ASVAB Electronics Information Practice Test 816900 Results

Your Results Global Average
Questions 5 5
Correct 0 3.55
Score 0% 71%

Review

1

The ohm is a unit of measurement for:

81% Answer Correctly

energy

capacitance

power

resistance


Solution

Resistance is opposition to the flow of current and is measured in ohms (Ω). One ohm is defined as the amount of resistance that will allow one ampere of current to flow if one volt of voltage is applied. As resistance increases, current decreases as resistance and current are inversely proportional.


2

An engineer who wants to document an electric circuit would create which of the following?

67% Answer Correctly

a matrix

a schematic

a blueprint

a layout


Solution

A schematic is the proper name for a drawing of an electric or electronic circuit.


3

Direct current flows from the __________ terminal of the voltage source to the __________ terminal.

62% Answer Correctly

negative, positive

positive, negative

negative, negative

positive, positive


Solution

Direct current flows in only one direction in a circuit, from the negative terminal of the voltage source to the positive. A common source of direct current (DC) is a battery.


4

The most common circuit configuration is:

58% Answer Correctly

parallel

series-parallel

series

household


Solution

Circuits are not limited to only series or only parallel configurations. Most circuits contain a mix of series and parallel segments. A good example is a household circuit breaker. Electrical outlets in each section of the house are wired in parallel with the circuit breaker for that section wired in series making it easy to cut off electricity to the parallel parts of the circuit when needed.


5 What's the overall power consumption of a piece of equipment that is rated for 3 amps at 30 volts?
80% Answer Correctly
180 W
90 W
89 W
45 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 3 amps (I) at 30 volts (V) so the equation becomes P = \( 30 \times 3 \) = 90 W