Module 1 Alternate Current h.w 2

An 11kW load draws 62 amps at 240 volts. What is the power factor of this circuit?An 11kW load draws 62 amps at 240 volts. What is the power factor of this circuit? 14% 74% 89% 45%

11000 W / 240 V = 45.833 A, 45.833 A / 62 A =0.7392, 0.7392 * 100 = 73.92 % round up to 74%

What is the ampere load of a 140kVA, 3-phase, load operating at 480-V? 16.8 A 292 A 168 A 0.292 A

140000 kW / (480 V *1.732) = 168 A

What is the MAXIMUM number of 150 watt lamps, rated at 130-volts, which can be connected on a dimmer which is rated at 1,800 watts? 11 12 18 19

1800 W / 150 W = 12

A single family residence has a general lighting load of 6000 VA. What is the minimum number of 15 A branch circuits needed?

6000 VA / 120 V = 50 A, 50 A / 15 A = 3.333 round up to 4

Calculate the voltage to ground for a 277/480V, 3-phase, 4-wire system.

277V

Determine the amperage of a single phase 1.5 HP 120 V motor. The motor has an efficiency rating of 90% and power factor of 80%.

(1.5 HP * 746 W/HP) / (120 V * .8 * .9) = 12.95 A

A range draws 40A when used on a circuit. A bad splice is discovered on the line at the range on one of the phase conductors. The splice has a measured resistance of 0.1Ω. How much power is lost at the splice?

(40 A)2 * 0.1 Ω = 160 W

Change 0.003A to milliamps.

0.003 A * 1000 = 3 mA

A florescent lighting bank in a commercial location operates for 8-hours per day. Each fixture has two ballasts rated at .8 amps each. How many lights can be wired to a 20-amp circuit breaker? 5 8 10 12

0.8 A + 0.8 A = 1.6 A, 1.6 A * 1.25 = 2 A, 20 A / 2 A = 10

Calculate the voltage drop on a three-phase, 480-V, branch circuit of a 1 AWG, THW, aluminum carrying 90 amps a distance of 150-ft. 3.73 Volts 4.34 Volts 5.91 Volts 6.79 Volts

1 AWG = 83690 cmil, KAL = 21.2, (1.732 * 21.2 * 90 A * 150') / 83690 cmil = 5.9 V

A 240V/1,600 watt resistance baseboard heater unit is installed in a construction trailer. The contractor complains the heater unit is not working correctly. Voltage measurements indicate the heater unit supply is 195-volts across the elements. How much heat in watts is the heater unit producing?

1,600 W / 240 V = 6.667 A, 240 V / 6.667 A = 36 Ω, 195 V / 36 Ω = 5.4167 A, 195 V * 5.4167 A = 1056.25 W OR (240 V)2 / 1,600 W = 36 Ω, (195 V)2 / 36 Ω = 1056.25 W

Given: A 120-volt, branch circuit has a load of 10 Amperes. The total circuit conductor length is 500 feet. The wire has a resistance of 1.75Ω per 1000 feet. What is the voltage drop in the conductors of this circuit? 9 Volts 12 Volts 18 Volts 120 Volts

1.75 Ω / 2 = 0.875, 0.875 Ω * 10A = 8.75 V round up to 9 V

Calculate the remote load voltage in the drawing to the below. 25 V 22.5 V 23 V 2 V

10 A * 0.1 Ω = 1 V 1 V * 2 = 2 V 25 V - 2 V = 23 V

Calculate the voltage drop on a 208-volt line, using 10 AWG copper conductor supplying 24 amperes to a resistive load 400-feet from the panel board. (Use KCU = 12.9.)

10 AWG = 10380 cmil, (2* 12.9 * 24 A *400') / 10380 cmil =23.86 V

A 2kW load is operated at 240-V and 80% power-factor. Find the current of this load.

10.4 A 2000 W / (240 V * 0.8) = 10.4 A

What is the turns ratio of a transformer with 240 volts on the primary and 24 volts on the secondary?

10:1

What is the turns ratio of a transformer with 240 volts on the primary and 24 volts on the secondary? 1:10 10:1 1:20 20:1

10:1

A single phase 120V, 16 ampere load is connected to a 12 AWG THHN conductor. The equipment nameplate requires no more than a 3% voltage drop on the branch circuit. What is the maximum wire length from the panel where the voltage drop will not exceed 3%? (K = 12.9)

12 AWG = 6530 cmil,120 V * 0.03 = 3.6 V,(6530 cmil *3.6 V) / (2 * 12.9 * 16) = 56.9' round down to 56' kcmil*V*0.03 (VD) = 3.6 (kcmil*V(3.6)) /(2*K*A)

Luminaires in a conference room are controlled by a dimmer switch rated at 8.34 amps at 120 volts. What is the MAXIMUM wattage total connected to the switch?

120 V * 8.34 A = 1000.8 W

Find the power-factor of the following circuit. 120 supplied V, 9-A, 1,000 W

120 V * 9 A = 1080 W, 1000 W/ 1080 W = 0.9259, 0.9259 * 100 = 92.59 % round up to 93%

How many 1.5 volt batteries are required to power a 13.5 volt radio and how are they connected? 10 in series-parallel 5 in parallel 14 in parallel-series 9 in series

13.5 V / 1.5 V = 9, Voltage is additive in a series circuit

What is the ampere capacity of a bus bar that is ½ inches thick and is 4 inches wide? 20 amperes 200 amperes 2,000 amperes 20,000 amperes

2,000

What is the horsepower of a motor 208V single-phase motor that draws 17.9 amperes? 3 hp 4 hp 5 hp 6 hp

208 V * 17.9 A = 3723.2 W, 3723.2 W / 746W/hp = 5 hp

A residential water pump has a FLC of 15 amps, a source voltage of 240 volts, and the voltage at the pump is 235 volts. What is the power loss in the conductors? 3600 watts 3450 watts 2400 watts 75 watts

240 V - 235 V = 5 V, 5 V * 15A= 75 W

If a single-phase electric kiln rated 5.2 amps at 240 volts is operated on 208 volts, how much power does it consume?

240 V / 5.2 A = 46.1538 Ω, 208 V * 46.1538 Ω = 9600 W, 9600 W / 1000 = 9.6 kW

If a single-phase electric kiln rated 5.2 amps at 240 volts is operated on 208 volts, how much power does it consume? 12.7 kW 11.0 kW 9.6 kW 8.0 kW

240 V / 5.2A = 46.1538Ω ,; 208 V * 46.1538Ω =9600 W, 9600 W / 1000 = 9.6 kW

What is the ampere load of a 242.76 kVA, 3-phase, load operating at 480-V?

242760 W / (480 V *1.732) = 292 A

A non-inductive heating coil operates at 12-volts using 3-watts. If the coil is connected to 24V, how much power will it use? 12 watts 6 watts 3 watts 1.5 watts

3 W / 12 V = 0.25 A. 12 V / 0.25 A = 48 Ω, 24 V / 48 Ω = 0.5 A, 24 V * 0.5 A = 12 W, OR (12 V)2 / 3 W = 48 Ω, (24 V)2 / 48 Ω = 12 W

Change 0.003A to milliamps. 30 mA 0.3 mA 3 mA 0.000003 mA

3 mA (0.003 A * 1000 = 3 mA)

A large room requires installation of 30 luminaires. Each luminaire has a nameplate power rating of 300 watts. The MAXIMUM wattage on any one circuit is 1,800 watts. How many circuits are needed to power the luminaires in this room? This installation is not considered a continuous load.

30 * 300 W = 9000 W, 9,000 W / 1,800 W= 5, OR 1,800 W / 300 W = 6 lights per circuit, 30 / 6 = 5

How many amperes will a 35-KVA, 480 Volt three-phase generator supply to a given three-phase load? 72.9 amperes 42.1 amperes 63.4 amperes 75.8 amperes

35000 W / (480 V *1.732) =42.1 A

A machine operating at 480V, 3-phase is connected with 4 AWG conductor. An ampere meter shows 26 amperes at the machine wire connection enclosure. The raceway is measured to be 300 feet from the panelboard to the machine. What is the voltage drop on the conductors? (K = 12.9)

4 AWG = 41740 cmil, (1.732 * 12.9 * 26 A *300') / 41740 cmil = 4.17 V

A heat trace cable is to be connected on a pipe at a jobsite. The cable is rated at 4.3-watts/foot. How many feet of cable can be installed on a 240V/20 ampere circuit? (The cable will operate for more than 2 hours at any one time.) 2.064 feet 1,032 feet 1,116 feet 112 feet

4.3 W / 240 V = 0.0179 A 20 A / 0.0179 A = 1116'

A 440 V, 3-phase branch circuit is to be installed with three, 4/0 AWG, THW, copper conductors serving a load of 250 amperes. Determine the maximum length of the conductors with the voltage drop not exceeding 2 percent.

4/0 AWG = 211600 cmil, 440 V * 0.02 = 8.8 V, KCU = 12.9, (211600 cmil * 8.8 V) / (1.732 * 12.9 * 250) = 333'

A three-phase 460-V generator delivers 52 amperes. Determine the kilovolt-amperes rating of the generator. 23.92KVA 41KVA 8.8KVA 25KVA

460 V *1.732 *52 A =41429.44W, 41429.44W / 1000 =41.429kW =41.429kVA

What is the true power of a 3 phase 480 V 200 A load with an 84% power factor? 96 kW 80 kW 166 kW 140 kW

480 V * 1.732 * .84 * 200 A = 139668.48 W, 139668.48 W / 1000 = 139.668 kW

What is the kVA of a three-phase load operating at 480 Volts and 168.4 amperes? 81 kVA 140 kVA 280 kVA 162 kVA

480 V *1.732 *168.4 A = 140001.024 W, 140001.024 / 1000 = 140 kW = 140 kVA

What is the monthly cost of operating a 240 volt, 5 kilowatt central electrical heater that operates 12 hours a day when the cost is 15 cents per kilowatt hour? $172 $270 $389 $598

5 kW * 12 hours = 60 kWh // 60kWh * $0.15 / kWh =$ 9.00 per day, $9 * 30 days/ month = $270

What is the month cost of operating a 208-volt, 5 kW electric kiln heater that operates 8-hours a day, 26-days per month at a cost of 12 cents per kilowatt hour?

5 kW * 8 hours = 40 kWh // 40kWh * $0.12/ kWh = $4.80 per day, $4.80*26 days / month = $124.80

How much voltage would be required at the source to provide 120 volts to a 500 watt load which is located 1,500 feet from the power source, and fed with 12 AWG copper wire? (12 AWG = 1.93Ω/1,000 ft) 24.79 V 128.8 V 130 V 144.69 V

500 W / 120 V = 4.166 A, (2 * 12.9 * 4.1667 A * 1500') / 6530 cmil = 24.69 V, 120 V + 24.69 V=144.69 V

A kitchen circuit is rated for 20 amperes. What is the total connected load on this circuit if a toaster rated at 550 watts, a coffee pot rated at 375 watts, and a toaster oven rated at 1,200 watts are operating on the circuit at the same time?

550 W + 375 W + 1200 W = 2125 W, 2125 W / 120 V = 17.7 A

A linear load heating coil operates at 24 Volts using 6 watts. If the coil is connected to 48 Volts, how much power will it use?

6 W / 24 V = 0.25 A, 24 V / 0.25 A = 96 Ω, 48 V / 96 Ω = 0.5 A, 48 V * 0.5 A = 24 W; OR (24 V)2 / 6 W = 96 Ω, (48 V)2 / 96 Ω = 24 W

An old ceiling fan is to be replaced with a new ceiling fan. The old ceiling fan is rated at 120 Volts, 80% efficiency, with a ½ hp motor. The new fan is rated at 120 Volts, 90% efficiency, with a ½ hp motor. What is the power savings of the new motor?

746W/HP, 746W/HP * ½ HP = 373W, 373W * 0.8 = 298.4W, 373W * 0.9 =335.7W, 335.7W-298.4W = 37.3W, 37.3W / 335.7W= 0.1111, 0.11 * 100 = 11%

An old ceiling fan is to be replaced with a new ceiling fan. The old ceiling fan is rated at 120 Volts, 80% efficiency, with a ½ hp motor. The new fan is rated at 120 Volts, 90% efficiency, with a ½ hp motor. What is the power savings of the new motor? 11% 20% 60% 80%

746W/HP/// 746W/HP * ½ HP = 373W, 373W * 0.8 = 298.4W // 373W * 0.9 =335.7W, 335.7W-298.4W = 37.3W, 37.3W / 335.7W= 0.1111, 0.11 * 100 = 11%

Two capacitors rated at 95 μF and 71 μF are connected in parallel. What is the total capacitance of the circuit?

95 µF + 71 µF = 166 µF

Current _____ in parallel. adds subtracts multiplies divides

Adds

Ammeters measure which of the following values?

Amperes

An incomplete current path is called ________. a closed circuit an open circuit an incomplete circuit a short circuit

An Open Circuit

A single Alternating Current sign wave is measured on an oscilloscope at 1/30th of a second. What is the frequency of this system?

Hz (frequency) = Number of alternating current sign waves in 1 second

What is the current of a 200 W circuit with an 8 ohm resistor? 0.20 A 0.32 A 5 A 25 A

I = √(P/R), √(200/8) = 5 A

Two lamps are installed in parallel on a circuit. With both lamps turned on, the total current is measured at 7.5 amperes. One lamp is marked 12V/30W. What is the wattage of the other lamp? 30 W 45 W 60 W 100 W

PT = 12 V * 7.5 A = 90 W, 90 W - 30 W = 60 W,

A watt-hour meter measures the consumption of ____. voltage ampere resistance power

Power

Consider the circuit in the figure below. The known voltages and currents are as indicated. Applying Kirchhoff's voltage and current laws, the value of the current flow through R1

R1 is in series with the source so IT = I1

A 125Ω resistor is connected in parallel with a 75-ohm resistor. The voltage source is providing 37 amps of current flow. What is the total resistance of the circuit?

RT = ((125 Ω)-1 + (75 Ω)-1)-1 = 46.875 Ω

Alternating current means the direction of the current flow is ______. continuing in the same direction changes direction during current flow reverses at intervals equivalent to frequency is intermittingly pulsating-no

Reverses at intervals equivalent to frequency

The nameplate on a 3-phase 150 KVA transformer gives a primary voltage of 480V and a secondary voltage of 208/120 V. This transformer is considered a ____ transformer. auto step-up step-down utility power distribution

Step-Down

The nameplate on a 3-phase 150 KVA transformer gives a primary voltage of 480V and a secondary voltage of 347/575 V. This transformer is considered a ____ transformer. auto step-up step-down utility power distribution

Step-up

With a load of 700 kW at 60% power factor, what kVA of capacitors is required to correct the power factor to 80%? 374 kVA 408 kVA 653 kVA 933 kVA

Ugly's page 27, Power factor correction multiplier = 0.583, 700 kW * 0.583 =408 kVA

Calculate the voltage between points "A" and "B" in the drawing below.

VPEAK = VRMS / 1/√2, 1/√2 = 0.7071, 120 V RMS / 0.7071 = 169.71 V, 169.71 V * 2 = 339.42 V

A 30-μF capacitor is connected into a 240-V, 1,000Hz. circuit. What is the current flow in this circuit? 45.24 amperes 57.45 amperes 67.34 amperes 80.00 amperes

XC = 1 / (2πƒC), XC = 1 / (2π(1000 Hz)(0.00003 F) = 5.305 Ω, 240 V / 5.305 Ω = 45.24 A

A capacitor with a capacitance of 107.6 μF is connected to a 480-V, 1000-Hz line. How much current will flow in the circuit?

XC = 1 / (2πƒC), XC = 1 / (2π(1000 Hz)(0.0001076 F) = 1.479 Ω, 480 V / 1.479 Ω = 324 A

An inductor with an inductance of 3.6H is connected to a 480-V, 60-Hz line. How much current will flow in the circuit? 0.354 A 0.706 A 0.845 A 1.07 A

XL = 2πƒL, XL = 2π(60 Hz * 3.6 H) = 1357.168 Ω, 480 V / 1357.168 Ω = 0.354 A

In a series circuit, the total current is equal to the________.

current in any other part of the circuit.

In a series circuit, the total current is equal to the________. sum of the voltages across all parts of the circuit. sum of the resistances in all parts of the circuit. power total divided by the total resistance. current in any other part of the circuit.

current in any other part of the circuit.

A home theater system requires the speakers to be protected by a fuse type overcurrent device. The speakers are rated at 200 watts / 8-ohms. What size fuse is required to protect each individual speaker?

√(200 W / 8 Ω) = 5 A

A fire alarm panel is installed with a source voltage of 24 VAC. An indicating device is located 1,500 feet from the panel using 2-wire copper conductor, size 12 AWG. The indicating device uses 1 ampere. The power factor is 100% (unity). What is the voltage at the indicating device terminals for this installation? 1 volt 6.2 volts 18.0 volts 24.0 volts

(2 * 12.9 * 1 A * 1500') / 6530 cmil = 6 V, 24 V - 6 V =18 V

Determine the amperage of a 3 phase 208 V 5 HP motor with 92% efficiency and a power factor of 90%. 12.05 A 11.25 A 10.35 A 21.66 A

(5 HP * 746 W/HP) / (208 *1.732 * .92 *.9) = 12.045 A

A range draws 50 amperes when used on a circuit. A bad splice is discovered on the line at the range on one of the phase conductors. The splice has a measured resistance of 0.4Ω. How much power is lost at the splice?

(50 A)2 / 0.4 Ω = 1000 W, 1000 W / 1000 = 1kW

A 14 AWG copper conductor has an area of 4,110cmil. The resistance of the conductor is rated at 3.07Ω per 1,000 feet. Determine the Kcu (specific resistance of copper) at 75°C.

(Resistance * Area) / Distance, (4110 cmil * 3.07 Ω) / 1000' = 12.61

A single family dwelling has a calculated load of 22800 VA. Voltage is 240 1 phase, the power feeder load is?

22800 VA / 240 V = 95 A

The nameplate of a 250W heating element requires a line voltage connection of 120V. The heating element is connected to circuit with 95 volts. How much heat, in watts, is the heater producing? 520 W 250 W 197.92 W 156.68 W

250 W / 120 V = 2.08333 A, 120 V / 2.08333 = 57.6 Ω, 95 V / 57.6 Ω = 1.649 A, 95 V * 1.649 A = 156.68 W, OR (95 V)2 / 57.6 Ω = 156.68 W

An electrician is to install thirty-five (35) luminaires in the ceiling of a large room. No blue prints are available and the number of circuits required must be determined. Each luminaire is rated 250 watts each. The maximum wattage allowed on each circuit is 1,800-watts. This installation is not considered a continuous load. How many circuits will be needed?

35 * 250 W = 8750 W, 8750 W / 1800 W = 4.86 round up to 5 OR 1800 W / 250 W = 7.2 lights per circuit round down to 7, 35/7= 5

Determine the THW aluminum wire size needed for branch circuit conductors supplying a 50 Amp, 120-V single-phase load at a distance of 430-ft. and the voltage drop held within the 3% Code recommendations. 4 AWG THW 1/0 THW 250 kcmil THW 300 kcmil THW

KAL = 21.2, 120 V * 0.03 = 3.6 V, (2 * 21.2 * 50 A * 430') / 3.6 V = 253222 cmil, round up to 300 kcmil

Determine the MAXIMUM load for a 120 V, single-phase circuit with 4 AWG solid, THW, aluminum conductors installed a distance of 240-ft. and the voltage drop is held to 3%. 14.7 amperes 17.1 amperes 21.2 amperes 29.4 amperes

KAL = 21.2, 4 AWG = 41740 cmil, 120 V * 0.03 = 3.6 V, (41740 cmil * 3.6) / (2 * 21.2 * 240') = 14.7 A

Calculate the voltage drop on 350 kcmil THHN, copper conductor supplying a 300 Ampere, single-phase load a distance of 160-ft.

KCU = 12.9, 350 kcmil = 350000 cmil, (2 * 12.9 * 300 A *160') / 350000 cmil = 3.5 V@ KCU = 12.9, 350 kcmil = 350000 cmil, (2 * 12.9 * 300 A *160') / 350000 cmil = 3.5 V

The following series circuit has resistors connected with the following values: R1=30 ohms, R2=170 ohms, R3=29.87 ohms, and R4=890 ohms. If the circuit voltage is connected at 240V, how much total power is consumed by resistors R1 and R2 in the circuit?

RT = 30 Ω + 170 Ω + 29.87 Ω + 890 Ω = 1119.87 Ω, IT = 240 V / 1119.87 Ω = 0.21431 A, IT = I1 = I2 = I3 = I4 = I1-2, R1-2 = 30 Ω + 170 Ω = 200 Ω, E1-2 = 200 Ω * 0.21431 A = 42.862 V, P1-2 = 42.862 V * 0.21431 A = 9.18 W, OR P1-2 = (0.21431 A)2 * 1119.87 Ω = 9.18 W

The following series circuit has resistors connected with the following values: R1=30 ohms, R2=170 ohms, R3=29.87 ohms, and R4=890 ohms. If the circuit voltage is connected at 240V, how much total power is consumed by the circuit?

RT = 30 Ω + 170 Ω + 29.87 Ω + 890 Ω = 1119.87 Ω, IT = 240 V / 1119.87 Ω = 0.21431 A, PT = 240 V * 0.21431 A = 51.4345 W, 51.4345 W / 1000 = 0.051 kW, OR PT = (240 V)2 / 1119.87 Ω = 51.4345 W, 51.4345 W / 1000 = 0.051 kW

The following series circuit has resistors connected with the following values: R1=30 ohms, R2=170 ohms, R3=29.87 ohms, and R4=890 ohms. If the circuit voltage is connected at 240V, how much total power is consumed by the circuit? 51 kW 1119.87 W 0.051 kW 0.22 MW

RT = 30 Ω + 170 Ω + 29.87 Ω + 890 Ω = 1119.87 Ω, IT = 240 V / 1119.87 Ω = 0.21431 A, PT = 240 V * 0.21431 A = 51.4345 W, 51.4345 W / 1000 = 0.051 kW, OR PT = (240 V)2 / 1119.87 Ω = 51.4345 W, 51.4345 W / 1000 = 0.051 kW

The following loads are connected in series, 5Ω, 20Ω, 10Ω, and 15Ω. The voltage total is 50-V. Find the voltage drop on the 10Ω resistor. 20V 15V 10V 5V

RT = 5 Ω + 20 Ω + 10 Ω + 15 Ω = 50 Ω, 50 V / 50 Ω = 1 A, 10 Ω * 1 A = 10 V