Final Exam Review
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EE301 Final Exam Review 25APR14 Final Exam Review 1) Determine the total resistance (π½π½πΆπΆ ) seen by the source and the currents (π°π°ππ ), (π°π°ππ ), and (π°π°ππ ) in the DC circuit below. 10 β¦ ππππ π½π½ + - π°π°ππ π°π°ππ ππ 15 β¦ 12 β¦ π°π°ππ 5β¦ 2) Find the power delivered by the source (π·π·ππ ) and the power absorbed by all of the resistors (π·π·ππππ ), (π·π·ππππ ), (π·π·ππππ ), and (π·π·ππ ). Does the power being delivered equal to the total power being absorbed? 3) Use nodal analysis to find the voltage at the node (π½π½). Now use Ohm’s Law to find the voltage across the 12 β¦ resistor. Do these voltages equal each other? 4) Determine the total impedance (πππΆπΆ ) seen by the source and the currents (π°π°ππ ), (π°π°ππ ), and (π°π°ππ ) in the AC circuit below. 8β¦ ππππβ‘ππππππ π½π½ + - π°π°ππ 14 β¦ π°π°ππ 12 β¦ -j10 β¦ π°π°ππ 6β¦ j18 β¦ οΏ½οΏ½οΏ½οΏ½βπΆπΆ ) and the Real/Reactive power of the 5) Find the complex power delivered by the source (πΊπΊ elements (π·π·ππ ), (π·π·ππππ ), (πΈ−ππππππ ), (π·π·ππππ ), (π·π·ππ ) and (πΈππππππ ). 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 (ππππ ) for the given Complex Power. Make sure to include if it is Leading or Lagging. a. πβ = 322β‘30.25π πππ΄ b. πβ = 259 − π324 πππ΄ 8) An AC circuit operates at a frequency of ππ. ππππππ ππ―π and its impedance is ππ = πππ + ππππππππ β¦. Is the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value? 9) An AC circuit operates at a frequency of ππππππ. ππππ ππ³π³π³π³/ππ and its impedance is ππ = ππππ − ππππππππ β¦. Is the impedance Inductive or Capacitive? What is the Inductive/Capacitive component value? 10) The following circuit is operating at a frequency of π = ππππππ ππ³π³π³π³/ππ. + π°π°ππ ππππππβ‘ππππππ π½π½ - ππππππ πΎπΎ ππππππ π½π½π½π½π½π½ 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 (π°π°ππ ). 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 (πͺπͺ) is determined by choosing a Capacitance Reactive Power (πΈππ ) equal to the Inductance Reactive Power (πΈπ³π³ ). 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 (π°π°ππ ) when a Capacitor is connected in parallel with the Load. Assume the Capacitance Reactive Power is −ππππππ π½π½π½π½π½π½. + π°π°ππ π°π°ππ ππππππβ‘ππππππ π½π½ ππππππ πΎπΎ π°π°ππ −ππππππ π½π½π½π½π½π½ ππππππ π½π½π½π½π½π½ - 11) Find the Primary (π°π°ππ ) and Secondary (π°π°ππ ) currents for the circuit shown below. ππππ β¦ ππππβ‘ππππ π½π½ + - π°π°ππ ππππππ β¦ 4:6 π°π°ππ −ππππ β¦ ππβ¦ 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 (πππ³π³πππ³π³π³π³ ) with components that will achieve Maximum Power Transfer? 13) Find the Thevenin Equivalent (π½π½πΆπΆπ―) and (πππΆπΆπ― ) for the circuit below. 220 β¦ j30 β¦ -j50 β¦ ππβ‘ππππππ π¨π¨ 100 β¦ πππ³π³π³π³π³π³π³π³ -j20 β¦ a. Determine the Load (πππ³π³πππ³π³π³π³ ) 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 β¦ 120 V + - 1 ππ 2 12 β¦ 20 β¦ + π½π½ππ - 30 β¦ πͺπͺ = ππππππ. ππππππππ 1 kβ¦ a. The switch is in the (ππ) position for a long time and then moves to position (ππ) 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 (π½π½ππ = 5.5 πππ), (π· = 8π), (π½π½π½π½ = 12 β¦), and (π³π³ = 3π). At (π‘ = 0) the switch is closed. Assume the sliding bar is initially at rest and there is no mechanical friction. πΉπΉπ¨π¨ π½π½ππ + - π°π°ππ + π¬π¬ππππππ π³π³ - a. Find the initial current and Force just after the switch is closed. b. Find the velocity of the bar and Force when (π°π°ππ = 50 π΄). 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 (π½π½π³π³ ), motor constant (π²π π ), and the Mechanical Torque Loss (πΆπΆπ³π³ππππππ) assuming it is independent of speed. Test 1: Applied (ππππ π½π½π«π«πͺπͺ ) and measured (π°π°π³π³ = ππ. ππ π½π½) at (πππππππ ππππ) with no load Test 2: Applied (ππππ π½π½π«π«πͺπͺ ) and measured (π°π°π³π³ = ππππ π½π½) at (ππππππππ ππππ) with a load πΉπΉππ π½π½π«π«π«π« + - π°π°ππ π¬π¬ππ + - π·π·πΆπΆπΆπΆπΆπΆ π·π·π π π π π π π·π·π°π°π°π° π·π·π¬π¬π¬π¬π¬π¬π¬π¬ π³π³π³π³π³π³π³π³ π·π·π΄π΄π΄π΄π΄π΄π΄π΄ π³π³π³π³π³π³π³π³ a. Find (π·π·π°π°π°π° ), (π·π·π¬π¬π¬π¬π¬π¬ππ π³π³ππππππ), (π·π·π³π³π¬π¬π π ), (π·π·π΄π΄π¬π¬πππ΄π΄ π³π³ππππππ ), (π·π·πΆπΆπΆπΆπΆπΆ ), and the efficiency (Ζ) of the Motor using the results from Test 2. Remember we are assuming ππΏππ π is independent of speed. 18) In the balanced 3-phase circuit find the Phase Voltage (π¬π¬π½π½π°π° ), the Line Current (π°π°ππ ), and the Total Complex Power at the Load (πΊπΊπΆπΆ ). Assume positive sequence and π¬π¬π½π½π©π© = ππππππ. ππππβ‘ππππππ π½π½. π¨π¨ + −ππππ β¦ ππππ β¦ π°π°ππ ππ π¬π¬π¨π¨π¨π¨ ππππππ β¦ - πͺπͺ π°π°ππ ππππ β¦ π©π© π°π°ππ −ππππ β¦ ππππ β¦ −ππππ β¦ ππππ β¦ ππππππ β¦ ππππ β¦ ππ ππππππ β¦ ππππ β¦ ππ 19) In the balanced 3-phase circuit find the Phase Impedance (ππβ ) and the Phase Current (π°π°π³π³ππ ). Also οΏ½οΏ½οΏ½οΏ½βπΆπΆ οΏ½ Power delivered by the find the Total Real (π·π·πΆπΆ ), Total Reactive (πΈπΆπΆ ), and Total Apparent οΏ½πΊπΊ οΏ½οΏ½οΏ½οΏ½β∅ = ππππππ − ππππππππ π½π½π½π½), Generator. The per-phase Complex Power at the Load is (πΊπΊ (π¬π¬π½π½π°π° = ππππβ‘ππππππ π½π½), and (π°π°ππ = ππππβ‘ππππππππ π½π½). Assume positive sequence. π¨π¨ + ππ β¦ π°π°ππ ππππππ β¦ ππ π¬π¬π¨π¨π¨π¨ - πͺπͺ π°π°ππ π©π© π°π°ππ ππ β¦ ππ β¦ ππππππ β¦ ππππππ β¦ ππ ππβ ππ 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¬πππππ³π³ ) 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 (π½π½πΊπΊ ) and mutual inductance of the stator winding (πΏπΏπΊπΊ ). The circuit diagram on the left is the Single Phase Circuit of the Generator. π°π°ππ + πΉπΉπΊπΊ πΏπΏπΊπΊ + + π·π·π°π°π°π° π·π·π¬π¬π¬π¬π¬π¬π¬π¬ π³π³π³π³π³π³π³π³ π°π°ππ π¬π¬π¨π¨π¨π¨ - - π΅π΅ π·π·π΄π΄π΄π΄π΄π΄π΄π΄ π³π³π³π³π³π³π³π³ π¨π¨ π¬π¬π¨π¨π¨π¨ π¬π¬ππππππ - π¨π¨ π΅π΅ π·π·πΆπΆπΆπΆπΆπΆ πͺπͺ Load π°π°ππ π©π© π°π°ππ 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 0.09β¦ 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 (ππππ)? b. Find the Line Current (π°π°π³π³ ). c. Find (π·π·πΆπΆπΆπΆπΆπΆ ), (π·π·π¬π¬π¬π¬π¬π¬ππ π³π³ππππππ), and (π·π·π°π°π°π° ) of the Generator. d. What is the efficiency (Ζ) of the generator? e. What is the prime mover torque?
