ASVAB Electronics Information Practice Test 425122 Results

Your Results Global Average
Questions 5 5
Correct 0 3.66
Score 0% 73%

Review

1

What type of current flows in only one direction in a circuit?

82% Answer Correctly

alternating

parallel

series

direct


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.


2

This circuit component symbol represents a(n):

65% Answer Correctly

capacitor

diode

potentiometer

fuse


Solution

Capacitors store electricity and are used in circuits as temporary batteries. Capacitors are charged by DC current (AC current passes through a capacitor) and that stored charge can later be dissipated into the circuit as needed. Capacitors are often used to maintain power within a system when it is disconnected from its primary power source or to smooth out or filter voltage within a circuit.


3

The watt is a unit of measurement for:

78% Answer Correctly

frequency

energy

resistance

power


Solution

Electrical power is measured in watts (W) and is calculated by multiplying the voltage (V) applied to a circuit by the resulting current (I) that flows in the circuit: P = IV. In addition to measuring production capacity, power also measures the rate of energy consumption and many loads are rated for their consumption capacity. For example, a 60W lightbulb utilizes 60W of energy to produce the equivalent of 60W of heat and light energy.


4

The valence shell of n insulator is how full of electrons?

56% Answer Correctly

less than half full

empty

half full

more than half full


Solution

Insulators have valence shells that are more than half full of electrons and, as such, are tightly bound to the nucleus and difficult to move from one atom to another.


5 What's the overall power consumption of a piece of equipment that is rated for 9 amps at 130 volts?
80% Answer Correctly
1170 W
1173 W
1287 W
585 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 9 amps (I) at 130 volts (V) so the equation becomes P = \( 130 \times 9 \) = 1170 W