ASVAB Electronics Information Practice Test 692602 Results

Your Results Global Average
Questions 5 5
Correct 0 3.75
Score 0% 75%

Review

1

Using a fuse with a current rating higher than that required by a circuit:

70% Answer Correctly

is required

makes no difference

is recommended

makes the circuit less safe


Solution

A fuse is a type of low resistance resistor that stops current flow in a circuit in response to a larger than intended electric current flow. Using a fuse with a higher current rating than required by a circuit is less safe as it could potentially allow overcurrent and risk a fire or heat-related equipment damage.


2

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

83% Answer Correctly

voltmeter

battery

ammeter

reostat


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

A __________ electric current produces a magnetic field proportional to the amount of current flow.

61% Answer Correctly

low voltage

moving

stationary

high voltage


Solution

A moving electric current produces a magnetic field proportional to the amount of current flow. This magnetic field can be made stronger by winding the wire into a coil and further enhanced if done around an iron containing (ferrous) core.


4

Which of these materials is not a good conductor of electricity?

79% Answer Correctly

air

gold

copper

tin


Solution

All conductors have resistance and the amount of resistance varies with the element. In general, metals make the best conductors of electricity and non-metals make the worst conductors of electricity.


5 Use Ohm's Law to calculate the value of resistance in this circuit if voltage is 150 volts and current is 7.5 amps.
81% Answer Correctly
22 Ω
20 Ω
15 Ω
28 Ω

Solution

Ohm's law specifies the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit: V = IR.

Solved for resistance, R = \( \frac{V}{I} \) = \( \frac{150}{7.5} \) = 20 Ω