ASVAB Electronics Information Practice Test 952061 Results

Your Results Global Average
Questions 5 5
Correct 0 3.17
Score 0% 63%

Review

1

What kind of diode allows current to flow in the opposite direction once a certain voltage threshold is reached?

51% Answer Correctly

zener diode

v-type diode

light emitting diode

standard diode


Solution

A diode allows current to pass easily in one direction and blocks current in the other direction. A zener diode is a diode which allows current to flow in one direction as normal and will also allow current flow in the reverse direction when the voltage is above a certain value. This value is called the breakdown voltage.


2

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

54% Answer Correctly

the primary voltage is higher than the secondary voltage

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

increases voltage

the secondary voltage is higher than the primary voltage


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.


3

From what energy do photovoltaic cells produce electrical energy?

68% Answer Correctly

sun

magnetic

nuclear

chemical


Solution

A photovoltaic cell (also known as a solar cell) converts energy from the sun into electrical energy.


4

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

52% Answer Correctly

more than half full

half full

full

less than half full


Solution

Conductors are elements that allow electrons to flow freely. Their valence shell is less than half full of electrons that are able to move easily from one atom to another.


5 What's the overall power consumption of a piece of equipment that is rated for 9 amps at 150 volts?
80% Answer Correctly
1345 W
1350 W
4050 W
1485 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 150 volts (V) so the equation becomes P = \( 150 \times 9 \) = 1350 W