ASVAB Electronics Information Practice Test 793421 Results

Your Results Global Average
Questions 5 5
Correct 0 3.34
Score 0% 67%

Review

1

A diode to an electronic circuit is like a _______________ to a city?

67% Answer Correctly

traffic light

highway

parking lot

one-way street


Solution

A diode allows current to pass easily in one direction and blocks current in the other direction.


2

This circuit component symbol represents a(n):

57% Answer Correctly

AC source

capacitor

transformer

DC source


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.


3

Which of the following is not a characteristic of a step-up transformer?

54% Answer Correctly

the secondary voltage is higher than the primary voltage

increases voltage

the primary voltage is higher than the secondary voltage

has more turns in the secondary winding than in the primary winding


Solution

As their names indicate, a step-up transformer is used to step up or increase voltage and a step-down transformer is used to step down or decrease voltage. In a step-up transformer, the secondary voltage is higher than the primary voltage and it has more turns in the secondary winding than in the primary winding.


4

Voltage and current are __________ proportional.

66% Answer Correctly

not

directly

inversely

indirectly


Solution

Voltage (V) is the electrical potential difference between two points. Electrons will flow as current from areas of high potential (concentration of electrons) to areas of low potential. Voltage and current are directly proportional in that the higher the voltage applied to a conductor the higher the current that will result.


5 What's the overall power consumption of a piece of equipment that is rated for 6 amps at 10 volts?
80% Answer Correctly
58 W
63 W
61.5 W
60 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 6 amps (I) at 10 volts (V) so the equation becomes P = \( 10 \times 6 \) = 60 W