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Transcript
EE301
Final Exam Review
25APR14
Final Exam Review
1) Determine the total resistance (𝑽𝑽𝑢𝑢 ) seen by the source and the currents (𝑰𝑰𝒔𝒔 ), (π‘°π‘°πŸπŸ ), and (π‘°π‘°πŸπŸ ) in the
DC circuit below.
10 Ω
πŸ‘πŸ‘πŸ‘πŸ‘ 𝑽𝑽
+
-
𝑰𝑰𝒔𝒔
π‘°π‘°πŸπŸ
𝑉𝑉
15 Ω
12 Ω
π‘°π‘°πŸπŸ
5Ω
2) Find the power delivered by the source (𝑷𝑷𝒔𝒔 ) and the power absorbed by all of the resistors
(π‘·π‘·πŸπŸπŸ‘πŸ‘ ), (π‘·π‘·πŸπŸπŸπŸ ), (π‘·π‘·πŸπŸπŸπŸ ), and (π‘·π‘·πŸπŸ ). Does the power being delivered equal to the total power being
absorbed?
3) Use nodal analysis to find the voltage at the node (𝑽𝑽). Now use Ohm’s Law to find the voltage
across the 12 Ω resistor. Do these voltages equal each other?
4) Determine the total impedance (𝒁𝒁𝑢𝑢 ) seen by the source and the currents (𝑰𝑰𝒔𝒔 ), (π‘°π‘°πŸπŸ ), and (π‘°π‘°πŸπŸ ) in
the AC circuit below.
8Ω
πŸπŸπŸπŸβˆ‘πŸπŸπŸπŸπ’π’ 𝑽𝑽
+
-
𝑰𝑰𝒔𝒔
14 Ω
π‘°π‘°πŸπŸ
12 Ω
-j10 Ω
π‘°π‘°πŸπŸ
6Ω
j18 Ω
����⃗𝑢𝑢 ) and the Real/Reactive power of the
5) Find the complex power delivered by the source (𝑺𝑺
elements (𝑷𝑷𝒋𝒋 ), (π‘·π‘·πŸπŸπŸπŸ ), (π‘Έβˆ’π’‹π’‹πŸπŸπŸ‘πŸ‘ ), (π‘·π‘·πŸπŸπŸ’πŸ’ ), (π‘·π‘·πŸπŸ ) and (π‘Έπ’‹π’‹πŸπŸπ’‹π’‹ ). Does the source complex power equal to
the total Real and Reactive power of the elements?
6) What is the difference between Apparent Power and Complex Power?
7) Find the power factor (𝝁𝝁𝒑𝒑 ) for the given Complex Power. Make sure to include if it is Leading or
Lagging.
a. 𝑆⃗ = 322∑30.25π‘œ 𝑉𝑉𝐴
b. 𝑆⃗ = 259 βˆ’ 𝑗324 𝑉𝑉𝐴
8) An AC circuit operates at a frequency of 𝟏𝟏. πŸ’πŸ’πŸπŸπŸπŸ π’Œπ‘―π’› and its impedance is 𝒁𝒁 = πŸ—πŸπŸ + π’‹π’‹πŸπŸπŸ‘πŸ‘πŸ‘πŸ‘ Ω. Is
the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value?
9) An AC circuit operates at a frequency of πŸ‘πŸ‘π’‹π’‹πŸ‘πŸ‘. πŸπŸπŸ’πŸ’ 𝒓𝑳𝑳𝑳𝑳/𝒔𝒔 and its impedance is 𝒁𝒁 = πŸ‘πŸ‘πŸπŸ βˆ’ π’‹π’‹πŸπŸπŸ’πŸ’πŸπŸ Ω.
Is the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value?
10) The following circuit is operating at a frequency of 𝝎 = πŸπŸπŸ‘πŸ‘πŸ‘πŸ‘ 𝒓𝑳𝑳𝑳𝑳/𝒔𝒔.
+
𝑰𝑰𝒔𝒔
πŸπŸπŸπŸπŸπŸβˆ‘πŸ’πŸ’πŸ’πŸ’π’π’ 𝑽𝑽
-
𝟏𝟏𝟏𝟏𝟏𝟏 𝑾𝑾
πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’ 𝑽𝑽𝑽𝑽𝑽𝑽
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 (π‘ͺπ‘ͺ) 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?
d. Find the source current (𝑰𝑰𝒔𝒔 ) when a Capacitor is connected in parallel with the Load.
Assume the Capacitance Reactive Power is βˆ’πŸ’πŸ’πŸ‘πŸ‘πŸ‘πŸ‘ 𝑽𝑽𝑽𝑽𝑽𝑽.
+
𝑰𝑰𝒔𝒔
π‘°π‘°πŸπŸ
πŸπŸπŸπŸπŸπŸβˆ‘πŸ’πŸ’πŸ’πŸ’π’π’ 𝑽𝑽
𝟏𝟏𝟏𝟏𝟏𝟏 𝑾𝑾
π‘°π‘°πŸπŸ
βˆ’πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’ 𝑽𝑽𝑽𝑽𝑽𝑽
πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’ 𝑽𝑽𝑽𝑽𝑽𝑽
-
11) Find the Primary (𝑰𝑰𝒑𝒑 ) and Secondary (𝑰𝑰𝒔𝒔 ) currents for the circuit shown below.
𝟏𝟏𝟏𝟏 Ω
πŸ”πŸ”πŸ”πŸ”βˆ‘πŸŽπŸŽπ’π’ 𝑽𝑽
+
-
𝑰𝑰𝒑𝒑
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
4:6
𝑰𝑰𝒔𝒔
βˆ’π’‹π’‹π’‹π’‹ Ω
πŸ”πŸ”β„¦
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?
13) Find the Thevenin Equivalent (𝑽𝑽𝑢𝑢𝑯) and (𝒁𝒁𝑢𝑢𝑯 ) for the circuit below.
220 Ω
j30 Ω
-j50 Ω
πŸ“πŸ“βˆ‘πŸ’πŸ’πŸ’πŸ’π’π’ 𝑨𝑨
100 Ω
𝒁𝒁𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
-j20 Ω
a. Determine the Load (𝒁𝒁𝑳𝑳𝒐𝒐𝑳𝑳𝑳𝑳 ) that will achieve Maximum Power Transfer and draw the
Thevenin Circuit with the Load.
14) Find the time constants for the circuit below if:
50 Ω
120 V
+
-
1
𝒏𝒏
2
12 Ω
20 Ω
+
𝑽𝑽𝒄𝒄
-
30 Ω
π‘ͺπ‘ͺ = πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘. πŸ‘πŸ‘πŸ‘πŸ‘ππππ
1 kΩ
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 π‘˜π‘‰π‘‰), (𝜷 = 8𝑇), (𝑽𝑽𝑽𝑽 = 12 Ω), and (𝑳𝑳 = 3π‘š). At (𝑑 = 0) the
switch is closed. Assume the sliding bar is initially at rest and there is no mechanical friction.
𝑹𝑹𝑨𝑨
𝑽𝑽𝒔𝒔
+
-
𝑰𝑰𝒔𝒔
+
π‘¬π‘¬π’Šπ’Šπ’Šπ’Šπ’Šπ’Š 𝑳𝑳
-
a. Find the initial current and Force just after the switch is closed.
b. Find the velocity of the bar and Force when (𝑰𝑰𝒔𝒔 = 50 𝐴).
c. When the bar begins to move why does the current decrease?
16) Match the following definitions.
___ Personnel Safety
A. Converts Electrical Energy into Mechanical Energy.
___ Faraday’s Law
B.
Converts Mechanical Energy into Electrical Energy.
___ Bus
C.
Magnetic field created by a current carrying wire interacts with
an existing magnetic field to exert a developed force on the wire.
___ Commutator
___ Equipment Reliability
___ Motor
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.
___ Circuit Breakers
___ Slip Ring
___ Lorentz Force Law
___ Generator
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.
17) A DC Motor was tested under two operating speeds. Find the armature resistance (𝑽𝑽𝑳𝑳 ), motor
constant (𝑲𝒅𝒅 ), and the Mechanical Torque Loss (𝑢𝑢𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔) assuming it is independent of speed.
Test 1: Applied (π’‹π’‹πŸ‘πŸ‘ 𝑽𝑽𝑫𝑫π‘ͺπ‘ͺ ) and measured (𝑰𝑰𝑳𝑳 = πŸ‘πŸ‘. 𝟐𝟐 𝑽𝑽) at (πŸπŸπŸ—πŸπŸπŸ‘πŸ‘ π’“π’‘π’‘π’Ž) with no load
Test 2: Applied (π’‹π’‹πŸ‘πŸ‘ 𝑽𝑽𝑫𝑫π‘ͺπ‘ͺ ) and measured (𝑰𝑰𝑳𝑳 = πŸπŸπ’‹π’‹ 𝑽𝑽) at (πŸπŸπŸ‘πŸ‘πŸπŸπŸ‘πŸ‘ π’“π’‘π’‘π’Ž) with a load
𝑹𝑹𝒂𝒂
𝑽𝑽𝑫𝑫𝑫𝑫
+
-
𝑰𝑰𝒂𝒂
𝑬𝑬𝒂𝒂
+
-
𝑷𝑷𝑢𝑢𝑢𝑢𝑢𝑢
𝑷𝑷𝒅𝒅𝒅𝒅𝒅𝒅
𝑷𝑷𝑰𝑰𝑰𝑰
𝑷𝑷𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
𝑷𝑷𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
a. Find (𝑷𝑷𝑰𝑰𝑰𝑰 ), (𝑷𝑷𝑬𝑬𝑬𝑬𝑬𝑬𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔), (𝑷𝑷𝑳𝑳𝑬𝑬𝒅𝒅 ), (𝑷𝑷𝑴𝑴𝑬𝑬𝒄𝒄𝑴𝑴 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔 ), (𝑷𝑷𝑢𝑢𝑢𝑢𝑢𝑢 ), and the efficiency (ƞ) of the Motor
using the results from Test 2. Remember we are assuming π‘‡πΏπ‘œπ‘ π‘  is independent of
speed.
18) In the balanced 3-phase circuit find the Phase Voltage (𝑬𝑬𝑽𝑽𝑰𝑰 ), the Line Current (𝑰𝑰𝒃𝒃 ), and the Total
Complex Power at the Load (𝑺𝑺𝑢𝑢 ). Assume positive sequence and 𝑬𝑬𝑽𝑽𝑩𝑩 = πŸ‘πŸ‘πŸ’πŸ’πŸπŸ. πŸ’πŸ’πŸπŸβˆ‘πŸ‘πŸ‘πŸ‘πŸ‘π’π’ 𝑽𝑽.
𝑨𝑨
+
βˆ’π’‹π’‹π’‹π’‹ Ω 𝟏𝟏𝟏𝟏 Ω
𝑰𝑰𝒂𝒂
𝒂𝒂
𝑬𝑬𝑨𝑨𝑨𝑨
π’‹π’‹πŸπŸπŸπŸ Ω
-
π‘ͺπ‘ͺ
𝑰𝑰𝒄𝒄
𝟏𝟏𝟏𝟏 Ω
𝑩𝑩
𝑰𝑰𝒃𝒃
βˆ’π’‹π’‹π’‹π’‹ Ω 𝟏𝟏𝟏𝟏 Ω
βˆ’π’‹π’‹π’‹π’‹ Ω 𝟏𝟏𝟏𝟏 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝟏𝟏𝟏𝟏 Ω
𝒃𝒃
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝟏𝟏𝟏𝟏 Ω
𝒄𝒄
19) In the balanced 3-phase circuit find the Phase Impedance (π’π’βˆ† ) and the Phase Current (𝑰𝑰𝑳𝑳𝒃𝒃 ). Also
����⃗𝑢𝑢 οΏ½ Power delivered by the
find the Total Real (𝑷𝑷𝑢𝑢 ), Total Reactive (𝑸𝑢𝑢 ), and Total Apparent �𝑺𝑺
οΏ½οΏ½οΏ½οΏ½βƒ—βˆ… = πŸ’πŸ’π’‹π’‹πŸ‘πŸ‘ βˆ’ π’‹π’‹πŸ‘πŸ‘πŸπŸπŸ‘πŸ‘ 𝑽𝑽𝑽𝑽),
Generator. The per-phase Complex Power at the Load is (𝑺𝑺
(𝑬𝑬𝑽𝑽𝑰𝑰 = π’‹π’‹πŸ‘πŸ‘βˆ‘πŸπŸπŸ‘πŸ‘π’π’ 𝑽𝑽), and (𝑰𝑰𝒄𝒄 = πŸπŸπŸπŸβˆ‘πŸπŸπŸ‘πŸ‘πŸ‘πŸ‘π’π’ 𝑽𝑽). Assume positive sequence.
𝑨𝑨
+
πŸ“πŸ“ Ω
𝑰𝑰𝒂𝒂
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒂𝒂
𝑬𝑬𝑨𝑨𝑨𝑨
-
π‘ͺπ‘ͺ
𝑰𝑰𝒄𝒄
𝑩𝑩
𝑰𝑰𝒃𝒃
πŸ“πŸ“ Ω
πŸ“πŸ“ Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒋𝒋𝒋𝒋𝒋𝒋 Ω
𝒄𝒄
π’π’βˆ†
𝒃𝒃
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 (π‘¬π‘¬π’Šπ’Šπ’π’π‘³π‘³ ) 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.
𝑰𝑰𝒂𝒂
+
𝑹𝑹𝑺𝑺
𝑿𝑿𝑺𝑺
+
+
𝑷𝑷𝑰𝑰𝑰𝑰
𝑷𝑷𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
𝑰𝑰𝒂𝒂
𝑬𝑬𝑨𝑨𝑨𝑨
-
-
𝑡𝑡
𝑷𝑷𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳
𝑨𝑨
𝑬𝑬𝑨𝑨𝑨𝑨
π‘¬π‘¬π’Šπ’Šπ’Šπ’Šπ’Šπ’Š
-
𝑨𝑨
𝑡𝑡
𝑷𝑷𝑢𝑢𝑢𝑢𝑢𝑢
π‘ͺπ‘ͺ
Load
𝑰𝑰𝒄𝒄
𝑩𝑩
𝑰𝑰𝒃𝒃
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 0.883 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.
a. At what speed does the shaft rotate (π’“π’‘π’‘π’Ž)?
b. Find the Line Current (𝑰𝑰𝑳𝑳 ).
c. Find (𝑷𝑷𝑢𝑢𝑢𝑢𝑢𝑢 ), (𝑷𝑷𝑬𝑬𝑬𝑬𝑬𝑬𝒄𝒄 𝑳𝑳𝒐𝒐𝒔𝒔𝒔𝒔), and (𝑷𝑷𝑰𝑰𝑰𝑰 ) of the Generator.
d. What is the efficiency (ƞ) of the generator?
e. What is the prime mover torque?