Electronics Information Practice Test 3 for the ASVAB

Answer: B

Correct answer: 1 C each second. The ampere is defined as a rate of charge flow, so 1 A means one coulomb of charge passes a point every second. The tempting wrong choice is joules per second, but that describes power, measured in watts. Keeping the base units straight helps: charge in coulombs, current in coulombs per second, voltage in joules per coulomb, and power in joules per second.

• A: $\frac{\mathrm{J}}{\mathrm{s}}$ is a watt, a unit of power.
• B (correct): By definition 1 A = 1 $\frac{C}{s}$.
• C: $\frac{\mathrm{V}}{\mathrm{s}}$ describes a rate of voltage change, not current.
• D: Ω/s is a rate of resistance change.

Answer: B

Correct answer: Light-emitting diode. Matching the function to the device: light output comes from an LED. When forward biased, electrons drop into holes at its junction and the released energy leaves as light.

• A: An IC combines many parts on one chip but is not the device that produces indicator light.
• B (correct): An LED is an output device that turns electrical energy into light when forward biased.
• C: A Zener diode holds a steady reference voltage in reverse breakdown; it does not give off useful light.
• D: A photodiode senses light that falls on it; it is an input device, not a light source.

Answer: C

Correct answer: Across the resistor, in parallel with it. A voltmeter measures the potential difference between two points, so its leads touch both ends of the resistor, side by side with it. Its very high resistance means it draws almost no current.

• A: The meter must touch both ends of the resistor to measure the difference between them.
• B: That is how an ammeter is connected; a high-resistance voltmeter in line would nearly stop the current.
• C (correct): A voltmeter compares the two ends of the part, so its leads go across it without breaking the circuit.
• D: That only gives the right value for identical resistors and does not measure the drop directly.

Answer: B

Correct answer: 72 W. Use

because voltage and current are given.

• A: This subtracts the current from the voltage, which has no physical meaning.
• B (correct): Multiplying 24 V by 3 A with $P=VI$ gives 72 W.
• C: This adds the voltage and current instead of multiplying them.
• D: This comes from dividing voltage by current, which gives resistance in ohms, not power.

Answer: B

Correct answer: Ground fault. The word fault signals that current is going somewhere it was never meant to, and ground tells you where: to a grounded part such as a metal enclosure or a person. Normal current leaves on the hot and returns on the neutral. In a ground fault, some or all of it returns through the grounding conductor or earth instead. A ground fault is the condition a GFCI is designed to detect, and the grounding conductor helps the breaker clear it.

• A: An accidental enclosure connection is a fault, not designed consumption.
• B (correct): Current travels outside the normal intended circuit to a grounded part.
• C: The metal enclosure is not an intended normal load-current return path.
• D: The hazard is an unintended conducting path, not an interrupted path.

Answer: B

Correct answer: Reluctance. Reluctance opposes flux the way resistance opposes current. Iron has low reluctance and air has high reluctance, so adding an air gap raises the reluctance of the path.

• A: Hysteresis is the lag of magnetization behind the driving field, not the opposition of the path.
• B (correct): Reluctance is the opposition a magnetic path offers to flux, and an air gap raises it sharply.
• C: Ampere-turns depend on the coil's turns and current, which the gap does not change.
• D: Air has far lower permeability than iron, so the gap lowers the path's ability to carry flux.

Answer: A

Correct answer: 960 W. For a fixed resistance,

so power scales with the square of the voltage ratio.

• A (correct): 96 V is 0.8 of 120 V, and 0.8 squared is 0.64, so the power is 1,500 times 0.64 = 960 W.
• B: This reduces power in proportion to voltage, but power depends on voltage squared.
• C: This is how much the power drops, not the power the heater still produces.
• D: The heater's resistance is fixed, so its power must fall when the voltage falls.

Answer: C

Correct answer: 6 × 10⁻⁴ N. The charge moves straight across the field, so

• A: This slips one power of ten in the multiplication.
• B: Only a charge at rest, or one moving along the field, feels no magnetic force; this one moves across it.
• C (correct): $F=qvB=3\times 10^{-6}\times 400\times 0.5=6\times 10^{-4}$ N.
• D: This divides by B instead of multiplying by it.

Answer: C

Correct answer: 500 µs. Convert 2 kHz to 2,000 Hz, then take the reciprocal.

• A: 0.0005 s is half a millisecond, not 500 milliseconds.
• B: This misplaces the decimal; 50 µs would be the period of a 20 kHz tone.
• C (correct): 2 kHz is 2,000 Hz, and 1/2,000 = 0.0005 s, which is 500 microseconds.
• D: This uses the frequency number as the period; 2 ms would belong to a 500 Hz tone.

Answer: A

Correct answer: It usually has high resistivity, but conditions and breakdown matter. Insulators such as rubber, glass and plastic hold their electrons tightly, so very little charge can move. That resistance is not unlimited: enough voltage can tear electrons loose and cause breakdown, as when lightning crosses air, and heat, moisture or damage can let current leak. The tempting idea that insulators contain no electrons is wrong; every material has electrons, and what differs is how freely they move.

• A (correct): A good insulator restricts charge transport but is not universally perfect.
• B: Insulators contain electrons; the key distinction is their limited mobility.
• C: Real insulating materials can leak current or break down under adverse conditions.
• D: Insulation alone does not guarantee safety around hazardous energized equipment.

Answer: A

Correct answer: 5 mA. Volts divided by kilohms gives milliamperes. The 4 kΩ branch carries

mA and the 6 kΩ branch carries

mA, so the total is 5 mA, or 0.005 A.

• A (correct): The branches carry 3 mA and 2 mA, which add to 5 mA.
• B: Treating kilohms as ohms makes the current a thousand times too large.
• C: 2.4 is the equivalent resistance in kilohms, not a current.
• D: 12 V divided by 10 kΩ treats the two resistors as if they were in series.

Answer: D

Correct answer: 12 V, the full battery voltage. A blown fuse acts like an open switch, so the loop current drops to zero. With no current, the lamp has no voltage across it, and the battery's full 12 V appears across the gap in the fuse.

• A: Splitting the voltage evenly assumes current still flows through two equal loads, which is not the case.
• B: Zero current does not mean zero voltage; the source voltage still exists and appears across the gap.
• C: An ideal source gives a steady voltage; the reading across the gap settles at the battery voltage.
• D (correct): With no current, the lamp drops no voltage, so the whole 12 V appears across the open fuse.

Answer: D

Correct answer: 10 times. When the shaft locks, back emf disappears and only the winding resistance limits current.

Comparing with the 3 A running current gives

times. (While running, back emf is

)

• A: This divides the supply voltage by the running current, which is not a current ratio.
• B: This compares the supply with the 54 V back emf; it is a voltage ratio, not the current ratio.
• C: Thirty is the stall current in amperes, not how many times it exceeds the running current.
• D (correct): Stall current is 60/2 = 30 A, and 30 A divided by the 3 A running current is 10.

Answer: A

Correct answer: 1 A. Two equal 12 ohm resistors in parallel act like 6 ohms, so the total is

and the source current is

The parallel pair drops 2 A × 6 ohms = 12 V, so each 12 ohm branch carries

• A (correct): The pair acts as 6 ohms, total is 11 ohms, the source current is 2 A, and each equal branch carries half.
• B: This treats all three resistors as being in series (29 ohms total).
• C: This places the full 22 V across one 12 ohm resistor, ignoring the drop across the 5 ohm resistor.
• D: This is the total source current; it splits equally between the two parallel branches.

Answer: C

Correct answer: They dissipate equal average power in the resistor. RMS voltage is defined precisely so that an AC value can be compared with DC: it is the steady DC voltage that would produce the same average heating in a resistor. A 12 V RMS sine therefore heats the resistor exactly as much as 12 V DC. The tempting error is to think AC averages to zero and so delivers no power, but power depends on voltage squared, which is never negative, so both halves of the cycle heat the resistor.

• A: DC can drive steady current and heating.
• B: Resistive power follows mean square voltage, not mean signed voltage.
• C (correct): A 12 V RMS AC source has the same resistive-heating effect as 12 V steady DC.
• D: The RMS comparison already accounts for varying AC voltage.