ASVAB Electronics Information Practice Test 899113 Results

Your Results Global Average
Questions 5 5
Correct 0 3.73
Score 0% 75%

Review

1

An inductor __________ changes in the electric current flowing through it.

61% Answer Correctly

doubles

enhances

resists

eliminates


Solution

An inductor is coiled wire that stores electric energy in the form of magnetic energy and resists changes in the electric current flowing through it. If current is increasing, the inductor produces a voltage that slows the increase and, if current is decreasing, the magnetic energy in the coil opposes the decrease to keep the current flowing longer. In contrast to capacitors, inductors allow DC to pass easily but resist the flow of AC.


2

The joule is a unit of measurement for:

67% Answer Correctly

resistance

power

capacitance

energy


Solution

The joule is a unit of measurement for energy.


3

All electricity is the movement of which subatomic particles?

89% Answer Correctly

neutrons

nuclei

protons

electrons


Solution

All electricity is the movement of electrons which are subatomic particles that orbit the nucleus of an atom. Electrons occupy various energy levels called shells and how well an element enables the flow of electrons depends on how many electrons occupy its outer (valence) electron shell.


4

This circuit component symbol represents a(n):

71% Answer Correctly

capacitor

transistor

diode

inductor


Solution

A diode allows current to pass easily in one direction and blocks current in the other direction. Diodes are commonly used for rectification which is the conversion of alternating current (AC) into direct current (DC). Because a diode only allows current flow in one direction, it will pass either the upper or lower half of AC waves (half-wave rectification) creating pulsating DC. Multiple diodes can be connected together to utilize both halves of the AC signal in full-wave rectification.


5 What's the overall power consumption of a piece of equipment that is rated for 4 amps at 130 volts?
80% Answer Correctly
572 W
520 W
522 W
1040 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 4 amps (I) at 130 volts (V) so the equation becomes P = \( 130 \times 4 \) = 520 W