Review 6. unlike poles cause the magnets to attract. like poles cause the magnets to repel.

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1 Review 6 1. The two characteristics of all magnets are: they attract and hold Iron, and, if free to move, they will assume roughly a south - north position. 2. Lines of flux always leave the north pole and enter the south pole. 3. The left-hand rule for conductors states that, when the left hand is placed on a current-carrying conductor with the thumb pointing in the direction of electron flow, the fingers will point in the Added Questions: unlike poles cause the magnets to attract. like poles cause the magnets to repel.

2 Review 7 1. A sine wave is the graphic representation of AC voltage or current values over time. 2. Each phase of three phase AC power is offset by 120 electrical degrees. 3. An AC generator produces one cycle(s) per revolution for each pair of poles. 4. What is the instantaneous voltage at 240 degrees for a sine wave with a peak voltage of 150 volts? e= E peak X sine Ѳ e= 150 X sine 240 e= 150 X e= What is the effective voltage for a sine wave with a peak voltage of 150 volts? RMS = E peak X.707 RMS = 150 X.707 RMS = Added Questions: 1. An AC generator produces 2 cycle(s) per revolution for 4 poles generator 2. The unit for frequency is hertz(hz) 3. A generator produce 80 cycles in 2 seconds, its frequency is 40 Hz. 4. If peak voltage is 100 volts, how much the peak to peak, 200 volts.

3 Review 8 1. The total inductance for this circuit is 10h. 4 h 2 h 3 h 1 h L 1 L 2 L 3 L 4 LT = L1 + L2 + L3 + L4 LT = 4h + 2h + 3h + 1h LT = 10h 2. The total inductance for this circuit is 2.5h. L 1 L 2 L 3 5h 10h 10h 1/LT = 1/L1+ 1/L2 + 1/L3 1/LT = 1/5 + 1/10 + 1/10 1/LT = 2/10 + 1/10 + 1/10 1/LT = 4/10 LT = 10/4 = 2.5h 3. The total capacitance for this circuit is 2.5F. 5 F 10 F 10 F C 1 C 2 C 3 1/CT = 1/C1+ 1/C2 + 1/C3 1/CT = 1/5 + 1/10 + 1/10 1/CT = 2/10 + 1/10 + 1/10 1/CT = 4/10 CT = 10/4 = 2.5F 4. The total capacitance for this circuit is 25F. C 1 C 2 C 3 CT = C1 + C2 + C3 CT = 5F + 10F + 10F CT = 25F 5 F 10 F 10 F

4 Added Questions: 1. What is the unit of measurement for inductance, henry (h). 2. The force that opposes the change in current is Inductance. 3. List things that affect inductance? number of turns in the coil coil diameter Length of coil core material. 4. The symbol that indicate inductor is 5. Find the total time for current to rise from zero to maximum t = 15/5 = 3 second total time = 3 X 5 = 15 second 6. What is the unit of measurement for capacitance, Farads (F). 7. List things that affect capacitance? Area of the plates Distance between the plates Type of dielectric material used. 8. The symbol that indicate capacitance is

5 9. Find the total time for current to rise from zero to maximum t = RC t = 2 X 10 = 20 seconds total time = 20 X 5 = 100 seconds

6 Review If the primary of a transformer has more turns than the secondary, it is a step down transformer. 2. If the primary of a transformer has fewer turns than the secondary, it is a step up transformer. 3. The secondary voltage of an iron-core transformer with 240 volts on the primary, 40 amps on the primary, and 20 amps on the secondary is 480 volts. Es = Ep X Ip /Is Es = 240 X 40/ 20 Es = 9600/20 = 480v 4. A transformer with a 480 volt, 10 amp primary, and a 240 volt, 20 amp secondary will be rated for 4.8 kva. Added Questions: kva = Volts X Amps / 1000 kva = 480 X 10/1000 = 4.8 kva or kva = 240 X 20/1000 = 4.8 kva Fined E s and I s Since the number of windings in the secondary is more than the number of windings in the primary, so it is a step-up transformer. E p = 120 I p = 10 The voltage E will increase and the current I will decrease. Because the proportion is 1 to 2 Then E s is two times E p = 240 Then I s is 1/2 I p = 5

7 Wire size and voltage drop The cross-sectional area of wire is measured in circular mils (CM). 1 mil = 1/1000 inch CM = squaring of mil Example 1: A piece of wire has a diameter of.008 of an inch..008 = 8/1000 = 8 mils CM = 8 2 = 64 Example 2: A piece of wire has a diameter of.064 of an inch..064 = 64/1000 = 64 mils CM = 64 2 = 4096 Example 3: Wire has 24 strands of wire and each strand is.012 inches in diameter. The circular mil area of stranded wire is determined by finding the area of one of the strands and then multiplying by the number of strands. One strand area =.012 = 12/1000 = 12 mils CM of one strand = 12 2 = 144 CM for entire conductor = 144 x 24 = 3456

8 American wire gauge (AWG) The table above gives the CM of different types of wire gauges: Example 4: From the table, what is the circular mils for the following: a. 14 AWG a. 10 AWG b. 30 AWG c. 20 AWG 14 AWG = AWG = AWG = AWG = 1022

9 Wire Material (K) The table below gives the resistance of different types of wire: Example 5: From the table, what is the K for the following wire types: a. Aluminum b. Copper c. Iron d. Silver Aluminum = 17 Copper = 10.4 Iron = 60 Silver = 9.6 Example 6: Find the resistance of a piece of #18 AWG copper wire 400 feet long? Solution: First, state the formula to be used K L CM Second, the value of K can be found in table K = 10.4 Third, CM of #18 AWG can be found in table CM = 1624 Fourth, substitute values in formula feet ohms

10 Example 7: Find the resistance of a piece of #12 AWG aluminum wire 250 feet long? Solution: First, state the formula to be used K L CM Second, the value of K can be found in table-1 K = 17 Third, CM of #12 AWG can be found in table-2 CM = 6530 Fourth, substitute values in formula feet ohms Wire size determined by two factors 1. Current drawn by the unit when operating ( A ). 2. The distance the current to be run ( L ). Example 8: A unit is to have a current draw of 24 amps and located 30 feet from the circuit-breaker. Using a #10 AWG copper wire, what is the amount of voltage drop the wire will have. K L CM The amount of voltage dropped by the wire can be found using Ohm s LAW E = I x R E = 24 x E = 1.44 volts

11 Example - 9: A unit is to have a current draw of 24 amps and located 100 feet from the circuit-breaker. Using a #14 AWG copper wire, what is the amount of voltage drop the wire will have. K L CM E = I x R E = 24 x E = volts Example - 10: Find the size of copper (constant 25) wire to carry a load of 40 amperes at 240 volts and distance of 500 feet with 2% voltage drop. Use the formula: material constant A L CM = Voltage Drop amps 500 feet CM = 4. 8 volts CM = circular mils More Examples: 1. Which of the following wires has the greatest cross-sectional area? e. 14 AWG f. 10 AWG g. 30 AWG h. 20 AWG 2. Which of the following wires has the smallest cross-sectional area? a. 14 AWG b. 10 AWG c. 30 AWG d. 20 AWG

12 3. A wire that has a diameter of inches. How much its circular mils? 1 mil = 1/1000 inch inches = 85/ /1000 inches = 85 mils Circular mils = 85 X 85 = If a current of 20 amps flows through 90 feet Copper wires that has a diameter of inches. How much voltage the wire will drop? The equations that we use is Resistance ( R ) = K x L CM Voltage Drop ( V ) = I x R ( K ) for copper is = 10.4 ( L ) wire length = 90x2 = 180 feet ( CM ) circular mils = 85 x 85 = 7225 ( I ) current = 20 Amps Then 10.4 x 180 = ohms 7225 V = I x 20 x = 5.18 volts

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