Electrical Calculations

26 amps. (Add up the continuous loads. 3,000 + 2,000 = 5,000. Continuous loads are always multiplied by 1.25. 5,000 x 1.25 = 6,250. Divide the 240 volts. 6,520/240 = 260.04 (26)

. Two electric heaters are installed. One heater is 3lw and the other is 2kw. Both run off 240 volts. The combined ampacity of the connected continuous load to the two heaters is?

6.894 amps

50KVA Transformer is fed from 7,200 volt utility, what would be the amps?

Answer 21.7 (Use formula found in pg. 50-52 in uglys. 250 x 2 x 21.2 x 60/52620 = 27.1)

A #3, 60 amp feeder is being supplied from a 240v service. The feeder runs a distance of 250 ft. How many potential volts of drop are possible on this feeder? Use 21.2 for the k factor.

Go to A2 125F =.76 in 90 degree column. Multiply 97 x.76 = 72. No it would not because it can only take 72F Experiment with different sizes in 310.15.b16. #2 = 130 x .76 = 98.8 so #2 would be the correct size.

A #4 THWN-2 conductor is rated for 95 amps before the adjustments are taken into consideration. If the ambient temperature is 125F would same size conductor work for the ambient temperature? If not, what size would work.

Answer: 39 amps (Check aluminum column in b16. = 55. Deration factor = .71. 55 x .71 = 39

A #6 XHHW-2 aluminum conductor is installed in an rea where the afternoon sun can raise the temperature to 140F. In this condition, the #6 conductor can carry an ampacity of:

1500 x 1.25 =1875/240 =7.8 (Only one circuit will work) 120.23

A 1500 watt/240v heater can carry how many 15 amp circuits?

Answer: d. (1250 X 4 + 750 X 1.25/240 = 29.94 (Correct)

A 240 volt fixed electric space heating branch circuit is installed in a dwelling unit, the circuit consists of #10 NM conductors and protected by a 30 amp over current device. Which of the following heater combos are allowed on this branch circuit? (1,000 W, 1,250 W, 1500 W, 2,000 W, 750W, 3,000W.) a. 2 ea. 3,000 Watts b. 2 ea. 1,500 watt heaters and 3 ea. 1250 watt heaters c. 3 ea. 1,000 watt heaters and 1 ea. 1500 watt heaters and 1 ea. 2,000 watt heater d. 4 each 1250 watts, and 1 ea. 750 wat heater.

Step 1: Find the starting ampacity of the conductor by looking at Table 310.15(B)(16). 10# is rated at 40 amps at 90 degrees Step 2: Find the ambient temperature de-rate of the conductor by looking at Table 310.15(B)(2)(a). 125 degrees shows a .76 correction factor. Step 3: Multiply the starting ampacity of the conductor (40-amps) found in T310.15(B)(16) by the de-rate (.76) found in Table 310.15(B)(2)(a). Answer: 40 x .76 = 30.4-amps.

A THWN-2 Conductor is sized at #10. The conductor is located in an area subject to an ambient temperature of 125F. Under this condition, this conductor can safely carry how many amps?

196 Amps. (THWN-2 is rated for 225 in the 90 C column in B16, In Table B2A it says that 108F temperature would have a deration factor of .87. Multiply .87 x 225 = 195.75 (196)

A THWN-2 conductor is sized at 3,0. The conductor is installed in an ambient temperature at 108 F. This conductor is rated to carry? '

Answer: #1 (Divide circular mils of phase conductors. 250,000/52620= 4.75. multiply no. 6 circular mils by 4.75 from t250.122. 16,510 x 4.75 = 78,422 = #1

A circuit is protected by a 100-amp overcurrent device. The phase conductors have been increased in size from #3 to 250 kcmil due to voltage drop. The egc conductor is sized?

#2 (Use formula found in pg. 50-52 in uglys. 21.2 x 2 x 300 x 30 = 381,000. (For copper switch 21.2 to 12.9) Experiment with different numbers of circular mils to divide 381,600 and then 240 volts with. Let's try #2 381,600/66360 = 7.25/240 = .023 (2.3%) any number higher would make it higher than 3%. OR 21.2 x 2 x 300 x 30/.03 = 12,720,000/240 = 53,000 = #2.

A detached structure is 300 ft. away from its source. Thre is a 240/120 volt, 30 amp feeder to the detached structure. Using XHHW-2 aluminum, provide the min. conductor size required in order to maintain a max. of 3% voltage drop.

Answer 1 : 1500 x 3x 1.25/240 go to 310.b16 to size conductor (#10) Answer 2: T240.6 Yes

A dwelling unit has 3 cadet heaters sized at 1500 rated for 250 volt what is the min. ampacity? Once you find the amps what standard size breaker would work?

Answer: #6 (multiply the 4 heaters by 2,000 and then multiply continuous load. Divide 240. 2 x 2,000 x 1.25/240 = 41.66. Go to b16. NM is 60C and no. 6 would work for 41.66

A heater circuit has four electric space heaters rating 2,000 watts @240v. What min. size NM cable is required to supply this circuit? Use copper conductors and this is a continuous load.

a. 2 x 12.9 x 465 x 60 divided by 26240 = 27.43 divided by 240 = .11 (11%) b. In table 8 divide the circular mils to find below 3% ( 1/0 = 719,820/105,600 = 6.816/240 = .028 (2.8%) So 1/0 would be the size of conductor. c. T250.122 says 60 amps is #10. Divide old and new size circ. mils in Table 8 for conductor. (105,600/26240 = 4.02 . multiply the circ. Mils for #10 10380 x 4.024 = 41727 the answer is #4

A house service is feeding a pole barn. The over current protection on the feeder is 60 amps. The barn is 465 feet away and we are using THHN copper. Normally, a # 6 would be the appropriate size for the feeder. a. What is the voltage drop for this installation? b. What size conductors are needed using THHN c. What size is the EGC for this installation

Step 1: Use Table 8 of Chapter 9 to determine circular mils of a #12 and a #6 conductor. Step 2: Divide the larger conductor cm (#6) by the smaller conductor cm (#12) to get a proportional increase in conductor size. 26,240 / 6,530 = 4.01 Step 3: Use Table 250.122 to determine the starting size of the EGC for a 20-amp overcurrent device. Which is a # 12 Step 4: Use Table 8 of Chapter 9 to determine circular mils of the #12 conductor (6,530cm). Step 5: Multiple the circular mils by the proportional increase of 4.01. 6,530 x 4.01 = 26,185cm Answer: In this case the EGC is increased in size to the same as the phase conductors. 26,185cm equates into a #6 conductor according to T.8 of Chapter 9.

A phase conductor has been increased in size from a #12 to a #6. The overcurrent device is sized at 20-amps. What size is the EGC required to be?

#2 (Divide cir. Mils of 350 and 3/0, 350,000/ 167,000 = 2.08. Multiply the cir. Mils of the #6 e.g.c by 2.08, 26,240 x 2.08 = 54,579. The cir. Mils of 54,579 would make it #2.

A phase conductor has been increased in size from a 3/0 to 350 cir. Mils. The original e.g.c. was sized #6. The new e.g.c size is?

240 x 240/5,000 = 11.52, 208 x 208/11.52 = 3,756 watts

A water heater is rated at 5,000 watts @ 240 volts of output does this water heater discharge if running on 208 V?

E / R = I 12 / 8 = 1.5

If R = 8 ohms and E = 12 V, what is the value of I?

E / R = I 120 / 12 = 10 amps

If a toaster produces 12 ohms of resistance in a 120 volt circuit, what is the amount of current in the circuit?

E / I = R 50 / 10 = 5 watts

In a circuit, the power source is supplying 50 volts and the current is 10 amps. What is the resistance?

E / R = I 6 / .3 = 20

Solve for I if E = 6volts and R = 0.3 ohms?

750 + 3000 + 1500 x 1.25 divided by 240 = 27.34 #10 size conductor 5,000 x 1.25 divided by 240= 26.04 #10 size conductor 4,000 + 4,500 x 1.25 divided by 240 = 44.23 #8

We have a house with three circuits that control all the baseboard heaters in the house. The heaters are a continuous load and 240 volts. Circuit 1 controls: (1) 750 w, (3) 1,000 w and (1) 1,500 w heaters Circuit 2 controls: (1) 5,000 w heater Circuit 3 controls: (2) 2,000 w and (3) 1,500 w heater a. What is the amps for each circuit? b. What size conductor size for each circuit based on the 75 degree column?

E / I = R 117 / 9 = 13 watts

What is the resistance if I = 9 A and E = 117 V?

E / I = R 110 / .25 = 440 watts

What is the resistance of a lamp if it draws 0.25 A from a 110 V line?

Use Table 250.66, a 3/0 AWG copper S/E conductor requires a GEC to a metal water line sized at #4.

What size grounding electrode conductor is needed for a service supplied by 3/0 AWG copper conductors, if the grounding electrode was an underground metal water pipe?

85. (Automatically go with the most restrictive temperature (160). The de-ration factor for 160 = .5. Go to B16 and THWN-2 is rated in the 90 column at 170. Multiply. 170 x .5 = 85.

a 1/0 AWG THWN-2 conductor is installed in an rea with an ambient temperature that ranges from 86 F - 160 F. What is the final re-rated ampacity of the conductor.