑆 = 259− 𝑗324 𝑉𝑉𝐴 𝐹𝑝 = 𝟑𝟑.𝟏𝟏𝟐𝟐𝟑𝟑 𝑳𝑳𝒆𝒆𝑳𝑳𝑳𝑳𝒊𝒊𝒏𝒏𝒈

8) An AC circuit operates at a frequency of 𝟏𝟏.𝟑𝟑𝟏𝟏𝟐𝟐 𝒌𝑻𝑻𝒛 and its impedance is 𝒁𝒁 = 𝟐𝟐𝟐𝟐 + 𝒋𝒋𝟏𝟏𝟑𝟑𝟑𝟑 Ω. Is the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value?

**Since the imaginary component is positive then the impedance is Inductive.

9) An AC circuit operates at a frequency of 𝟑𝟑𝟔𝟔𝟑𝟑.𝟏𝟏𝟑𝟑 𝒓𝑳𝑳𝑳𝑳/𝒔𝒔 and its impedance is 𝒁𝒁 = 𝟑𝟑𝟐𝟐 − 𝒋𝒋𝟏𝟏𝟑𝟑𝟐𝟐 Ω.

Is the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value? **Since the imaginary component is negative then the impedance is Capacitive.

𝑋𝐿 = 𝜔𝐿

𝐿 =𝑋𝐿𝜔

=130

2𝜋(6.465𝑥103)= 𝟑𝟑.𝟐𝟐 𝒎𝑻𝑻

𝑋𝑐 =1𝜔𝐶

𝐶 =1

𝜔(𝑋𝑐)=

1(383.14)(145)

= 𝟏𝟏𝟔𝟔 𝝁𝝁𝝁𝝁

−𝟑𝟑𝟐𝟐𝟑𝟑 𝑽𝑽𝑽𝑽𝑽𝑽

𝟐𝟐𝟐𝟐𝟐𝟐 𝑾𝑾

𝑺𝑺

𝜽𝜽 tan𝜃 =

𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡

=𝑄𝑃

𝜃 = tan−1 −324259

= −51.36𝑜

𝑆 = 𝑆∡−51.36𝑜

𝐹𝑝 = cos(−51.36) = 0.624

**Since the Complex Power Angle is Negative the Load is Capacitive (𝑰𝑰𝑪𝑪𝑬𝑬) so current Leads voltage.

𝐹𝑝 = cos𝜃 = cos(30.25) = 0.864

**Since the Complex Power Angle is Positive the Load is Inductive and using the acronym 𝑬𝑬𝑳𝑳𝑰𝑰 current Lags voltage, therefore it is a Lagging Power Factor.

10) The following circuit is operating at a frequency of 𝝎 = 𝟐𝟐𝟑𝟑𝟑𝟑 𝒓𝑳𝑳𝑳𝑳/𝒔𝒔.

a. Draw the Power Triangle for the Load and label the Apparent, Real, and Reactive Power. 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 connecting a Capacitor in parallel to an Inductive Load. The Capacitor component value (𝑪𝑪) is determined by choosing a Capacitance Reactive Power (𝑸𝒄𝒄) equal to the Inductance Reactive Power (𝑸𝑳𝑳). Find the Capacitor component value so all of the Reactive Power at the Load is cancelled. In other words, what component value will correct the power factor to unity?

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.

11) Find the Primary (𝑰𝑰𝒑𝒑) and Secondary (𝑰𝑰𝒔𝒔) currents for the circuit shown below.

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

𝟏𝟏𝟑𝟑∡𝟑𝟑𝒐𝒐 𝑽𝑽

𝑰𝑰𝒑𝒑

+ -

𝟏𝟏𝟑𝟑 Ω 𝒋𝒋𝟏𝟏𝟐𝟐 Ω

−𝒋𝒋𝟔𝟔 Ω

𝟏𝟏Ω

𝑰𝑰𝒔𝒔

4 : 6

𝟏𝟏𝟑𝟑∡𝟑𝟑𝒐𝒐 𝑽𝑽

𝑰𝑰𝒑𝒑

+

-

𝟏𝟏𝟑𝟑 Ω 𝒋𝒋𝟏𝟏𝟐𝟐 Ω

𝒁𝒁−𝒋𝒋𝟔𝟔

𝒁𝒁𝟏𝟏

**Reflecting everything to the Primary side. 𝑎 =𝑁𝑝𝑁𝑠

=46

=23

𝑍−𝑗8 = 𝑎2(𝑍𝑠) = 232

(−𝑗8) = −𝑗3.56 Ω

𝑍6 = 232

(6) = 2.67 Ω

𝐼𝑝 =𝑉𝑉𝑠𝑍𝑇

=60∡0𝑜

10 + 𝑗15 − 𝑗3.56 + 2.67= 𝟑𝟑.𝟐𝟐𝟏𝟏∡−𝟑𝟑𝟐𝟐.𝟑𝟑𝟔𝟔𝒐𝒐 𝑽𝑽

𝐼𝑠 = 𝑎𝐼𝑝 =23

(3.51∡−42.08𝑜) = 𝟐𝟐.𝟑𝟑𝟑𝟑∡−𝟑𝟑𝟐𝟐.𝟑𝟑𝟔𝟔𝒐𝒐 𝑽𝑽

12) Refer to the circuit in problem 11. What component values do we need if we treat the secondary side as the Load and can replace the Load impedances (𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳) with components that will achieve Maximum Power Transfer?

𝑍𝐿𝑜𝑎𝑑 =(𝑍𝑇𝐻)∗

𝑎2=

10 − 𝑗15

23

2 = 𝟐𝟐𝟐𝟐.𝟐𝟐 − 𝒋𝒋𝟑𝟑𝟑𝟑.𝟑𝟑𝟐𝟐 Ω

13) Find the Thevenin Equivalent (𝑽𝑽𝑻𝑻𝑻𝑻) and (𝒁𝒁𝑻𝑻𝑻𝑻) for the circuit below.

a. Determine the Load (𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳) that will achieve Maximum Power Transfer and draw the Thevenin Circuit with the Load.

𝟐𝟐∡𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽

100 Ω 220 Ω

𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳

j30 Ω

-j50 Ω

-j20 Ω

100 Ω 220 Ω j30 Ω

-j50 Ω

-j20 Ω 𝒁𝒁𝑻𝑻𝑻𝑻

First remove the Load. Next turn off the source (Current Source = Open) and find the Thevenin Impedance from the Load perspective. Note the 220Ω and the –j20Ω are not included because they do not provide a complete path.

𝑍𝑇𝐻 = 100 + 𝑗30 − 𝑗50 = 𝟏𝟏𝟑𝟑𝟑𝟑 − 𝒋𝒋𝟐𝟐𝟑𝟑 Ω

Now turn on the source and find the Thevenin Voltage at the open terminals where the Load was. Note the j30Ω and 100Ω are on an open branch so no current travels through them and no voltage drop occurs.

100 Ω 220 Ω j30 Ω

-j50 Ω

-j20 Ω

𝑽𝑽𝑻𝑻𝑻𝑻 +

- 𝟐𝟐∡𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽

𝑉𝑉𝑇𝐻 = (𝐼𝑠)(𝑍−𝑗50) = (5∡46𝑜)(50∡−90𝑜) = 𝟐𝟐𝟐𝟐𝟑𝟑∡−𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽

𝑍𝐿𝑜𝑎𝑑 = (𝑍𝑇𝐻)∗ = 100 + 𝑗20 Ω

𝟐𝟐𝟐𝟐𝟑𝟑∡−𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽 + -

𝟏𝟏𝟑𝟑𝟑𝟑 Ω −𝒋𝒋𝟐𝟐𝟑𝟑 Ω

𝟏𝟏𝟑𝟑𝟑𝟑 Ω

𝒋𝒋𝟐𝟐𝟑𝟑 Ω

14) Find the time constants for the circuit below if:

a. The switch is in the (𝒏𝒏) position for a long time and then moves to position (𝟏𝟏) until steady-state is reached. Is the Capacitor Charging or Discharging?

b. The switch is in the (1) position for a long time and then moves to position (2) until steady-state is reached. Is the Capacitor Charging or Discharging?

15) In the Linear Motor below (𝑽𝑽𝒔𝒔 = 5.5 𝑘𝑉𝑉), (𝜷 = 8𝑇), (𝑽𝑽𝑽𝑽 = 12 Ω), and (𝑳𝑳 = 3𝑚). At (𝑡 = 0) the switch is closed. Assume the sliding bar is initially at rest and there is no mechanical friction.

a. Find the initial current and Force just after the switch is closed.

120 V 30 Ω

1 2 𝑪𝑪 = 𝟑𝟑𝟑𝟑𝟑𝟑.𝟑𝟑𝟑𝟑𝝁𝝁𝝁𝝁

+ -

50 Ω

+

- 𝑽𝑽𝒄𝒄

𝒏𝒏

1 kΩ 20 Ω

12 Ω

𝑅𝑒𝑞 =(50)(30)50 + 30

= 18.75 Ω

𝜏𝑛→1 = (𝑅𝑒𝑞)(𝐶) = (18.75)(373.33𝑥10−6) = 𝟑𝟑 𝒎𝒔𝒔

𝑅𝑒𝑞 = 12 + 20 = 32 Ω

𝜏1→2 = (𝑅𝑒𝑞)(𝐶) = (32)(373.33𝑥10−6) = 𝟏𝟏𝟏𝟏.𝟐𝟐𝟐𝟐 𝒎𝒔𝒔

Charging

Discharging

+ -

𝑽𝑽𝒔𝒔

𝑽𝑽𝑽𝑽

𝑬𝑬𝒊𝒊𝒏𝒏𝑳𝑳

+

-

𝑳𝑳

𝑰𝑰𝒔𝒔

𝐾𝑉𝑉𝐿: − 𝑉𝑉𝑠 + (𝐼𝑠)(𝑅𝐴) + 𝐸𝑖𝑛𝑑 = 0

𝐼𝑠 =𝑉𝑉𝑠𝑅𝐴

=5.5𝑥103

12= 𝟑𝟑𝟐𝟐𝟔𝟔.𝟑𝟑𝟑𝟑 𝑽𝑽

**Initial 𝐸𝑖𝑛𝑑 = 0 because there is no motion. 𝐹 = 𝐼𝐿𝛽 = (458.33)(3)(8)

𝐹 = 𝟏𝟏𝟏𝟏 𝒌𝑰𝑰

30 Ω

1 50 Ω

+

- 𝑽𝑽𝒆𝒆𝒆𝒆

2

+

-

20 Ω

12 Ω

𝑽𝑽𝒆𝒆𝒆𝒆

b. Find the velocity of the bar and Force when (𝑰𝑰𝒔𝒔 = 50 𝐴).

c. When the bar begins to move why does the total current (𝑰𝑰𝒔𝒔) in the circuit decrease?

16) Match the following definitions.

𝐾𝑉𝑉𝐿: − 𝑉𝑉𝑠 + (𝐼𝑠)(𝑅𝐴) + 𝑢𝛽𝐿 = 0

𝑢 =𝑉𝑉𝑠 − (𝐼𝑠)(𝑅𝐴)

𝛽𝐿=

(5.5𝑥103)− (50)(12)(8)(3)

= 𝟐𝟐𝟑𝟑𝟑𝟑.𝟏𝟏𝟑𝟑 𝒎/𝒔𝒔

𝐹 = 𝐼𝐿𝛽 = (50)(3)(8) = 𝟏𝟏.𝟐𝟐 𝒌𝑰𝑰

** 𝐸𝑖𝑛𝑑 = 𝑢𝛽𝐿

The current decreases because the induced voltage (𝑬𝑬𝒊𝒊𝒏𝒏𝑳𝑳) across the sliding bar will increase creating a current opposing the source current (𝑰𝑰𝒔𝒔).

A. Converts Electrical Energy into Mechanical Energy.

B. Converts Mechanical Energy into Electrical Energy.

C. Magnetic field created by a current carrying wire interacts with an existing magnetic field to exert a developed force on the wire.

D. Movement of a conductor in a magnetic field that will induce a voltage.

E. A segmented device commonly found in DC motors that are used with brushes to reverse the direction of the applied current on the rotor.

F. A device commonly found in AC generators that are used with brushes to pass a DC current to create an electromagnet on the rotor.

G. A heavy gauge conductor that connects multiple circuits or loads to a common voltage supply.

H. The main reason an ungrounded system is used on Navy Ships.

I. A critical downside of using an ungrounded system on Navy Ships.

J. A device designed to trip when overcurrent or high currents are reached.

___ Personnel Safety

___ Faraday’s Law

___ Bus

___ Commutator

___ Equipment Reliability

___ Motor

___ Circuit Breakers

___ Slip Ring

___ Lorentz Force Law

___ Generator

H

I

A

J

F

E

G

C

D

B

17) A DC Motor was tested under two operating speeds. Find the armature resistance (𝑽𝑽𝑳𝑳), motor constant (𝑲𝒅𝒅), and the Mechanical Torque Loss (𝑻𝑻𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔). Test 1: Applied (𝟔𝟔𝟑𝟑 𝑽𝑽𝑫𝑫𝑪𝑪) and measured (𝑰𝑰𝑳𝑳 = 𝟑𝟑.𝟐𝟐 𝑽𝑽) at (𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑 𝒓𝒑𝒑𝒎) with no load Test 2: Applied (𝟔𝟔𝟑𝟑 𝑽𝑽𝑫𝑫𝑪𝑪) and measured (𝑰𝑰𝑳𝑳 = 𝟏𝟏𝟔𝟔 𝑽𝑽) at (𝟏𝟏𝟑𝟑𝟐𝟐𝟑𝟑 𝒓𝒑𝒑𝒎) with a load

+ -

𝑽𝑽𝑫𝑫𝑪𝑪

𝑽𝑽𝑳𝑳

𝑰𝑰𝑳𝑳

+ -

𝑬𝑬𝑳𝑳

𝑷𝑷𝑰𝑰𝑰𝑰 𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻

𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 𝑷𝑷𝑴𝑴𝒆𝒆𝒄𝒄𝑴𝑴 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔

𝑷𝑷𝑳𝑳𝒆𝒆𝒅𝒅

𝐸𝑎 = (𝐾𝑣)𝜔

−𝑉𝑉𝐷𝐶 + (𝐼𝑎)(𝑅𝑎) + (𝐾𝑣)𝜔 = 0

Applying KVL to the circuit to the right we get:

−𝑉𝑉𝐷𝐶 + (𝐼𝑎)(𝑅𝑎) + 𝐸𝑎 = 0

788.8− (68)𝑅𝑎 + (𝐾𝑣)(−1813.42) = 0

701.8 + (0)𝑅𝑎 + (𝐾𝑣)(−1633.3) = 0

𝐾𝑣 =701.8

1633.3= 𝟑𝟑.𝟑𝟑𝟐𝟐𝟐𝟐𝟑𝟑 𝑽𝑽 − 𝒔𝒔

Test 1: −87 + (7.5)𝑅𝑎 + (𝐾𝑣)(200.01) = 0 Test 2: −87 + (68)𝑅𝑎 + (𝐾𝑣)(180.12) = 0 Multiply the Test 1 Equation by − 68

7.5 and add

the two equation together to solve for 𝑲𝒅𝒅.

−87 + (68)𝑅𝑎 + (𝐾𝑣)(180.12) = 0

𝜔1 = 2𝜋 𝑅𝑃𝑀

60 = 2𝜋

191060

= 200.01 𝑟𝑎𝑑/𝑠

𝜔2 = 2𝜋 𝑅𝑃𝑀

60 = 2𝜋

172060

= 180.12 𝑟𝑎𝑑/𝑠

+

−87 + (7.5)𝑅𝑎 + (0.4297)(200.01) = 0

𝑅𝑎 =87 − (0.4297)(200.01)

7.5= 𝟑𝟑.𝟏𝟏𝟑𝟑𝟏𝟏 Ω

Now substitute 𝐾𝑣 into Test 1 Equation to solve for 𝑽𝑽𝑳𝑳.

𝑇𝑑𝑒𝑣 = (𝐾𝑣)(𝐼𝑎) = 𝑇𝐿𝑜𝑠𝑠 + 𝑇𝐿𝑜𝑎𝑑

𝑇𝐿𝑜𝑠𝑠 = (0.4297)(7.5) = 𝟑𝟑.𝟐𝟐𝟐𝟐𝟑𝟑 𝑰𝑰−𝒎

Since Test 1 is unloaded than we can set 𝑇𝐿𝑜𝑎𝑑 = 0.

a. Find (𝑷𝑷𝑰𝑰𝑰𝑰), (𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔), (𝑷𝑷𝑳𝑳𝒆𝒆𝒅𝒅), (𝑷𝑷𝑴𝑴𝒆𝒆𝒄𝒄𝑴𝑴 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔), (𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻), and the efficiency (ƞ) of the Motor using the results from Test 2. Assume 𝑇𝐿𝑜𝑠𝑠 is constant.

18) In the balanced 3-phase circuit find the Phase Voltage (𝑬𝑬𝑽𝑽𝑰𝑰), the Line Current (𝑰𝑰𝒃𝒃), and the Total Complex Power at the Load (𝑺𝑺𝑻𝑻). Assume positive sequence and 𝑬𝑬𝑽𝑽𝑩𝑩 = 𝟑𝟑𝟑𝟑𝟏𝟏.𝟑𝟑𝟏𝟏∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽. Typically the objective in 3-phase problems is to redraw the circuit as a Single Phase with reference to the neutral line. Phase 𝑽𝑽 is commonly chosen as the phase to work with and this means we need to know the Line to Neutral Voltage (𝑬𝑬𝑽𝑽𝑰𝑰) and make sure our Load is in the Y-formation.

𝑃𝐼𝑁 = (𝐼𝑎)(𝑉𝑉𝐷𝐶) = (68)(87) = 𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏 𝑾𝑾

𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = (𝐼𝑎)2(𝑅𝑎) = (68)2(0.141) = 𝟏𝟏𝟐𝟐𝟏𝟏.𝟐𝟐𝟐𝟐 𝑾𝑾

𝑃𝑑𝑒𝑣 = (𝐼𝑎)(𝐾𝑣)(𝜔) = (68)(0.4297)(180.12) = 𝟐𝟐𝟐𝟐𝟏𝟏𝟑𝟑.𝟑𝟑𝟑𝟑 𝑾𝑾

𝐴𝑙𝑠𝑜 𝑃𝑑𝑒𝑣 = 𝑃𝐼𝑁 − 𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = 5916 − 651.99 = 𝟐𝟐𝟐𝟐𝟏𝟏𝟑𝟑.𝟑𝟑𝟏𝟏 𝑾𝑾

𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 = (𝑇𝐿𝑜𝑠𝑠)(𝜔) = (3.223)(180.12) = 𝟐𝟐𝟔𝟔𝟑𝟑.𝟐𝟐𝟑𝟑 𝑾𝑾

𝑃𝑂𝑈𝑇 = 𝑃𝐼𝑁 − 𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 − 𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 = 5916 − 651.99 − 580.53 = 𝟑𝟑𝟏𝟏𝟔𝟔𝟑𝟑.𝟏𝟏𝟔𝟔 𝑾𝑾

ƞ =𝑃𝑂𝑈𝑇𝑃𝐼𝑁

(100) =4683.68

5916(100) = 𝟑𝟑𝟐𝟐.𝟏𝟏𝟑𝟑 %

𝐸𝐴𝐵 = 𝐸𝐴𝑁(√3 ∡30𝑜)

𝐸𝐴𝑁 =𝐸𝐴𝐵

√3 ∡30𝑜=

346.41∡30𝑜

√3 ∡30𝑜= 𝟐𝟐𝟑𝟑𝟑𝟑∡𝟑𝟑𝒐𝒐 𝑽𝑽

𝟏𝟏𝟑𝟑 Ω

𝒋𝒋𝟏𝟏𝟐𝟐 Ω

𝒋𝒋𝟏𝟏𝟐𝟐 Ω 𝒋𝒋𝟏𝟏𝟐𝟐 Ω

𝟏𝟏𝟑𝟑 Ω 𝟏𝟏𝟑𝟑 Ω

𝟏𝟏𝟑𝟑 Ω −𝒋𝒋𝟔𝟔 Ω

𝟏𝟏𝟑𝟑 Ω −𝒋𝒋𝟔𝟔 Ω

𝟏𝟏𝟑𝟑 Ω −𝒋𝒋𝟔𝟔 Ω

𝑽𝑽

𝑩𝑩 𝑪𝑪

𝑳𝑳

𝒄𝒄 𝒃𝒃

𝑰𝑰𝑳𝑳

𝑰𝑰𝒃𝒃 𝑰𝑰𝒄𝒄

𝑬𝑬𝑽𝑽 +

-

Single Phase Circuit

19) In the balanced 3-phase circuit find the Phase Impedance (𝒁𝒁∆) and the Phase Current (𝑰𝑰𝑳𝑳𝒃𝒃). Alsofind the Total Real (𝑷𝑷𝑻𝑻), Total Reactive (𝑸𝑻𝑻), and Total Apparent 𝑺𝑺𝑻𝑻 Power delivered by the

Generator. The per-phase Complex Power at the Load is (𝑺𝑺∅ = 𝟑𝟑𝟔𝟔𝟑𝟑 − 𝒋𝒋𝟑𝟑𝟏𝟏𝟑𝟑 𝑽𝑽𝑽𝑽), and (𝑰𝑰𝒄𝒄 = 𝟏𝟏𝟐𝟐∡𝟏𝟏𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽). Assume positive sequence.

𝟏𝟏𝟑𝟑 Ω

𝒋𝒋𝟏𝟏𝟐𝟐 Ω

𝟏𝟏𝟑𝟑 Ω−𝒋𝒋𝟔𝟔 Ω 𝑽𝑽 𝑳𝑳 𝑰𝑰𝑳𝑳

𝑬𝑬𝑽𝑽𝑰𝑰 +

-

𝑰𝑰 𝒏𝒏

𝐼𝑎 =𝐸𝐴𝑁𝑍𝑇

=200∡0𝑜

−𝑗8 + 14 + 10 + 𝑗15= 8∡−16.26𝑜 𝐴

𝐼𝑏 = 𝐼𝑎(∡−120𝑜) = 8∡(−16.26𝑜 − 120𝑜) = 𝟔𝟔∡ − 𝟏𝟏𝟑𝟑𝟏𝟏.𝟐𝟐𝟏𝟏𝒐𝒐 𝑽𝑽

𝑆∅ = (𝐼)2(𝑍) = (8)2(10 + 𝑗15) = 640 + 𝑗960 𝑉𝑉𝐴

𝑆𝑇 = (3)𝑆∅ = 1920 + 𝑗2880 = 𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏.𝟑𝟑𝟑𝟑∡𝟐𝟐𝟏𝟏.𝟑𝟑𝟏𝟏𝒐𝒐 𝑽𝑽𝑽𝑽

𝒋𝒋𝟐𝟐𝟐𝟐 Ω 𝟐𝟐 Ω 𝑽𝑽

𝑩𝑩 𝑪𝑪

𝑳𝑳

𝒃𝒃 𝒄𝒄

𝑰𝑰𝑳𝑳

𝑰𝑰𝒃𝒃 𝑰𝑰𝒄𝒄

𝑬𝑬𝑽𝑽𝑰𝑰 +

-

𝒁𝒁∆

𝒋𝒋𝟐𝟐𝟐𝟐 Ω 𝟐𝟐 Ω

𝒋𝒋𝟐𝟐𝟐𝟐 Ω 𝟐𝟐 Ω

Transform the (∆ − 𝐿𝑜𝑎𝑑) to a (𝑌 − 𝐿𝑜𝑎𝑑) so you can construct a Single Phase Circuit.

Single Phase Circuit

𝑳𝑳

𝒃𝒃 𝒄𝒄 𝒁𝒁∆

𝑳𝑳

𝒃𝒃 𝒄𝒄

𝒁𝒁𝒀𝒀 𝑍𝑌 =𝑍∆3

𝒋𝒋𝟐𝟐𝟐𝟐 Ω 𝟐𝟐 Ω 𝑽𝑽 𝑳𝑳 𝑰𝑰𝑳𝑳

𝑬𝑬𝑽𝑽𝑰𝑰 +

-

𝒁𝒁𝒀𝒀

𝑰𝑰 𝒏𝒏

𝐼𝑐 = 𝐼𝑎(∡+120𝑜)

𝐼𝑎 =𝐼𝑐

∡1200=12∡130𝑜

∡120𝑜= 12∡10𝑜 𝐴

𝐼𝑎 = 𝐼𝑎𝑏(√3 ∡−30𝑜)

𝐼𝑎𝑏 =𝐼𝑎

√3 ∡−30𝑜=

12∡10𝑜

√3 ∡−30𝑜= 𝟏𝟏.𝟐𝟐𝟑𝟑∡𝟑𝟑𝟑𝟑𝒐𝒐 𝑽𝑽

𝑆∅ = (𝐼𝑎)2(𝑍𝑌)

𝑍𝑌 =𝑆∅

(𝐼𝑎)2=

480 − 𝑗310(12)2

= 3.33 − 𝑗2.15 Ω

𝑍∆ = (3)𝑍𝑌 = (3)(3.33 − 𝑗2.15) = 𝟏𝟏𝟑𝟑 − 𝒋𝒋𝟏𝟏.𝟑𝟑𝟐𝟐 Ω

Single Phase Complex Power at the Load is

𝑆∅ = 480− 𝑗310 𝑉𝑉𝐴

Complex Power delivered by the Generator.

In previous 3-phase problems we have been only interested in the Line or Phase variables that the Generator (source) has produced to solve for the 3-phase system characteristics. In AC Generator problems we will step back to a more detail look of the internal parameters of the Generator. Remember a Generator converts Mechanical Energy from a prime mover (an axle) into Electrical Energy. An axle will rotate with an established constant magnetic field on the rotor that will induce a voltage (𝑬𝑬𝒊𝒊𝒏𝒏𝑳𝑳) on the stator windings. Before the voltage potential reaches the outer terminals of the Generator (Line Voltages) the induced voltage will suffer internal losses due to resistance of the stator windings (𝑽𝑽𝑺𝑺) and mutual inductance of the stator winding (𝑿𝑿𝑺𝑺). The circuit diagram on the left is the Single Phase Circuit of the Generator.

20) A 3-phase, Y-connected, 8-pole, 2.88 KV, 50 Hz synchronous generator is rated to deliver 4.5 MVA at a power factor of 0.8829 Lagging. The per-phase stator resistance is 0.09Ω and the synchronous reactance is negligible. The AC Generator is operating at the rated Load with Mechanical Losses at 410 kW. To simplify the problem first list all the given variables.

𝑽𝑽

𝑩𝑩 𝑪𝑪

𝑰𝑰𝑳𝑳

𝑰𝑰𝒃𝒃 𝑰𝑰𝒄𝒄

𝑬𝑬𝑽𝑽𝑰𝑰 +

-

𝑰𝑰 Load

𝑽𝑽 𝑰𝑰𝑳𝑳

𝑬𝑬𝒊𝒊𝒏𝒏𝑳𝑳

𝑰𝑰

𝑬𝑬𝑽𝑽𝑰𝑰

+

-

+ -

𝑽𝑽𝑺𝑺 𝑿𝑿𝑺𝑺

𝑷𝑷𝑰𝑰𝑰𝑰 𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻

𝑷𝑷𝑴𝑴𝒆𝒆𝒄𝒄𝑴𝑴 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔

3 − 𝑝ℎ𝑎𝑠𝑒 𝑠𝑦𝑡𝑒𝑚

𝑌 − 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑

8 − 𝑝𝑜𝑙𝑒

𝑓 = 50 𝐻𝑧

𝑅𝑎𝑡𝑒𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 𝐿𝑖𝑛𝑒 𝑡𝑜 𝐿𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 2.88 𝑘𝑉𝑉

𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 = 410 𝑘𝑊

𝑅𝑎𝑡𝑒𝑑 𝑝𝑜𝑤𝑒𝑟 = 3 − 𝑝ℎ𝑎𝑠𝑒 𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟

𝑆𝑇 = 4.5 𝑀𝑉𝑉𝐴

𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝐶𝑜𝑚𝑝𝑙𝑒𝑥 𝑃𝑜𝑤𝑒𝑟

𝑅𝑆 = 0.09 Ω

𝑋𝑆 = 0 Ω

𝜃 = cos−1(0.8829) = 28𝑜 *Lagging = positive angle

a. At what speed does the shaft rotate (𝒓𝒑𝒑𝒎)?

𝑁𝑟𝑝𝑚 =120(𝑓)𝑃𝑜𝑙𝑒𝑠

=120(50)

8= 𝟑𝟑𝟐𝟐𝟑𝟑 𝒓𝒑𝒑𝒎

b. Find the Line Current (𝑰𝑰𝑳𝑳).

Single Phase Circuit

c. Find (𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻), (𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔), and (𝑷𝑷𝑰𝑰𝑰𝑰) of the Generator.

𝑽𝑽 𝑰𝑰𝑳𝑳

𝑬𝑬𝒊𝒊𝒏𝒏𝑳𝑳

𝑰𝑰

𝑬𝑬𝑽𝑽𝑰𝑰

+

-

+ -

𝑽𝑽𝑺𝑺 𝑿𝑿𝑺𝑺

Rated Single Phase Complex Power

𝑆𝑇 = 4.5∡28𝑜 𝑀𝑉𝑉𝐴

𝑆∅ =𝑆𝑇3

=4.5∡28𝑜 𝑀𝑉𝑉𝐴

3= 1.5∡28𝑜 𝑀𝑉𝑉𝐴

𝐸𝐴𝐵 = 2.88∡𝜃𝐴𝐵𝑜 𝑘𝑉𝑉

𝐸𝐴𝑁 =𝐸𝐴𝐵

√3 ∡30𝑜=

2.88∡𝜃𝐴𝐵𝑜 𝑘𝑉𝑉√3 ∡30𝑜

= 1.663∡0𝑜 𝑘𝑉𝑉

𝑆∅ = (𝐸𝐴𝑁)(𝐼𝑎)∗

(𝐼𝑎)∗ =𝑆∅𝐸𝐴𝑁

=1.5∡28𝑜 𝑀𝑉𝑉𝐴1.663∡0𝑜 𝑘𝑉𝑉

= 901.98∡28𝑜 𝐴

𝐼𝑎 = 𝟐𝟐𝟑𝟑𝟏𝟏.𝟐𝟐𝟔𝟔∡−𝟐𝟐𝟔𝟔𝒐𝒐 𝑽𝑽

𝑺𝑺∅

Note the given rated Line Voltage (𝑬𝑬𝑽𝑽𝑩𝑩) does not include an angle so we can choose any angle. It is common for the ease of calculations to set (𝑬𝑬𝑽𝑽𝑰𝑰) as the reference angle (𝜽𝜽𝑽𝑽𝑰𝑰 = 𝟑𝟑𝒐𝒐) or you can think of it as setting (𝜽𝜽𝑽𝑽𝑩𝑩 = 𝟑𝟑𝟑𝟑𝒐𝒐).

𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = 3(𝐼𝑎)2(𝑅𝑆) = 3(901.98)2(0.09) = 𝟐𝟐𝟏𝟏𝟐𝟐.𝟏𝟏𝟏𝟏 𝒌𝑾𝑾

𝑃𝐼𝑁 = 𝑃𝑂𝑈𝑇 + 𝑃𝑀𝑒𝑐ℎ 𝐿𝑜𝑠𝑠 + 𝑃𝐸𝑙𝑒𝑐 𝐿𝑜𝑠𝑠 = 3.973 + 0.41 + 0.21966 = 𝟑𝟑.𝟏𝟏𝟑𝟑𝟑𝟑 𝑴𝑴𝑾𝑾

𝑷𝑷𝑶𝑶𝑶𝑶𝑻𝑻 is the real component of the rated Total Complex Power.

𝑃𝑂𝑈𝑇 = (4.5𝑥106) cos(28𝑜) = 𝟑𝟑.𝟐𝟐𝟑𝟑𝟑𝟑 𝑴𝑴𝑾𝑾

𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 is caused only by the stator resistance. Remember the given 𝑽𝑽𝑺𝑺 and the current we found (𝑰𝑰𝑳𝑳) is for single phase, so the 𝑷𝑷𝑬𝑬𝑬𝑬𝒆𝒆𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 needs to be multiplied by 3.

d. What is the efficiency (ƞ) of the generator?

e. What is the prime mover torque?

ƞ =𝑃𝑂𝑈𝑇𝑃𝐼𝑁

(100) =3.973 𝑀𝑊4.603 𝑀𝑊

(100) = 𝟔𝟔𝟏𝟏.𝟑𝟑𝟏𝟏%

𝜔 = 2𝜋 𝑅𝑃𝑀

60 = 2𝜋

75060

= 78.54 𝑟𝑎𝑑/𝑠

𝑃𝐼𝑁 = (𝑇𝑃𝑀)(𝜔)

𝑇𝑃𝑀 =𝑃𝐼𝑁𝜔

=4.603𝑥106

78.54= 𝟐𝟐𝟔𝟔.𝟏𝟏𝟏𝟏 𝒌𝑰𝑰 −𝒎

The prime mover will induce a synchronous Voltage which delivers the Power (𝑷𝑷𝑰𝑰𝑰𝑰) into the Generator system.