ASVAB Electronics Information Practice Test 536962 Results

Your Results Global Average
Questions 5 5
Correct 0 3.64
Score 0% 73%

Review

1

This circuit component symbol represents a(n):

65% Answer Correctly

fuse

diode

potentiometer

capacitor


Solution

Capacitors store electricity and are used in circuits as temporary batteries. Capacitors are charged by DC current (AC current passes through a capacitor) and that stored charge can later be dissipated into the circuit as needed. Capacitors are often used to maintain power within a system when it is disconnected from its primary power source or to smooth out or filter voltage within a circuit.


2

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

83% Answer Correctly

battery

voltmeter

reostat

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

An engineer who wants to document an electric circuit would create which of the following?

67% Answer Correctly

a matrix

a blueprint

a schematic

a layout


Solution

A schematic is the proper name for a drawing of an electric or electronic circuit.


4

This circuit diagram represents a(n):

69% Answer Correctly

parallel circuit

open circuit

series circuit

series-parallel circuit


Solution

A series circuit has only one path for current to flow. In a series circuit, current (I) is the same throughout the circuit and is equal to the total voltage (V) applied to the circuit divided by the total resistance (R) of the loads in the circuit. The sum of the voltage drops across each resistor in the circuit will equal the total voltage applied to the circuit.


5 Use Ohm's Law to calculate the value of current in this circuit if voltage is 75 volts and resistance is 30 Ω.
81% Answer Correctly
4 A
3.75 A
2.5 A
0.83 A

Solution

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

Solved for current, I = \( \frac{V}{R} \) = \( \frac{75}{30} \) = 2.5 A