ASVAB Electronics Information Practice Test 6

Answer: C

Correct answer: A motor. Ask what goes in and what comes out. Electrical energy from the battery goes in and a turning fan comes out, so the device converts electrical energy to mechanical energy, which is the job of a motor.

• A: A generator does the reverse job, taking in motion and putting out electricity.
• B: A transformer only changes AC voltage levels and has no moving parts.
• C (correct): It takes in electrical energy from the battery and delivers mechanical energy as a spinning fan.
• D: A rectifier converts AC to DC with diodes; it does not produce motion.

Answer: D

Correct answer: Emitter, base, and collector. A BJT is built from three layers of doped silicon. Its terminals are the emitter, the base (the control terminal), and the collector, which usually connects toward the load.

• A: This mixes a MOSFET terminal name into the BJT list; the control terminal of a BJT is the base.
• B: Anode and cathode name the two diode terminals; a BJT uses different names.
• C: These are the terminals of a field-effect transistor such as a MOSFET.
• D (correct): A BJT has three doped layers whose terminals are named emitter, base, and collector.

Answer: A

Correct answer: kilowatt-hours. Utilities charge for energy, which is power multiplied by the time it is used. One kilowatt-hour is the energy of a 1,000 W load running for one hour, or a 100 W load running for ten hours. The most tempting wrong choice is kilowatts, but that is a rate, like speed, while kilowatt-hours are the total, like distance traveled. A meter tallies kilowatt-hours over the month, and the bill multiplies that total by a price per kWh.

• A (correct): kWh measures energy accumulated over time.
• B: Ampere-hours measure electric charge, as on battery ratings, not billed energy.
• C: V measures electrical potential difference.
• D: Kilovolt-amperes rate apparent power for equipment such as transformers, not energy used over time.

Answer: C

Correct answer: It stays dark. Induction requires a change in magnetic flux. With the magnet held still, the flux through the coil does not change, so no emf is induced and the lamp stays dark.

• A: This treats the magnet as a battery that runs down; a still magnet produces no current at all.
• B: Flickering would need a changing flux, but nothing in this setup is changing.
• C (correct): The flux through the coil is large but steady, and a steady flux induces no emf.
• D: A steady glow would need a continuous source of energy; a still magnet does not supply one.

Answer: A

Correct answer: 36 Ω. In series, the ratio of drops equals the ratio of resistances.

• A (correct): The current is 3 ÷ 12 = 0.25 A, and 9 V ÷ 0.25 A = 36 Ω.
• B: This inverts the ratio; the resistor with the larger drop must have the larger resistance.
• C: This multiplies 12 by 9 without dividing by the 3 V drop of the known resistor.
• D: This multiplies the drop by the current instead of dividing the drop by the current.

Answer: C

Correct answer: A large filter capacitor. Each power-supply stage has its own job. The rectifier makes one-direction pulses, a filter (usually a large capacitor) smooths them with some ripple left over, and a regulator then holds the output at a set voltage.

• A: A photodiode senses light and has no role in smoothing a supply.
• B: A switching transistor turns a load on and off; it does not store charge to fill in the gaps between pulses.
• C (correct): A filter capacitor smooths the rectifier's pulses into a fairly steady voltage with some remaining ripple.
• D: The bridge is the rectifier itself; it creates the pulses rather than smoothing them.

Answer: A

Correct answer: Mobile charges generally move more readily through metal. Metals have free electrons that drift easily, while rubber binds its electrons tightly to its molecules, so very few can move. The difference in conductivity is enormous, which is why rubber and plastic coat wires and tools. A common misconception is that conductors cannot hold static charge; an insulated metal object can, but the charge spreads quickly over its surface.

• A (correct): Mobile charges travel more readily in most metals.
• B: Metal objects can hold net charge under some conditions.
• C: Their conductivities differ enormously under normal conditions.
• D: Rubber is not an electrical source merely because it is an insulator.

Answer: A

Correct answer: The designed overcurrent interruption is defeated. A fuse is meant to be the weakest link, melting before the wiring overheats. A plain wire has a much higher melting current, so in an overload or short the wiring insulation can char and start a fire before anything opens. The circuit may seem to work normally, which is why the practice is tempting and dangerous. A blown fuse should be replaced only with one of the same type and rating, after finding why it blew.

• A (correct): A wire bridge can permit fault current beyond the intended protective limit.
• B: The bridge creates a path and may permit dangerous current.
• C: A wire bridge does not detect ground leakage.
• D: Bypassing a fuse does not regulate voltage.

Answer: A

Correct answer: 2,640 µC. Charge equals capacitance times voltage.

• A (correct): Q = CV = 220 µF × 12 V = 2,640 µC.
• B: This misplaces the decimal; 220 × 12 is 2,640, not 26.4.
• C: This divides capacitance by voltage instead of multiplying them.
• D: This adds the two numbers instead of multiplying them.

Answer: A

Correct answer: 30 Ω. On a current-versus-voltage graph the slope is $1/R$, so take voltage over current at the point.

• A (correct): Resistance is voltage divided by current: 12 V ÷ 0.4 A = 30 ohms.
• B: This drops a decimal place in the current, dividing 12 by 4 instead of 0.4.
• C: This is the slope, 0.4/12, which on this graph is 1/R in siemens, not the resistance.
• D: This multiplies the voltage by the current instead of dividing.

Answer: A

Correct answer: The meter can show an equivalent resistance lower than its value. An ohmmeter sends a small test current and measures whatever path that current can take. With another resistor still connected in parallel, the test current splits between both, and parallel resistance is always less than the smallest branch. Turning the power off removes the hazard but not the extra path. To read a component's true value, disconnect at least one of its leads from the circuit first.

• A (correct): A parallel path can reduce the effective resistance seen by the ohmmeter; proper safe isolation matters.
• B: Removing power does not open the parallel path, so the meter still sees a finite combined resistance.
• C: An ohmmeter reports resistance, not voltage, so a 0 V reading does not answer the question.
• D: An in-circuit meter also measures the parallel path, so the reading can differ from the label.

Answer: A

Correct answer: It falls below 6 V. A voltage divider assumes no current is drawn from its midpoint. A load connected across the lower resistor is in parallel with it, so the lower section's effective resistance falls, its fraction of the total falls, and the output drops below the unloaded 6 V.

• A (correct): The motor sits in parallel with the lower resistor, lowering that section's effective resistance and its share of the voltage.
• B: The divider formula assumes nothing draws current from the midpoint, which is no longer true.
• C: The motor is across only the lower resistor, and the upper resistor still limits the current.
• D: A load across one resistor is in parallel with it, which lowers the resistance there rather than adding to it.

Answer: C

Correct answer: 30 Hz. Convert speed to rotations per second:

Since the two-pole machine gives 60 Hz at 60 rotations per second, it makes one cycle per rotation, so at 30 rotations per second it produces 30 Hz.

• A: A four-pole machine gives 60 Hz at 1,800 rpm, but this generator has only two poles.
• B: Slowing the generator lowers its frequency; it cannot double it.
• C (correct): For a two-pole machine, each rotation makes one cycle, and 1,800 rpm ÷ 60 = 30 rotations per second.
• D: This treats rotations per minute as cycles per second without dividing by 60.

Answer: A

Correct answer: Total current rises while the current in each original resistor stays the same. Each new branch adds a current path between the same two nodes. The original branches still have the full source voltage, so they carry exactly what they did, and the new branch's current is added on top. More current at the same voltage means a lower equivalent resistance.

• A (correct): The original branches keep the same voltage and current, and the new branch adds its own current to the total.
• B: An ideal source holds its voltage, so adding a branch does not lower the voltage on the others.
• C: With an ideal source, the original branches keep the full voltage, so their currents do not change.
• D: Adding a parallel path always lowers the equivalent resistance.

Answer: D

Correct answer: 6.25 times. With R fixed, heat is proportional to $I^2$.

So the wire produces 6.25 times as much heat.

• A: This subtracts the currents, but heat depends on the ratio of currents squared, not the difference.
• B: This squares only the new current and forgets to compare it with the old current squared, which is 16.
• C: This uses the current ratio directly; heating grows with the square of the current.
• D (correct): Heat follows I squared R, so the factor is (10/4) squared = 2.5 squared = 6.25.