Electronics Information Practice Test 5 (ASVAB)

Answer: C

Correct answer: The positive terminal. In a cell symbol the longer line is the positive terminal and the shorter line is the negative terminal. A handy memory aid is to picture the long line as a plus sign with a tall vertical stroke.

• A: Ground has its own symbols, such as stacked lines of shrinking length; a cell plate is a source terminal.
• B: Internal resistance is not drawn as a plate; the two lines only mark the two terminals of the cell.
• C (correct): The longer plate of a cell symbol marks the positive side, where conventional current leaves the source.
• D: This reverses the convention; the short, often thicker plate is the negative side.

Answer: D

Correct answer: Electric potential difference. One volt means one joule of energy for each coulomb of charge moved between two points, so voltage describes the electrical push behind current, not the current itself. Current, the charge passing per second, is measured in amperes, and resistance is measured in ohms. Remember that voltage is always measured between two points, which is why a voltmeter has two leads placed across a component.

• A: Resistance in a conductor is measured in ohms, not volts.
• B: Charge passing per second is current, measured in amperes.
• C: Current in a circuit is measured in amperes, not volts.
• D (correct): Voltage represents energy difference per unit charge.

Answer: D

Correct answer: Gold resists corrosion, so the contact surface stays clean. Gold plating is used on connectors because it resists corrosion. Silver is actually the best ordinary metal conductor, and copper is a close second.

• A: Silver, not gold, is the best ordinary metal conductor, with copper close behind.
• B: Gold is a metal conductor; a contact plating must conduct, not insulate.
• C: Gold costs far more than copper; low cost is not the reason it is used.
• D (correct): Gold is chosen mainly because it does not corrode, which keeps contact resistance low over time.

Answer: A

Correct answer: 4 A. Rearrange the power formula to find current.

• A (correct): Dividing 480 W by 120 V with $I=P/V$ gives 4 A.
• B: This divides voltage by power, which flips the formula upside down.
• C: This is a decimal slip; 480 divided by 120 is 4, not 40.
• D: This subtracts the voltage from the power, which mixes two different units.

Answer: B

Correct answer: 5 Ω. Ohm's law,

rearranges to

when you need resistance. Dividing 15 V by 3 A gives 5 Ω, meaning 5 volts are needed for each ampere of current. The common trap is multiplying voltage by current, which gives power in watts rather than resistance. A useful memory aid is the triangle with V on top and I and R below; cover the unknown to see the formula.

• A: $\frac{3}{15}$ reverses the ratio and gives conductance in siemens.
• B (correct): R = $\frac{V}{I}$ = $\frac{15}{3}$ = 5 Ω under these specified conditions.
• C: Multiplying V by I gives watts, not the requested R in ohms.
• D: Adding volt and ampere numerically is not an electrical resistance formula.

Answer: A

Correct answer: Humps of one polarity during half of each cycle, with zero between them. This is a half-wave rectifier. During the half-cycle that forward biases the diode, the load sees a positive hump; during the other half, the diode is reverse biased and the load voltage is zero.

• A (correct): The diode conducts only during the half-cycle when its anode is positive and blocks during the other half.
• B: The diode blocks one polarity, so the load does not see a full sine wave of any size.
• C: A rectifier alone gives pulses; a filter and regulator are needed for smooth DC.
• D: Gap-free humps come from a full-wave bridge, which uses both half-cycles; a single diode uses only one.

Answer: D

Correct answer: 400 Ω. Convert first: 15 mA = 0.015 A. Then use

• A: This divides the current by the voltage, which is upside down.
• B: This divides 6 by 15 without converting milliamperes to amperes.
• C: This multiplies voltage by current instead of dividing.
• D (correct): Converting 15 mA to 0.015 A and dividing gives 6 ÷ 0.015 = 400 ohms.

Answer: C

Correct answer: The current falls and the lamp gets less voltage. The bad connection is an extra resistance in series. Total resistance increases, so the loop current falls, and the clip drops some voltage itself, leaving less for the lamp. The bad connection may also get hot.

• A: The current is equal at every point, but its value drops when total resistance increases.
• B: The clip takes its own drop, so the lamp's share of the voltage shrinks.
• C (correct): The extra series resistance raises the total, lowering the current and taking part of the voltage away from the lamp.
• D: Adding resistance in series lowers current; it cannot raise it.

Answer: D

Correct answer: NAND. An AND gate gives 1 only when both inputs are 1. Inverting that output gives a gate that gives 0 only when both inputs are 1, which is the NAND gate.

• A: An OR gate outputs 0 only when both inputs are 0.
• B: An AND gate outputs 1 only when both inputs are 1, which is the opposite of this behavior.
• C: A NOR gate outputs 1 only when both inputs are 0.
• D (correct): A NAND gate is an AND followed by an inverter, so it outputs 0 only when both inputs are 1.

Answer: B

Correct answer: 20 V. Find the loop current from the known drop:

The total resistance is

so the source is

Check: the 30 Ω drops 12 V, and

• A: This adds 8 volts to 30 ohms, mixing two different quantities.
• B (correct): The current is 8 ÷ 20 = 0.4 A, and 0.4 A × 50 Ω = 20 V.
• C: This assumes the source equals one resistor's drop, ignoring the other resistor.
• D: This is only the drop across the 30 Ω resistor; the source supplies both drops.

Answer: A

Correct answer: The lock physically holds the switch off, while a tag can only warn. A tag is a warning sign. A lockout device physically holds the energy-isolating device in the off position, usually with a padlock the worker controls, so no one can re-energize the equipment while it is being serviced. A full program also isolates every source, releases stored energy and verifies the result.

• A (correct): A tag communicates but cannot stop anyone from restoring power; a lock physically prevents it.
• B: Releasing stored energy is a separate step; a padlock on a switch does nothing to capacitors.
• C: Verification is a separate step done by qualified people with rated test equipment.
• D: A lock changes nothing electrically; it only keeps the isolating device in the off position.

Answer: A

Correct answer: 200 W. For fixed delivered power,

so doubling V halves I. Line loss is

so it becomes

as large: 800

• A (correct): Doubling voltage halves the current, and since loss depends on current squared, the loss falls to one fourth: 800 ÷ 4 = 200 W.
• B: This applies the square factor in the wrong direction; the loss shrinks fourfold rather than growing.
• C: This cuts the loss in proportion to current, but loss goes with the square of current.
• D: Higher voltage lowers line current, so the loss must fall, not rise.

Answer: A

Correct answer: 480 W. Branch currents add in the shared wiring. The microwave draws

A and the lighting draws

for 11 A. The breaker allows 15 A, so 4 A remains, and 120 V times 4 A is 480 W.

• A (correct): The existing loads draw 8 A plus 3 A, or 11 A, leaving 4 A, and 4 A at 120 V is 480 W.
• B: Subtracting only the microwave's 8 A forgets that the lighting also draws current through the breaker.
• C: 1,320 W is the power the existing loads already use, not the remaining room.
• D: 1,800 W is the whole capacity of the circuit; the existing loads already use part of it.

Answer: D

Correct answer: 28.3 V. RMS is the equivalent heating value of a sine wave, and for a sine the peak is √2 (about 1.414) times the RMS value. So 10 V RMS reaches about 14.14 V at its positive peak and −14.14 V at its negative peak. Peak-to-peak spans both extremes, so it equals 2 × 1.414 × RMS, or about 2.83 × RMS. A common slip is stopping at 14.1 V, which is only the peak, or dividing by √2 when you should multiply.

• A: Twice RMS is not the correct sine peak-to-peak conversion.
• B: That divides RMS by √2 rather than converting it to peak.
• C: That is the positive peak only.
• D (correct): Peak is 10√2 ≈ 14.14 V, and peak-to-peak is twice that.

Answer: A

Correct answer: 3 A. Coil X:

ampere-turns. Twice that is 240 ampere-turns.

• A (correct): Coil X gives 120 ampere-turns, so twice that is 240 ampere-turns, and 240 divided by 80 turns is 3 A.
• B: This doubles the target twice, aiming for four times coil X's strength.
• C: This scales the current down with the turns instead of up.
• D: This only matches coil X's strength and forgets to double it.