ASVAB Electronics Information Test 4

Answer: A

Correct answer: The opposition a material or component presents to electric current. Resistance is the opposition to current flow, measured in ohms. Voltage is the push and current is the flow; resistance is a property of the part itself.

• A (correct): Resistance is the property of a path that opposes the flow of charge, measured in ohms.
• B: Storing charge is not what resistance does; a resistor converts energy to heat rather than storing it.
• C: That describes current, measured in amperes, which flows through a part.
• D: That describes voltage, which is applied across a part, not a property of the part.

Answer: D

Correct answer: A solar panel. Solar panels and batteries have a fixed positive and negative terminal, so current flows one way and they supply DC. Outlets, utility generators and inverter outputs supply AC.

• A: Wall outlets in U.S. homes supply AC at 60 Hz.
• B: An inverter's job is to turn DC into AC, so its output is AC.
• C: A rotating coil generator naturally produces AC for the grid.
• D (correct): Light frees charges that are driven one way, so a panel has fixed positive and negative leads and supplies DC.

Answer: C

Correct answer: N-type silicon with free electrons as majority carriers. Silicon has four valence electrons. A dopant with five supplies one extra, loosely bound electron, so it acts as a donor and creates n-type material in which electrons are the majority carriers.

• A: Doping raises conductivity by adding carriers; it does not turn silicon into an insulator.
• B: This is what a three-electron acceptor such as boron produces, not a five-electron donor.
• C (correct): Antimony donates a loosely held fifth electron, so electrons become the main mobile carriers and the material is n-type.
• D: The type is right, but in n-type material holes are the minority carriers, not the majority.

Answer: A

Correct answer: 0 V. A closed ideal switch has zero resistance. With current I flowing, its drop is

and the entire 12 V is shared by the two resistors.

• A (correct): An ideal closed switch acts like a plain wire with zero resistance, so it has no voltage drop.
• B: This splits the source among three parts, but a closed switch adds no resistance and takes no share.
• C: The full source voltage appears across a switch only when it is open.
• D: This splits the source between the switch and the resistors as if the switch had resistance.

Answer: A

Correct answer: That is an unsafe and inappropriate resistance measurement. An ohmmeter measures resistance by pushing a small current from its own battery and reading the result, so it expects a dead circuit. Connected to a live outlet, the outlet's 120 V overwhelms that tiny test source, producing meaningless readings and possibly damaging the meter or causing an arc. To check a live outlet, a technician switches to the voltage function and uses rated leads; to measure resistance, the power must be off first.

• A (correct): An ohmmeter uses its own test source and is not connected to energized circuits for resistance testing.
• B: A screen indication alone cannot validate safety or the selected mode.
• C: Internal batteries do not remove an external circuit hazard.
• D: Voltage measurement is a different function and requires qualified practices.

Answer: D

Correct answer: 2 A. Reduce in stages. Two equal 12 Ω resistors in parallel act as 6 Ω. Add the 3 Ω series resistor for 9 Ω total, so the source current is 18 V divided by 9 Ω, or 2 A.

• A: Adding all three resistors to get 27 Ω treats the parallel pair as if it were in series.
• B: 1 A is the current in each 12 Ω branch, which is half of the source current.
• C: Dividing 18 V by only the 3 Ω series resistor ignores the parallel block.
• D (correct): The parallel pair is 6 Ω, the total is 9 Ω, and 18 V divided by 9 Ω is 2 A.

Answer: A

Correct answer: 10 V. Apply Faraday's law for the average emf.

• A (correct): Average emf is turns times flux change over time: 50 times 0.02/0.1 equals 10 V.
• B: This multiplies turns by flux change but leaves out the division by time.
• C: This multiplies by the time instead of dividing by it.
• D: This is the rate of flux change for one turn; it still must be multiplied by the 50 turns.

Answer: A

Correct answer: 2 A. Average current is the total charge moved divided by the time taken,

Four coulombs delivered over 2 s means 2 coulombs per second on average, which is 2 A. Writing units makes the operation clear: coulombs over seconds gives amperes. Multiplying charge and time or dividing time by charge produces units that are not current, so those results can be ruled out.

• A (correct): I = $\frac{Q}{t}$ = $\frac{4}{2}$ = 2 A. Additional check: A current measured in A must equal transferred coulombs divided by seconds, not their product.
• B: Adding values lacks a unit basis.
• C: Multiplication gives the wrong operation.
• D: $\frac{t}{Q}$ inverts the relation.

Answer: D

Correct answer: 36 V. Emf is proportional to $\Delta \Phi /\Delta t$. Cutting the time from 0.3 s to 0.1 s makes the rate three times larger, so the emf becomes

• A: This divides by three, as if a faster change made a smaller emf.
• B: This doubles the emf, but the time was cut to one third, not one half.
• C: This ignores the rate; Faraday's law depends on how fast the flux changes.
• D (correct): The change happens three times faster, so the emf is three times larger.

Answer: D

Correct answer: 108 V. The supply equals back emf plus the drop across the winding:

• A: This adds the winding drop to the supply; the back emf must be the supply minus the drop.
• B: This divides the supply by the resistance and treats the answer as a voltage, which mixes up the relationships.
• C: Twelve volts is the drop across the winding resistance, not the back emf.
• D (correct): The winding drop is 6 times 2 = 12 V, so the back emf is 120 - 12 = 108 V.

Answer: B

Correct answer: 0.5 A. Divide voltage by current in each row: 3/0.1, 6/0.2 and 9/0.3 all equal 30 ohms, so the resistance is constant.

• A: This multiplies 15 V by the first current reading instead of dividing by the resistance.
• B (correct): Every reading gives V/I = 30 ohms, so the part is ohmic, and 15 V divided by 30 ohms is 0.5 A.
• C: This slips a decimal place; 15 divided by 30 is 0.5, not 0.05.
• D: This is the next step in the table, at 12 V; 15 V is two steps beyond 9 V, not one.

Answer: D

Correct answer: 18 W. For a fixed resistor,

so power scales with the square of current.

• A: This cubes the factor of 3, but current is squared in the power formula.
• B: This assumes a fixed resistance fixes the power, but power grows with current squared.
• C: This scales power directly with current, which holds only at fixed voltage, not for a fixed resistor.
• D (correct): With $P=I^{2}R$, tripling the current multiplies power by 9, and 2 W times 9 is 18 W.

Answer: C

Correct answer: 9 V. Total resistance is

kΩ, so the current is 12 V ÷ 6 kΩ = 2 mA. The voltmeter across the 4.5 kΩ resistor reads 2 mA × 4.5 kΩ = 9 V.

• A: This is the voltage across the 1.5 kΩ resistor, not the 4.5 kΩ one.
• B: This splits the voltage equally, which would happen only if the resistors were equal.
• C (correct): The current is 12 V ÷ 6 kΩ = 2 mA, and 2 mA × 4.5 kΩ = 9 V.
• D: This is the whole battery voltage; the voltmeter is across only one of two series parts.

Answer: C

Correct answer: The magnet is repelled, because the top face of the loop acts as a north pole. Curl the right-hand fingers counterclockwise, the way the current flows as seen from above; the thumb points up, so the top face of the loop is a north pole. Like poles repel, so the magnet's north end is pushed away.

• A: A current loop has poles like any magnet; whether it attracts or repels depends on which pole faces it.
• B: A current loop makes its own magnetic field and acts like a short magnet even without iron.
• C (correct): Curling right-hand fingers counterclockwise puts the thumb upward, so the top face is north, and north repels north.
• D: This reverses the coil rule; counterclockwise current seen from above makes the top face north.

Answer: A

Correct answer: 0.6 W. Convert first: 20 mA = 0.02 A and 1.5 kilohms = 1,500 ohms.

Check:

and 30 V × 0.02 A = 0.6 W.

• A (correct): Converting to 0.02 A and 1,500 ohms gives P = 0.0004 × 1,500 = 0.6 W.
• B: This slips a decimal place after squaring the current.
• C: This is I × R, which gives the 30 V drop across the resistor, not the power.
• D: This squares 20 and multiplies by 1.5 without converting the milli and kilo prefixes.