Journeyman Theory Practice Test

A 35 μF capacitor is connected to a 60-Hz, 120 V supply. What is the XC of the capacitor? a. 75.79 Ω b. 80.33 Ω c. 83.2 Ω d. 93.43 Ω

a. 75.79 Ω 1 / (2 x 3.1416 x Frequency x Capacitance) 2 x 3.1416 x 60 x .000035 = .01319472 1 / .01319472 = 75.79 ohms

A 2kW load is operated at 240-V and 80% power-factor. Find the current of this load. a. 10.4 A b. 33A c. 25A d. 3 A

a. 10.4 A 2,000W ÷ (240V x 0.8) = 10.4A

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? a. 11% b. 20% c. 60% d. 80%

a. 11% (.5hp x 746) ÷ (120 Volts x .8) = 3.88 amperes(.5hp x 746) ÷ (120 Volts x .9) = 3.45 amperes3.45 amperes ÷ 3.88 amperes x 100 = 89%100% - 89% = 11%

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

a. 14.7 amperes Calculate voltage drop at 3% 120V x .03 = 3.6Vd maximumI = Acmil x Vd ÷ 2 x K x L41,740cm x 3.6 VD ÷ 2 x 21.2 x 240-ft = 14.7 amperes

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? a. 324 amperes b. 706 amperes c. 845 amperes d. 1,007 amperes

a. 324 amperes 1÷(6.28 x 1000 x .0001076 = 1.48Ω480V ÷ 1.48Ω = 324.349A

What is the peak voltage of a 277 V RMS AC voltage? a. 391.68 V b. 434.85 V c. 576.45 V d. 695.90 V

a. 391.68 V 277 volts x 1.414 = 391.68 volts

.A 30-µF capacitor is connected into a 240-V, 1,000Hz. circuit. What is the current flow in this circuit? a. 45.28 amperes b. 57.45 amperes c. 67.34 amperes d. 80.00 amperes

a. 45.28 amperes 1 ÷ (2 x 3.14 x Frequency x Capacitance)1 ÷( 6.28 x 1,000 Hz x 30-µ farads) = 5.3Ω240 Volts ÷ 5.3Ω = 45.28 amperes

An electric motor is running on 120 V. The current is measured to be 2 A. How many ohms of resistance is the motor? a. 60 b. 118 c. 122 d. 240

a. 60 E / I = R

What is the ohmic value of R-1? a. 75 b. 10 c. 14.4 d. 50

a. 75 Series circuit. ET - (E1 + E2) = E1 36V - 6V = 30V E2 ÷ R2 = I2 4V ÷ 10Ω = .4A 30V ÷ .4A = 75Ω

What is the monthly cost of operating a 240V, 5kW central electric heater that operates 12 hours per day when the cost is 15 cents per kilowatt-hour (kWhr)? a. $ 270 b. $ 87 c. $ 390 d. $ 313

a. $ 270 2020 Ugly's Pg. 117 Cost = Watts x hours used x rate per kWhr/1000 (5000 x 360 x 0.15) / 1000 = 270

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? a. .354 amperes b. .706 amperes c. .845 amperes d. 1.07 amperes

a. .354 amperes 2 x 3.14 x F x L = XLE ÷ XL = I2 x 3.14 x 60Hz x 3.6H = 1,356.48Ω480 Volts ÷ 1,356.48Ω = .354 amperes

What's the direct-current resistance of a 6 AWG copper conductor that's 400 ft long? a. 0.20 ohms b. 0.30 ohms c. 0.40 ohms d. 0.50 ohms

a. 0.20 ohms 6 AWG resistance = 0.491 ohms/1,000 ft [Chapter 9, Table 8] (0.491 ohms/1,000 ft) x 400 ft = 0.1964 ohms

Given: A series circuit consisting of a battery and a resistor has the following values:Battery voltage = 1.5 voltTotal current = 2 ampsWhat is the resistance of this circuit? a. 0.75 ohm b. 1.33.ohms c. 3.00 ohms d. 6.00 ohms

a. 0.75 ohm 1.5 V ÷ 2 A = 0.75

Potential difference best describes ______. a. current b. resistance c. voltage d. power

c. voltage

A wye-delta transformer has a primary line voltage of 7200 V and a secondary line voltage of 480 V. What is the turns ratio? a. 8.66:1 b. 9: 66 c. 10:15 d. 15:1

a. 8.66:1 turns ratio must be calculated using phase voltages. Wye line of 7200 v is 7200 / 1.732 = 4,157.04 phase volts Delta line of 480 v is 480 phase volts 4,157.04 / 480 = 8.66

In an electrical circuit the relationship of voltage and current is ______ ______ to the circuitsresistance. a. Directly proportional b. Inversely Proportional c. Equal Value

a. Directly proportional

In an inductive circuit ____________________ leads ______________________. a. Voltage, Current b. Current, Voltage c. Power, Resistance d. Resistance, Power

a. Voltage, Current ELI the ICE man

A complete circuit is often referred to as a(n) _____ circuit. a. closed b. open c. shorted d. grounded

a. closed

The rate of current flow is equal to electromotive force divided by resistance represents a value of_____. a. coulombs b. impedance c. voltage d. power

a. coulombs

The source voltage to a load is 120-V. An electrician measures a voltage drop of 3-1/3% at the load.The load carries 25 amperes. What is the resistance of the wire to the load? a. .13Ω b. .16Ω c. .18Ω d. .25Ω

b. .16Ω 120 volts x .033333 = 4 Vd

The source voltage to a load is 120-V. An electrician measures a voltage drop of 3-1/3% at the load.The load carries 25 amperes. What is the resistance of the wire to the load? a. .13Ω b. .16Ω c. .18Ω d. .25Ω

b. .16Ω 120 x .033333 = 3.9 V; 4V / 25A = .16Ω

The voltage measured at the panel is 120-V. The voltage measured at the sign is 117-V. Anelectrician measured the current on the black wire at 35 amps. What is the total ohms calculated onthe TOTAL length of wire? a. 0.04285 Ω b. 0.0857 Ω c. 0.35 Ω d. 3.428 Ω

b. 0.0857 Ω 120V - 117V = 3 V3V ÷ 35A = 0.0857Ω

What's the direct-current resistance of a 6 AWG copper conductor that's 650 ft long? a. 0.22 ohms b. 0.32 ohms c. 0.42 ohms d. 0.52 ohms

b. 0.32 ohms

R1 = 75 Ohms, R2 = 250 Ohms, and R3 = 650 Ohms, 600 V is applied from A to B. What is the current flow through R3? a. 1.25 A b. 0.65 A c. 3 A d. 400 A

b. 0.65 A Add R2 and R3 in parallel = 180.56 Ohms Add R1 and R2,3 in series = 255.56 Ohms Use 600V / 255.56 Ohms to get a total current of 2.35 A Voltage at R1 is 75 Ohms x 2.35 A = 176.25V Voltage at R2,3 is found by removing E1(176.25V) from Et(600V) = 423.75V Voltage is the same at every point in parallel, so divide E3(423.75) by R3(650 Ohms) = .65 A

A two-branch parallel circuit has a 30 Ω and a 15 Ω resistor. What is the total resistance? a. 2 Ω b. 10 Ω c. 45 Ω d. 15 Ω

b. 10 Ω (1/30) + (1/15) = (1/0.1) = 10 ohms

An RLC series circuit has an applied voltage of 240 volts. R = 48 Ω, XL = 100 Ω, XC = 36 Ω, and Z = 80 Ω. What is the voltage drop across the capacitor? a. 40 V b. 108 V c. 300 V d. 444 V

b. 108 V Current is the same at any point in series. Find total current using Z. 240 volts / 80 ohms = 3 amps Xc (36 ohms) x 3 amps = 108 volts

What is the turns ratio of a transformer with 240 volts on the primary and 24 volts on thesecondary? a. 1:10 b. 10:1 c. 1:20 d. 20:1

b. 10:1 Ep ÷Es = Np ÷ Ns 240Vp ÷ 24Vs = 10 Ratio is 10:1

The source voltage is 120V. The load carries 5 amps. The line resistance of the conductor is 0.9.Calculate the line voltage at the load. a. 125V b. 115.5V c. 110V d. 100V

b. 115.5V 0.9 x 5A = 4.5 V; 120V - 4.5V = 115.5V

A 14 AWG copper conductor has a diameter of 4,110cmil. The resistance of the conductor is ratedat 3.07Ω per 1,000 feet. Determine the Kcu (specific resistance of copper) at 75°C. a. 10.4 b. 12.6 c. 17 d. 21.2

b. 12.6 KCU = (Cmil x R) ÷ LKCU = (4,110 x 3.07) ÷ 1,000 = 12.6

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

b. 140 kVA E x I x 1.73 ÷ 1000 = kVA480 Volts x 168.4 amperes x 1.732 ÷ 1,000 = 139.84 kVA

An inductor has an inductive reactance of 12 Ω, and a wire resistance of 16 Ω. What is the impedance of the inductor? a. 4 Ω b. 20 Ω c. 28 Ω d. 400 Ω

b. 20 Ω Z = square root ( 16 squared + 12 squared) = 20 ohms

Refer to the diagram above and solve using the values provided. R1 = 30 Ω, R2 = 600 Ω, and R3 = 1800 Ω, and 240 V is applied between A and B. What is the voltage across R2? a. 1.5 V b. 225 V c. 240 V d. 900 V

b. 225 V Add R2 and R3 in parallel = 450 ohms Add that to R1 (30 ohms) and Rt= 480 ohms 240 volts / 480 ohms = .5 A (It) 30 ohms x .5 A = 15 volts (E1) Et (240 volts) - E1 (15 volts) = E2,3 (225 volts) Voltage is the same at any point in parallel

A linear load heating coil operates at 24 Volts using 6 watts. If the coil is connected to 48 Volts, howmuch power will it use? a. 12 W b. 24 W c. 36 W d. 48 W

b. 24 W E2 ÷ P = OhmsE2 ÷ Ohms = Watts24 Volts2 ÷ 6 Watts = 96Ω48 Volts2 ÷ 96Ω = 24 Watts

Given:- The DC resistance of 4/0 uncoated copper conductors is 0.0608Ω per 1,000 feet.- Four (4) - 4/0 copper conductors are connected in parallel.- Each conductor has a length of 274-feet.- The total current of the single-phase equipment connected at the load is measured at 547 amperes.Using the DC resistance, calculate the total voltage drop in the paralleled conductors. a. 2.28V b. 3.89V c. 4.35V d. 4.6 V

b. 3.89V R = 0.0608 ÷ 1000 x 274-ft = 0.0166Ω0.0166Ω /4 = 0.00416Ω (Parallel circuit "Law of Equal Resistance" divide total resistance bynumber of conductors resistance.)0.00416Ω x 547 amperes = 2.27825 volts

Calculate the Power loss on the conductors for on a 100-ft., 120-V, Single-phase, branch circuitcarrying 60 amperes, using 6 AWG, THW, copper conductors. a. 115 Watts b. 354 Watts c. 487 Watts d. 600 Watts

b. 354 Watts Vd = 2 x K x L x I ÷ Acmil2 x 12.9 x 100-ft x 60 amperes ÷ 26,240 cm = 5.9 VdE x I = W5.9 Vd x 60 amperes = 354 Watts

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

b. 42.1 amperes 35 VA x 1,000 ÷ 480 Volts x 1.73 = 42.1 amperes

Three loads are connected in series, RT = 225W, R1 = 100W, R2 = 75W. Find the value of R3. a. 52.9W b. 50W c. 39.334W d. 25W

b. 50W 225W - 100W - 75W = 50W

An installation on a 208-V three-phase system consists of a 750 kcmil, THW, aluminum feedersupplying 400 amps to a panel located 300-ft. from the service. A three-phase branch circuit of 1/0AWG, THW, aluminum, carrying 120 amps to a three-phase machine located 157-ft. from the loadcenter panel. FIND the voltage drop percentage for the feeder and the branch circuit. a. 5% b. 6% c. 7% d. 8%

b. 6% Vd = 1.73 x K x L x I ÷ AcmilCalculate feeder Vd: 1.73 x 21.2 x 300-ft x 400A ÷ 750,000cm = 5.87-VdCalculate Branch Circuit Vd: 1,732 x 21.2 x 157-ft x 120A ÷ 105,600cm = 6.54-VdAdd Feeder and Branch Circuit Vd together and divide by line-line Voltage:5.87V + 6.54V = 12.41V12.41V ÷ 208V x 100 = 5.96%

What is the MINIMUM size copper conductor needed for a 208V, 3-phase machine located 140 feetfrom the power source? (Machine draws 27 amperes, 3% voltage drop, use 12.9K.) a. 10 AWG b. 8 AWG c. 6 AWG d. 4 AWG

b. 8 AWG

In a capacitive circuit ____________________ leads ______________________. a. Voltage, Current b. Current, Voltage c. Power, Resistance d. Resistance, Power

b. Current, Voltage ELI the ICE man

Line voltage and phase voltage are the same in the _____ connection. a. Wye b. Delta

b. Delta

During normal operation of a typical 120-volt appliance circuit, current flows through the hot, grounding, and neutral conductors. a. True b. False

b. False

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

b. step-up

If the resistance of copper conductor is 12.9 ohms per mil foot at 75°C and 12 AWG copperconductor has a circular-mill area of 6,530 cmil, what is the resistance of 1,000-ft of 12 AWG copper? a. 0.002Ω b. 0.02Ω c. 1.92Ω d. 19.2Ω

c. 1.92Ω R = (KCU x L) ÷ CmilR = (12.9 x 1,000) ÷ 6,530 cmil = 1.92Ω

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

c. 10 20A ÷ 1.25 = 16 A 0.8 x 2 (ballasts per fixture) = 1.6 A 16A ÷ 1.6A = 10 fixtures

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

c. step-down

What's the approximate distance that a 240V, 31A, single-phase load can be located from the panelboard so the voltage drop doesn't exceed three percent? The load is wired with 8 AWG copper. a. 55 ft b. 110 ft c. 145 ft d. 220 ft

c. 145 ft D = (Cmil x VD)/(2 x K x I) Cmil = 16,510 VD = 7.20V (240V x 0.03) K = 12.90 I = 31A D = (16,510 cmil x 7.20V)/(2 x 12.90 x 31A) D = 118,872/806 D = 148.60 or approximately 145 ft

What is the ampere load of a 140kVA, 3-phase, load operating at 480-V? a. 16.8 b. 292 c. 168 d. .292

c. 168 (140 x 1000) ÷ (480 x 1.732) = 168.4A

A delta-connected set of windings has a phase current of 10 A. What is the line current? a. 5.77 A b. 10 A c. 17.32 A d. 30 A

c. 17.32 A Iline = Iphase x 1.732

Given: A fire alarm indicating appliance is installed as follows:Source voltage: 24 volts, AC Distance between alarm panel and the indicating appliance:1,500 feet (3,000 feet of conductor). Conductor: Size 12 AWG uncoated copper.Current drawn by the indicating appliance: 1 amp.Power factor: Consider power factor as 1 (unity).What is the voltage at the appliance terminals for the installation described above? a. 1 Volts b. 6.2 Volts c. 18.0 Volts d. 24.0 Volts

c. 18.0 Volts Vd = 2 x K x L x I ÷ Acmil2 x 12.9 x 1,500-ft x 1 ampere ÷ 6,530cm = 5.9 Vd24 Volts - 5.9 Volts = 18.1 Volts

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

c. 18.0 volts

R1 = 40 Ohms, R2 = 700 Ohms, and R3 = 1200 Ohms, 277V is applied from A to B. What is the voltage across R2? a. 1.5 V b. 225 V c. 254 V d. 900V

c. 254 V Add R2 & R3 in parallel = 442.1 Ohms Add that to R1(40 Ohms) = 482.1 Ohms 277V / 482.1 Ohms = .57 A (It) 40 Ohms x .57 A = 22.8V (E1) Et(277V) - E1(22.8V) = E2,3(254.2V) Voltage in parallel is the same at any point

An RLC series circuit has an applied voltage of 240 volts. R = 48 Ω, XL = 100 Ω, XC = 36 Ω, and Z = 80 Ω. What is the total current? a. 1 A b. 2 A c. 3 A d. 4 A

c. 3 A 240/80 = 3

A 100 Watt incandescent lamp is to be replaced with a 15 Watt, LED equivalent. If the cost per kWhr is 15 cents, how many hours would the new lamp need to operate to pay for itself? Lamp cost is $4. Energy saved is 85 Watts. a. 1307 Hours b. 9 Hours c. 314 Hours d. 173 Hours

c. 314 Hours 2020 Ugly's Pg. 117 Hours = Lamp cost x 1000 / Watts saved x kWhr (4 x 1000) / (85 x 0.15) = 313.73

A machine operating at 480V, 3-phase is connected with 4 AWG conductor. An ampere metershows 26 amperes at the machine wire connection enclosure. The raceway is measured to be 300 feetfrom the panelboard to the machine. What is the voltage drop on the conductors? a. 3.16 Vd b. 3.89 Vd c. 4.17 Vd d. 5.16 Vd

c. 4.17 Vd (1.73 x 12.9 x 26A x 300-ft) ÷ 41,740cmil = 4.17VD

A transformer has a 240 V primary and a 120 V secondary. With a 30 ohm load connected, what is the primary volt-amps? a. 4 b. 8 c. 480 d. 960

c. 480 volt-amps stay the same from winding to winding 120 volts / 30 ohms = 4 amps 120 v x 4 amps = 480 volt-amps

The voltage at a load center is measured at 116V. A hair dryer is plugged into the furthestreceptacle outlet on 15 ampere circuit. With the hair dryer operating the voltage measurement at thereceptacle outlet is 109V. What is the percentage of voltage drop on the circuit? a. 3.4% b. 5% c. 6% d. 94%

c. 6% 116V - 109V ÷ 116 x 100 = 6%

In a balanced three-phase system, the conductors need be only about _____ the size of conductors for a single-phase, two-wire system of the same kilovolt-ampere (kVA) rating. a. 25% b. 50% c. 75% d. 90%

c. 75%

If a single-phase electric kiln rated 53 amps at 240 volts is operated on 208 volts, how much power does it consume? a. 12.7 kW b. 11.0 kW c. 9.6 kW d. 8.0 kW

c. 9.6 kW E ÷ I = RE2 ÷ R = W240 Volts ÷ 53 amperes = 4.53Ω2082 Volts ÷ 4.53Ω = 9,571.68W9,600W ÷ 1000 = 9.6 kW

A three-phase motor is running in the clockwise direction, how is this motor direction changed tocounter-clockwise? a. Add a capacitor to the start windings. b. Connect the grounded conductor to the stator pole c. Reverse motor leads T-1 & T-3 with their connected line conductors d. Increase the voltage with a step-up transformer

c. Reverse motor leads T-1 & T-3 with their connected line conductors

What is the proper procedure for reversing a DC motor? a. reverse the polarity of the exciter winding b. reverse the polartiy of the stator winding c. reverse the field connections d. reverse the control switch

c. reverse the field connections

To measure current a meter must be connected in _____. a. combination b. parallel c. series d. series parallel

c. series

Refer to the diagram above and solve using the values provided. R1 = 50 Ω, R2 = 200 Ω, and R3 = 600 Ω, and 600 V is applied between A and B. What is the current flow through R3? a. 400 A b. 3 A c. 1.25 A d. 0.75 A

d. 0.75 A Add R2 and R3 ohms in parallel, R2,3= 150 ohms Add R1 and R2,3 in series Rt= 200 ohms Use 600v / 200 ohms to get total current of 3A Voltage at R1 is 50 ohms x 3A = 150 volts Voltage at R2,3 is found by removing E1 (150 volts) from Et (600 volts) =450 volts Voltage is the same at every point in parallel, so divide E3 (450 volts) by R3 (600 ohms) = .75 amps

A range draws 50 amperes when used on a circuit. A bad splice is discovered on the line at the rangeon one of the phase conductors. The splice has a measured resistance of 0.4Ω. How much power is lostat the splice? a. 9.6 kW b. .96 kW c. 1.6 kW d. 1.0 kW

d. 1.0 kW I x R = ElossEloss x I = Power loss50 amperes x 0.4Ω = 20 Voltsloss 20 Voltsloss x 50 amperes = 1,000 W ÷ 1,000 = 1 kW

An inductor with an inductance of three henrys (H) is to be connected to a 60 Hz circuit. What will the inductive reactance be? a. 180 Ω b. 360 Ω c. 565 Ω d. 1131 Ω

d. 1131 Ω Xl = 2 x 3.1416 x Frequency x Inductance 2 x 3.1416 x 60 x 3 = 1130.98 ohms

Two capacitors rated at 95 μF and 71 μF are connected in parallel. What is the total capacitance of the circuit? a. 0.0246F b. 41 F c. 83 μF d. 166 μF

d. 166 μF 95 μF + 71 μF = 166 μF

A kitchen circuit is rated for 20 amperes. What is the total connected load on this circuit if a toasterrated at 550 watts, a coffee pot rated at 375 watts, and a toaster oven rated at 1,200 watts areoperating on the circuit at the same time? a. 12.3 amperes b. 14.7 amperes c. 16 amperes d. 17.7 amperes

d. 17.7 amperes

What's the approximate distance that a 277V, 31A, single-phase load can be located from the panelboard so the voltage drop doesn't exceed three percent? The load is wired with a 8 AWG copper. a. 55 ft b. 110 ft c. 145 ft d. 170 ft

d. 170 ft D = (Cmil x VD)/(2 x K x I) Cmil = 16,510 VD = 8.31V (277V x 0.03) K = 12.90 I = 31A D = (16,510 x 8.31)/(2 x 12.9 x 31) D = 118,872/806 D = 171.54 or approximately 170 ft

A 240/120-Volt panel has a load of 7,960 watts connected to line one and 3,420 watts connected toline two. Assuming all loads are 120-V, what is the current measured on the grounded conductor of thissystem? a. 45A b. 65A c. 19A d. 38A

d. 38A 7,960W - 3,420W = 4,540W4.540 ÷ 120V = 37.83Amperes

What is the MINIMUM size copper conductor needed for a sign located 350-feet from a panel tomaintain a 3% voltage drop? (Sign nameplate: 120V, 14 amperes.) a. 10 AWG b. 8 AWG c. 6 AWG d. 4 AWG

d. 4 AWG

What size resistor is necessary to limit the current flow to 0.47 amperes on a 24V circuit? a. 0.0196Ω b. 11.2Ω c. 24.47Ω d. 51Ω

d. 51Ω

What is the peak voltage of a 480 V RMS AC voltage? a. 391.68 b. 434.85 c. 576.45 d. 678.72

d. 678.72 480 volts x 1.414 = 678.72 volts

A 240V, 24A, single-phase load is located 160 ft from the panelboard. The load is wired with 10 AWG conductors. What's the approximate voltage drop of the branch-circuit conductors? a. 3.2V b. 4.25V c. 5.9V d. 9.5V

d. 9.5V VD = (2 x K x I x D)/Cmil K = 12.90, copper I = 24A D = 160 ft Cmil = 10,380 VD = (2 wires x 12.90 x 24A x 160 ft)/10,380 cmil VD = 9.50V

The magnetic field strength of an electromagnet can be increased by completing which of the following: a. Increase the number of windings in the coil b. Increase the voltage c. Put an iron core inside the coil of wire d. All of the above will increase the magnetic field

d. All of the above will increase the magnetic field