ASVAB Electronics Information Practice Test 733757 Results

Your Results Global Average
Questions 5 5
Correct 0 3.52
Score 0% 70%

Review

1

This circuit diagram represents a(n):

69% Answer Correctly

series circuit

parallel circuit

series-parallel circuit

open 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.

2

This circuit component symbol represents a(n):

70% Answer Correctly

capacitor

fuse

diode

resistor
Solution

Fuses are thin wires that melt when the current in a circuit exceeds a preset amount. They help prevent short circuits from damaging circuit components when an unusually large current is applied to the circuit, either through component failure or spikes in applied voltage.

3 A circuit with a 120-volt power supply is protected by a 20-amp circuit breaker. What is the largest number of watts loads on this circuit can safely use?
84% Answer Correctly
2400 W
2405 W
2640 W
2393 W
Solution
Wattage is current multiplied by voltage: W = IV. So, the maximum amount of power a 120-volt circuit with a 20A circuit breaker would allow is 120V x 20A = 2400 W.
4

In metals, increasing the temperature typically _____________ conductivity.

50% Answer Correctly

eliminates

increases

does nothing to

decreases
Solution

All conductors have resistance and the amount of resistance varies with the element. Metals exhibit increased resistance (and, therefore, lower conductivity) as their temperature increases.

5 Use Ohm's Law to calculate the value of current in this circuit if voltage is 320 volts and resistance is 80 Ω.
81% Answer Correctly
4 A
8 A
-2 A
4.4 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 = \( \frac{V}{R} \) = \( \frac{320}{80} \) = 4 A