ASVAB Electronics Information Electron Flow Practice Test 840782 Results

Your Results Global Average
Questions 5 5
Correct 0 4.59
Score 0% 92%

Review

1 What's the overall power consumption of a piece of equipment that is rated for 5 amps at 140 volts?
92% Answer Correctly
1400 W
630 W
700 W
350 W
Solution
Power is measured in watts (W) and 1 watt equals 1 ampere multiplied by 1 volt: P = \( V \times I \). For this problem, the equipment is rated for 5 amps (I) at 140 volts (V) so the equation becomes P = \( 140 \times 5 \) = 700 W
2 What's the overall power consumption of a piece of equipment that is rated for 5 amps at 140 volts?
92% Answer Correctly
1400 W
630 W
700 W
350 W
Solution
Power is measured in watts (W) and 1 watt equals 1 ampere multiplied by 1 volt: P = \( V \times I \). For this problem, the equipment is rated for 5 amps (I) at 140 volts (V) so the equation becomes P = \( 140 \times 5 \) = 700 W
3 What's the overall power consumption of a piece of equipment that is rated for 5 amps at 140 volts?
92% Answer Correctly
1400 W
630 W
700 W
350 W
Solution
Power is measured in watts (W) and 1 watt equals 1 ampere multiplied by 1 volt: P = \( V \times I \). For this problem, the equipment is rated for 5 amps (I) at 140 volts (V) so the equation becomes P = \( 140 \times 5 \) = 700 W
4 What's the overall power consumption of a piece of equipment that is rated for 5 amps at 140 volts?
92% Answer Correctly
1400 W
630 W
700 W
350 W
Solution
Power is measured in watts (W) and 1 watt equals 1 ampere multiplied by 1 volt: P = \( V \times I \). For this problem, the equipment is rated for 5 amps (I) at 140 volts (V) so the equation becomes P = \( 140 \times 5 \) = 700 W
5 What's the overall power consumption of a piece of equipment that is rated for 5 amps at 140 volts?
92% Answer Correctly
1400 W
630 W
700 W
350 W
Solution
Power is measured in watts (W) and 1 watt equals 1 ampere multiplied by 1 volt: P = \( V \times I \). For this problem, the equipment is rated for 5 amps (I) at 140 volts (V) so the equation becomes P = \( 140 \times 5 \) = 700 W