ASVAB Electronics Information Practice Test 54387 Results

Your Results Global Average
Questions 5 5
Correct 0 3.98
Score 0% 80%

Review

1

The valence shell of a semiconductor is how full of electrons?

67% Answer Correctly

half full

empty

more than half full

less than half full


Solution

Semiconductors have valence shells that are exacly half full and can conduct electricity under some conditions but not others. This property makes them useful for the control of electrical current.


2

You would measure the amount of voltage between two points in a circuit with a(n):

83% Answer Correctly

reostat

battery

voltmeter

ammeter


Solution

Voltage (V) is the electrical potential difference between two points. A voltmeter is used to measure the voltage between two points in a circuit.


3

This circuit diagram represents a(n):

65% Answer Correctly

series circuit

rectifier

parallel circuit

series-parallel circuit


Solution

In a parallel circuit, each load occupies a separate parallel path in the circuit and the input voltage is fully applied to each path. Unlike a series circuit where current (I) is the same at all points in the circuit, in a parallel circuit, voltage (V) is the same across each parallel branch of the circuit but current differs in each branch depending on the load (resistance) present.


4

Resistance is measured in:

86% Answer Correctly

coulombs

ohms

volts

amperes


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.


5 What's the overall power consumption of a piece of equipment that is rated for 2 amps at 60 volts?
92% Answer Correctly
120 W
122 W
360 W
108 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 2 amps (I) at 60 volts (V) so the equation becomes P = \( 60 \times 2 \) = 120 W