ASVAB Electronics Information Practice Test 686039 Results

Your Results Global Average
Questions 5 5
Correct 0 3.40
Score 0% 68%

Review

1

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

70% Answer Correctly

makes the circuit less safe

makes no difference

is required

is recommended


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

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.


3

The formula specifying Ohm's law is which of the following?

76% Answer Correctly

\(V = {I \over R}\)

V = IR

V = I2R

\(V = {R \over I}\)


Solution

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


4 Suppose you have 10 [12V 25A] batteries that you can connect together in series, in parallel, or in series-parallel. Which of the following voltage and ampere combinations cannot be attained using these 10 batteries?
46% Answer Correctly
12V 250A
10V 300A
120V 25A
60V 125A

Solution

Connecting the 10 batteries in series multiplies their voltage while keeping their current the same yielding a 120V 25A configuration. Connecting the 10 batteries in parallel multiplies their current while keeping their voltage the same yielding a 12V 250A configuration. Using a series-parallel connection, 5 batteries can be connected in series and 5 can be connected in parallel resulting in a 60V 125A configuration.


5 Use Ohm's Law to calculate the value of resistance in this circuit if voltage is 140 volts and current is 7 amps.
80% Answer Correctly
6 Ω
20 Ω
21.5 Ω
18 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 = \( \frac{V}{I} \) = \( \frac{140}{7} \) = 20 Ω