General Knowledge Flashcards
Operation at a substantially constant load for an indefinitely longtime.
Continuous duty
Voltage drop in a series circuit
3 100W @ 120V light bulbs are connected in series. What is the voltage drop on each lamp?
R = 120V x 120V / 100W = 144ohm
Rt = 144 + 144 +144 = 432 ohm
I = 120V / 432ohm = .2777
R1 = .2777 x 144 = 40V
R2 = .2777 x 288 = 79.97V
R3 = .2777 x 432 = 119.96V
- find total resistance (series) Rt =
1. Find resistance R = E x E / W
A clamp ammeter is used to measure the current flowing in each ungrounded leg of 120/240V, 3wire service. The current flowing on leg A is 27A and the current flowing on leg B is 19A. What is the current flowing on the nuetral conductor?
I n = Ia - Ib
In = 27-19A = 8A
An apartment building is supplied power from a 3 phase, 4wire 208/120V electrical system with a 3wire, 208/120V feeder with two ungrounded conductors and a neutral conductor to each living unit. Only 120V loads are operating and the loads are balanced. The current on each ungrounded conductor is measured at 42A. What is the current flowing on the neutral?
In a 3 phase system, the current will never be smaller than the smallest phase current or larger than the largest phase current. A good estimate is the average of the two currents which, in this case is 42A.
An electrical conductor is made up of 19 strands each with a diameter of 0.0837 inches. What is the approximate circular mil area of the conductor?
- Convert inches to mils to get circular mils
- That gives the area of one strand
- Multiply that number by 19 strands
1inch = 1/1000 of a mil
0.0837 = 83.7 mil
cmil = mil squared = 1 strand
83.7mils x 83.7mils = 7005.7 cmils /per strand
19strands x 7005.7cmils = 133,108 cmil
An electrician installs two 20 ampere, 120 volt circuits to supply 16 fluorescent lighting
fixtures. Each fixture draws 1.4 amperes. In addition there are two 120 volt fans on
individual circuits each drawing 5.6 amperes. If the building is served by a 120/240 volt,
single-phase, 3-wire service, the minimum neutral current to serve these loads is:
A. 0 amperes. C. 5.6 amperes. E. 33.6 amperes.
B. 2.8 amperes. D. 16.8 amperes.
A 0 amperes
It is expected the electrician will attempt to balance the loads on the ungrounded
conductors and minimize the current on the neutral conductor. In this case, it is possible
to balance the loads by placing 8 lighting fixtures on one circuit and 8 fixtures on the other
circuit as follows:
PH A = 1.4 x 8 =
+ 5.6 A fan
PH B = 1.4 x 8 =
+ 5.6 A fan
A 3-phase 480 volt capacitor bank rated 90 kVARs is connected near the main service of a
building to correct the power factor. The minimum permitted ampere rating of the
conductors to the capacitor bank is:
A. 115 amperes. C. 146 amperes. E. 260 amperes.
B. 130 amperes. D. 198 amperes.
C 146 Amperes
First the current must be determined that will flow to the capacitor bank when it is
energized. The same formula is used as would be used to determine the full load current
of a transformer except in this case kVAR is used instead of kVA.
Icap = kVAR × 1000 / 1.73 × V
Icap = 90 x 1000 / 1.73 x 480 = 108.4A
The question is what is the minimum ampere rating of the conductors to the capacitor
bank, therefore, follow the rule in 460.8(A). Multiply the capacitor current by 1.35.
I CONDUCTOR = 108.4 A × 1.35 = 146 Amperes
The total connected load in watts of the following electric heaters rated 1200 watts at 120 volts, 1600 watts at 120 volts, and 1800 watts at 240 volts is:
A. 4600 watts. B. 5200 watts. C. 5800 watts. D. 6200 watts. E. 7400 watts.
A 4600 watts.
The voltage makes no difference. Just add the wattage of each heater.
1200 + 1600 + 1800 = 4600
A 3-phase electrical heater drawing 10 amperes at 240 volts, line-to-line, with a power factor of 1.0 produces:
A. 2101 watts of heat. B. 2400 watts of heat.
C. 2985 watts of heat. D. 4152 watts of heat.
E. 4315 watts of heat.
D 4152 watts of heat
Power = 1.73 × Volts × Amperes × power factor = 1.73 × 240 V × 10 A × 1.0 = 4152 W
If three resistors are connected in series across a 120 volt source, the first 4 ohms, the second 6 ohms and the third resistor 12 ohms. The total resistance of the circuit will be:
A. 11 ohm. B. 18 ohm. C. 22 ohm. D. 24 ohm.
E. 20 ohm.
A 96 volts
The voltage drop will be in the same proportion as each resistance in series is to the total circuit resistance. 28 ohms is 80% of the total resistance (35 ohms) so 80% of the voltage will be across the 28 ohms resistor.
(0.8 × 120 V = 96 V)
Wherever practical, electrical motors with the highest voltage rating are used to:
A. obtain higher motor rpm. B. get more power from the motor. C. reduce the size of supply conductors required. D. improve electrical safety. E. lower the cost of operation.
C reduce the size of supply conductors required
Check current of a particular motor at two different voltages in Table 430.148 or Table 430.150.
Two 100 watt incandescent lamps operating for 18 hours where the average cost of electrical energy is 11 cents per kWh ($0.11 / kWh). Total cost of operating these lamps will be:
A. $0.11. B. $0.40. C. $1.28. D. $2.70. E. $5.62.
**B $0.40 **
200 W = 0.2 kW
0.2 kw × 18 hrs. = 3.6 kWh
3.6 kWh × $0.11 = $0.40
If the current flowing in each “hot” conductor of a single-phase, 120/240 volt, 3-wire service entrance is 28 amperes on leg A and 42 amperes on leg B, the current flowing in the neutral will be:
A. 28 amperes. B. 42 amperes. C. 14 amperes. D. 35 amperes. E. 0 amperes.
C. 14A
42A - 28A = 14A on the Neutral
If the resistance of a copper conductor is 0.410 ohms/k ft, the total resistance of the circuit conductors for a single-phase load 125 ft from the current supply is:
A. 0.0514 ohms. B. 0.1025 ohms. C. 0.1135 ohms. D. 0.5125 ohms. E. 1.0250 ohms.
B 0.1025 ohms
There are two wires in the 125 ft run so the total length of wire is 250 ft.
250 ft is 25% of 1000 ft so the resistance will be 0.1025 ohm.
(0.25 × 0.410 ohm = 0.1025 ohm)
If the diameter of a solid copper conductor is 0.125 in., the area of the conductor is:
A. 12,265 cmil. B. 15,625 cmil. C. 25,000 cmil. D. 37,500 cmil. E. 49,063 cmil.
. B 15,625 cmil
0.125 in. = 125 mils
125 mils x 125 mils = 15,625cmil
