ASVAB Electronics Information Practice Test 2

Answer: B

Correct answer: Qualified personnel using suitable equipment and procedures. Testing live or possibly live equipment exposes the tester to shock and arc flash, so safety depends on three things together: training to recognize hazards, equipment rated for the circuit, and a procedure such as verifying the meter before and after use. A qualified person is someone trained in those hazards and practices. Theory knowledge or simply owning a meter covers only one piece, which is why those answers are tempting but incomplete.

• A: Hearing a symptom provides no qualification or safe isolation.
• B (correct): OSHA restricts testing work on electrical circuits or equipment to qualified persons.
• C: Owning a meter does not establish qualification or show that the device is suitable for the circuit.
• D: A formula and drawings do not provide the training, ratings, or precautions needed for real work.

Answer: A

Correct answer: 18 V. In a simple parallel group, each branch spans the same two nodes as the source. So the 6 Ω resistor has the full 18 V across it, the same as the 3 Ω and 9 Ω resistors.

• A (correct): Every branch connected across the same two rails has the full source voltage across it.
• B: Branch voltages in parallel do not add up; multiplying 18 V by three branches gives a value larger than the source can supply.
• C: 18 divided by 6 gives the branch current in amperes, not the voltage across the branch.
• D: Dividing the 18 V equally among three resistors treats the circuit as if voltage were split, which happens in series, not parallel.

Answer: B

Correct answer: Henry. When current through a coil changes, its changing magnetic field induces a voltage that pushes against the change; this self-induction is called inductance. One henry means a current change of one ampere per second induces one volt. The likely confusion is with resistance in ohms, which opposes steady current, or capacitance in farads, which stores charge. Think of an inductor as electrical inertia: it resists speeding up or slowing down the current, not the current itself.

• A: Farads measure capacitance, which stores energy in an electric field.
• B (correct): The property is inductance, and inductance is measured in henries.
• C: Ohms measure resistance, which opposes current itself rather than changes in current.
• D: Teslas measure magnetic flux density, not a coil's opposition to current change.

Answer: A

Correct answer: 2 kWh. Energy equals power times time, and keeping power in kilowatts and time in hours gives kilowatt-hours directly. A kilowatt-hour is defined as one kilowatt running for one hour, so two hours at that rate is twice as much. The tempting answer 2 kW has the right number but the wrong unit: kilowatts describe the heater's rate of use, while kilowatt-hours describe the total energy that accumulates while it runs.

• A (correct): 1 × 2 = 2 kWh.
• B: kW alone measures power, not accumulated energy.
• C: This accounts for only one hour.
• D: This would be for half an hour at 1 kW.

Answer: D

Correct answer: 2 V: 0.05 A; 4 V: 0.10 A; 8 V: 0.20 A. Divide each voltage by its current. Only one set gives the same ratio every time:

A constant V/I ratio means a straight line through the origin, which defines an ohmic part.

• A: The V/I ratio climbs from 40 to 50 to about 67 ohms, the way a heating lamp filament behaves.
• B: The V/I ratio falls from 40 to about 27 to 20 ohms, so the resistance is not constant.
• C: The current stays fixed while voltage changes, so V/I goes from 20 to 80 ohms and is not constant.
• D (correct): Every reading gives V/I = 40 ohms, so the resistance is constant and the part is ohmic.

Answer: A

Correct answer: Both provide overcurrent interruption; many circuit breakers are resettable. Both devices sit in series with a circuit and open it when current exceeds a safe level, protecting the wiring from overheating. A fuse does this by melting a thin metal link, so it must be replaced with one of the same rating. A breaker uses a thermal or magnetic trip mechanism that opens contacts, which can then be closed again with the handle. Neither one is a meter or a source; they are protective switches.

• A (correct): The fuse is generally sacrificial while many breakers can be reset after correcting the issue.
• B: Protection devices interrupt faults rather than primarily display measurements.
• C: Those describe neither ordinary device.
• D: Neither creates its own source voltage.

Answer: C

Correct answer: The 5 kΩ resistor. With a common current, each drop

is proportional to resistance. The 5 kΩ resistor takes 5/10, or half, of the source voltage, more than either of the others.

• A: Its position in the loop does not matter; only its resistance relative to the others does.
• B: Equal current does not mean equal drops; drops are proportional to resistance.
• C (correct): Every resistor carries the same current, so the largest resistance has the largest IR drop.
• D: This reverses the rule; the smallest resistor takes the smallest share of the voltage.

Answer: C

Correct answer: The meter supplies its own test current, so outside voltage gives false readings and can damage it. An ohmmeter sends a small known current from its internal battery through the part and calculates resistance from the resulting voltage. Voltage from a live circuit adds to that and produces false readings, and it can damage the meter.

• A: An ohmmeter connects across the isolated part; series connection describes how an ammeter is used.
• B: A resistor still has resistance while operating; the problem is interference with the meter, not a zero value.
• C (correct): An ohmmeter computes resistance from its own small internal current, so any circuit voltage spoils the result and may harm the meter.
• D: Switching off does not change the resistor's rated value; it protects the meter and the reading.

Answer: D

Correct answer: Part of the voltage drops across the battery's own internal resistance. A real battery behaves like an ideal source in series with a small internal resistance. Under load, current flows through that resistance and drops some voltage inside the battery, so the terminal voltage falls below the no-load value.

• A: Current is the same all around a series loop; it is not used up by loads.
• B: The reading went down, and a heavier load lowers terminal voltage rather than raising it.
• C: An ideal voltmeter draws almost no current, so it does not explain the drop.
• D (correct): A real battery acts like an ideal source in series with a small internal resistance that drops voltage when current flows.

Answer: A

Correct answer: 20 W. Power can be found in two steps,

or in one step with P = V² ÷ R, which gives

Both routes must agree, which makes a useful self-check. A resistor on steady DC converts electrical energy to heat continuously. The tempting slips are reporting the current, 2, as the power, or multiplying voltage by resistance, which is not a valid power formula.

• A (correct): Current is 10/5 = 2 A, so P = 10 × 2 = 20 W.
• B: A DC resistor can dissipate real power continuously.
• C: This multiplies voltage by resistance.
• D: Two amperes is the current, not watts.

Answer: A

Correct answer: They have the same resistivity, but the long, thin wire has more resistance. Resistivity is a property of the material, so both copper wires share it. Resistance belongs to the object and rises with length and falls with area, so the long, thin wire has more resistance.

• A (correct): Resistivity belongs to the material, while resistance also depends on length and area.
• B: Being the same material gives the same resistivity, not the same resistance; size and shape still matter.
• C: Resistivity does not change with an object's size or shape, only resistance does.
• D: More copper in the cross section gives more paths for current, which lowers resistance.

Answer: D

Correct answer: 18 C. Charge equals current times time, but the units must match first: amperes are coulombs per second, so time must be in seconds and current in amperes. Convert 150 mA to 0.150 A and 2 minutes to 120 s, then multiply to get 18 C. The most common errors are leaving time in minutes or current in milliamperes, which gives answers off by factors of 60 or 1,000.

• A: 0.150 A × 2 s ignores conversion of minutes to seconds.
• B: This treats 150 mA as 150 A or mishandles time.
• C: This is an unrelated numerical ratio.
• D (correct): Q = 0.150 A × 120 s = 18 C.

Answer: A

Correct answer: The result is inconclusive because 2 V is below the LED's forward drop. Blue and white LEDs drop about 3 V or more. A meter that supplies only about 2 V cannot forward bias the LED, so it shows open both ways even if the LED is good. A test with a higher voltage, such as a supply and a series resistor, is needed.

• A (correct): Blue LEDs drop about 3 V or more, so a 2 V test cannot forward bias the LED, and OL both ways is expected even from a good part.
• B: The meter showed OL in both directions, so lead orientation does not explain the result.
• C: A blue LED needs about 3 V or more to conduct, so an open reading from a 2 V test is not proof of failure.
• D: A shorted part would read near zero, not OL, in both directions.

Answer: C

Correct answer: 0.15 A. Work from the secondary back to the primary.

• A: This divides the 3 A by the turns ratio twice; the current ratio uses the turns ratio only once.
• B: This multiplies the 3 A secondary current by the 20:1 ratio, inverting the current relationship.
• C (correct): Secondary voltage is 12 V, so 3 A flows in the load; 36 W divided by 240 V gives 0.15 A.
• D: This is the secondary current; the primary is on the high-voltage side and carries less current.

Answer: B

Correct answer: 4 µA. Capital M means mega, 10⁶, so 1.5 MΩ = 1,500,000 Ω.

• A: This ignores the prefix and divides 6 V by just 1.5 Ω.
• B (correct): 1.5 MΩ is 1,500,000 Ω, and 6 ÷ 1,500,000 = 0.000004 A = 4 µA.
• C: This treats capital M as kilo; 4 mA would flow through 1.5 kΩ.
• D: This treats the prefix as 10⁹ instead of mega, 10⁶.