ASVAB Electronics Information (EI) Practice Test

ASVAB Electronics Information (EI) Practice Test 2026

Q001. A circuit has 12 V across a 6 Ω resistor. What is the current through the resistor?

Correct Answer: B
Explanation: Ohm’s law: I = V/R = 12/6 = 2 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law (I = V/R).

Q002. Two resistors, 4 Ω and 8 Ω, are connected in series. What is the total resistance?

Correct Answer: B
Explanation: Series resistances add: 4 + 8 = 12 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: series resistance addition.

Q003. Two resistors, 6 Ω and 3 Ω, are connected in parallel. What is the equivalent resistance?

Correct Answer: A
Explanation: Parallel: 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel resistance (reciprocal sum).

Q004. A 9 V battery powers a device drawing 0.5 A. Approximately how much power is used?

Correct Answer: A
Explanation: Power P = VI = 9 × 0.5 = 4.5 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: electrical power (P = VI).

Q005. If voltage across a fixed resistor is doubled, the current through it will:

Correct Answer: B
Explanation: With constant resistance, I = V/R, so doubling V doubles I.
Citation: ASVAB content domain — Electronics Information; underlying principle: linear V–I relationship for resistors.

Q006. Which device is designed primarily to allow current to flow in only one direction?

Correct Answer: B
Explanation: A diode conducts forward and blocks reverse (ideally).
Citation: ASVAB content domain — Electronics Information; underlying principle: diode rectification.

Q007. In a basic LED circuit, which component is most commonly added in series to limit current?

Correct Answer: B
Explanation: A series resistor drops voltage and limits LED current.
Citation: ASVAB content domain — Electronics Information; underlying principle: current limiting with resistors.

Q008. A capacitor’s stored energy increases most directly when its:

Correct Answer: B
Explanation: Energy stored is U = ½CV², so increasing V raises energy strongly.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor energy (½CV²).

Q009. Which quantity is measured in coulombs (C)?

Correct Answer: B
Explanation: The coulomb is the SI unit of electric charge.
Citation: ASVAB content domain — Electronics Information; underlying principle: electric charge units.

Q010. A 10 µF capacitor is connected to a 5 V source. Approximately how much charge does it store?

Correct Answer: B
Explanation: Q = CV = 10 µF × 5 V = 50 µC.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor charge (Q = CV).

Q011. Which component opposes changes in current and stores energy in a magnetic field?

Correct Answer: C
Explanation: Inductors resist current change and store energy magnetically.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductance and magnetic energy storage.

Q012. AC voltage is stepped up most directly using a:

Correct Answer: B
Explanation: Transformers use mutual induction to change AC voltage levels.
Citation: ASVAB content domain — Electronics Information; underlying principle: transformer action (mutual induction).

Q013. A transformer has 200 primary turns and 50 secondary turns. If 120 V AC is applied to the primary, the secondary voltage is closest to:

Correct Answer: A
Explanation: V_s/V_p = N_s/N_p = 50/200 = 1/4, so V_s = 30 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: transformer turns ratio.

Q014. Which instrument is typically used to measure electrical resistance directly?

Correct Answer: C
Explanation: An ohmmeter measures resistance (often within a multimeter).
Citation: ASVAB content domain — Electronics Information; underlying principle: basic electrical measurement instruments.

Q015. In a series circuit, if one component opens (fails open), what happens to current in the circuit?

Correct Answer: C
Explanation: A series circuit needs a complete path; an open makes current zero.
Citation: ASVAB content domain — Electronics Information; underlying principle: series circuit continuity.

Q016. Which waveform is characterized by a single frequency with smooth periodic oscillation?

Correct Answer: C
Explanation: A sine wave is the fundamental smooth periodic waveform.
Citation: ASVAB content domain — Electronics Information; underlying principle: AC waveform basics.

Q017. Frequency is measured in:

Correct Answer: B
Explanation: Hertz (Hz) is cycles per second.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency units.

Q018. A 60 Hz AC signal completes how many cycles in 2 seconds?

Correct Answer: C
Explanation: Cycles = frequency × time = 60 × 2 = 120 cycles.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency-time relationship.

Q019. Which material is generally a good electrical conductor?

Correct Answer: C
Explanation: Metals like copper conduct well due to free electrons.
Citation: ASVAB content domain — Electronics Information; underlying principle: conductors vs insulators.

Q020. A fuse in a circuit is intended to protect equipment primarily by:

Correct Answer: B
Explanation: A fuse opens the circuit when current exceeds its rating.
Citation: ASVAB content domain — Electronics Information; underlying principle: overcurrent protection.

Q021. If a circuit draws 3 A from a 120 V source, how much electrical power is consumed?

Correct Answer: C
Explanation: P = VI = 120 × 3 = 360 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power calculation (P = VI).

Q022. Which term describes opposition to current flow in an AC circuit due to capacitors or inductors?

Correct Answer: B
Explanation: Reactance is the frequency-dependent opposition from capacitors/inductors.
Citation: ASVAB content domain — Electronics Information; underlying principle: AC reactance concept.

Q023. In a simple RC circuit, increasing capacitance (with R fixed) will generally:

Correct Answer: B
Explanation: Time constant τ = RC; increasing C increases τ.
Citation: ASVAB content domain — Electronics Information; underlying principle: RC time constant.

Q024. A silicon diode typically drops about how much voltage when forward-biased at moderate current?

Correct Answer: B
Explanation: A common approximation for silicon diode forward drop is about 0.7 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: diode forward voltage (silicon).

Q025. In a typical NPN transistor used as a switch, a small base current primarily allows a larger current to flow between:

Correct Answer: B
Explanation: In an NPN transistor, base drive controls a much larger collector-to-emitter current.
Citation: ASVAB content domain — Electronics Information; underlying principle: transistor current control (switching/amplification).

Q026. A 8 Ω resistor is connected across 24 V DC. What current flows and what power is dissipated (closest)?

Correct Answer: A
Explanation: First find current: I = V/R = 24/8 = 3.0 A. Then power: P = VI = 24×3.0 = 72 W (equivalently V²/R).
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law and power (I = V/R, P = VI).

Q027. Three resistors 5 Ω, 7 Ω, and 8 Ω are in series across 20 V. What is the circuit current (closest)?

Correct Answer: A
Explanation: Series resistance adds: R_total = 5+7+8 = 20 Ω. Current I = V/R_total = 20/20 ≈ 1.00 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: series resistance and Ohm’s law.

Q028. Resistors 6 Ω, 12 Ω, and 24 Ω are connected in parallel. The equivalent resistance is closest to:

Correct Answer: A
Explanation: For parallel resistors, 1/R_eq = 1/6 + 1/12 + 1/24 = 4/24 + 2/24 + 1/24 = 7/24, so R_eq = 24/7 ≈ 3.43 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel resistance (reciprocal sum).

Q029. A voltage divider has 3 kΩ on top and 6 kΩ on bottom across 18 V. What is the output voltage across the bottom resistor?

Correct Answer: A
Explanation: V_out = V_in × R_bottom/(R_top+R_bottom) = 18×6000/(3000+6000) = 18×(2/3) = 12.0 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: voltage divider (series resistors).

Q030. On a resistor color code, what does the tolerance band indicate?

Correct Answer: B
Explanation: The tolerance band specifies how far the actual resistance may vary from the stated nominal value (e.g., ±5%).
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor tolerance concept.

Q031. A 22 µF capacitor is charged to 9 V. Approximately how much charge is stored?

Correct Answer: A
Explanation: Q = CV. Using µF×V gives µC: Q ≈ 22×9 = 198 µC.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor charge (Q = CV).

Q032. A 47 µF capacitor is charged to 12 V. Approximately how much energy is stored?

Correct Answer: A
Explanation: Energy in a capacitor is U = ½CV² = 0.5×47×10⁻⁶×12² ≈ 0.003384 J = 3.38 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor energy (½CV²).

Q033. An inductor of 20 mH carries 3 A. Approximately how much magnetic energy is stored?

Correct Answer: A
Explanation: Inductor energy is U = ½LI² = 0.5×0.020×3² = 0.5×0.020×9 = 0.09 J.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductor energy (½LI²).

Q034. An inductor of 15 mH is used at 400 Hz. Its inductive reactance is closest to:

Correct Answer: A
Explanation: X_L = 2πfL = 2π×400×0.015 ≈ 37.7 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance (X_L = 2πfL).

Q035. A 0.10 µF capacitor is used at 1,000 Hz. Its capacitive reactance is closest to:

Correct Answer: A
Explanation: X_C = 1/(2πfC) = 1/(2π×1000×0.10×10⁻⁶) ≈ 1592 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance (X_C = 1/(2πfC)).

Q036. If the frequency of an AC signal increases while capacitance stays constant, capacitive reactance X_C will:

Correct Answer: B
Explanation: X_C = 1/(2πfC). As frequency f increases, the denominator increases, so X_C decreases.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency dependence of capacitive reactance.

Q037. If the frequency of an AC signal increases while inductance stays constant, inductive reactance X_L will:

Correct Answer: A
Explanation: X_L = 2πfL, so it increases directly with frequency.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency dependence of inductive reactance.

Q038. For a sine-wave AC source, “RMS voltage” is most directly useful because it equals the DC voltage that would produce the same:

Correct Answer: C
Explanation: RMS is defined so that P = V_rms²/R matches the heating effect of an equivalent DC voltage on a resistor.
Citation: ASVAB content domain — Electronics Information; underlying principle: RMS equivalence for resistive power.

Q039. A silicon diode is placed in series with a 5 V supply feeding a resistor. When forward-biased, the diode will most commonly cause the resistor’s voltage to be approximately:

Correct Answer: B
Explanation: A forward-biased silicon diode typically drops about 0.7 V, reducing the voltage available to the resistor by that amount.
Citation: ASVAB content domain — Electronics Information; underlying principle: diode forward drop (silicon approximation).

Q040. In a basic LED circuit, the LED will not light if it is installed backward because it is:

Correct Answer: B
Explanation: An LED is a diode; reverse bias blocks current (until breakdown), so the LED will not conduct normally.
Citation: ASVAB content domain — Electronics Information; underlying principle: diode polarity (forward vs reverse bias).

Q041. Three identical LEDs are wired in series with one current-limiting resistor. Compared with using one LED (same supply), the series string will generally require:

Correct Answer: A
Explanation: Series LEDs drop more total voltage, so the resistor must drop less voltage for the same current, meaning a smaller resistance is needed (within safe design limits).
Citation: ASVAB content domain — Electronics Information; underlying principle: series voltage drops and current limiting.

Q042. Two identical resistors are connected in parallel across the same DC source. Compared with one resistor alone, the source current will be:

Correct Answer: C
Explanation: Parallel halves the equivalent resistance, so I = V/R increases to about twice for identical resistors.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel circuits and equivalent resistance.

Q043. A resistor’s wattage rating specifies the maximum:

Correct Answer: C
Explanation: The wattage rating indicates how much heat (power) the resistor can handle without overheating or failing.
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor power dissipation limits.

Q044. A standard iron-core transformer will not properly step up a pure DC voltage because transformers require:

Correct Answer: B
Explanation: Transformer action depends on changing flux (AC or switching). Steady DC creates no changing magnetic field after transients.
Citation: ASVAB content domain — Electronics Information; underlying principle: transformer requires changing flux (Faraday’s law).

Q045. A transformer has 300 primary turns and 75 secondary turns. With 240 V AC on the primary, the secondary voltage is closest to:

Correct Answer: A
Explanation: Voltage ratio follows turns ratio: V_s/V_p = N_s/N_p = 75/300 = 0.25, so V_s ≈ 240×0.25 = 60 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: transformer turns ratio (V proportional to turns).

Q046. A fuse is placed in series with a circuit. Under normal operation it should behave like:

Correct Answer: B
Explanation: A properly sized fuse adds minimal resistance; it opens only when current exceeds its rating.
Citation: ASVAB content domain — Electronics Information; underlying principle: overcurrent protection (fuse operation).

Q047. In many DC circuits, “ground” is best described as:

Correct Answer: B
Explanation: Ground is often a reference point (0 V) for measuring other node voltages; it may or may not be tied to earth ground.
Citation: ASVAB content domain — Electronics Information; underlying principle: voltage reference (ground) concept.

Q048. Which statement best distinguishes AC from DC in terms of frequency?

Correct Answer: B
Explanation: AC varies and alternates over time (nonzero frequency). Ideal DC is constant, equivalent to 0 Hz.
Citation: ASVAB content domain — Electronics Information; underlying principle: time-varying signals and frequency.

Q049. A capacitor placed in series with a signal line is commonly used to block:

Correct Answer: B
Explanation: A capacitor’s reactance is very high at DC (f=0), so it blocks DC but can pass changing (AC) signals depending on frequency.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive coupling (blocking DC).

Q050. An inductor placed in series with a power line is commonly used to reduce high-frequency noise because an inductor’s reactance:

Correct Answer: B
Explanation: Inductive reactance X_L = 2πfL rises with frequency, so it impedes high-frequency noise more than low-frequency current.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance increases with frequency.

Q051. A current of 1.10 A flows through a 5 Ω resistor. Approximately how much power is dissipated?

Correct Answer: A
Explanation: Power is P = I²R = (1.10)²×5 ≈ 6.05 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = I²R).

Q052. Two resistors 3 Ω and 12 Ω are in series across 24 V. What is the voltage drop across the 3 Ω resistor (closest)?

Correct Answer: A
Explanation: Series current I = V/(R1+R2) = 24/(3+12) ≈ 1.60 A. Then V_R1 = IR1 ≈ 4.80 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: series circuits (voltage division via IR drops).

Q053. Resistors 5 Ω and 7 Ω are connected in parallel across 15 V. What is the total current drawn (closest)?

Correct Answer: A
Explanation: Each branch has 15 V. I1 = 15/5 ≈ 3.00 A and I2 = 15/7 ≈ 2.14 A. Total is 5.14 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel circuits (branch currents add).

Q054. Two capacitors, 12 µF and 18 µF, are connected in series. The equivalent capacitance is closest to:

Correct Answer: A
Explanation: C_eq = (C1·C2)/(C1+C2) = (12·18)/(30) ≈ 7.20 µF.
Citation: ASVAB content domain — Electronics Information; underlying principle: series capacitance (reciprocal sum).

Q055. An inductor of 14 mH is used at 380 Hz. Its inductive reactance X_L is closest to:

Correct Answer: A
Explanation: X_L = 2πfL. L = 14 mH = 0.014 H, so X_L ≈ 33.43 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance (X_L = 2πfL).

Q056. A capacitor of 0.52 µF is used at 355 Hz. Its capacitive reactance X_C is closest to:

Correct Answer: A
Explanation: X_C = 1/(2πfC). With C = 0.52 µF, X_C ≈ 862 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance (X_C = 1/(2πfC)).

Q057. An ideal transformer has 400 primary turns and 50 secondary turns. If primary current is 0.98 A, the secondary current is closest to:

Correct Answer: A
Explanation: I_s/I_p = N_p/N_s, so I_s = 0.98×(400/50) ≈ 7.84 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer current ratio (inverse turns).

Q058. In a full-wave rectified DC supply, the capacitor connected across the load mainly serves to:

Correct Answer: B
Explanation: The capacitor charges near waveform peaks and then supplies current between peaks, reducing ripple.
Citation: ASVAB content domain — Electronics Information; underlying principle: filter capacitor smoothing (reducing ripple).

Q059. A Zener diode is most commonly used in circuits to provide:

Correct Answer: B
Explanation: A Zener holds near-constant voltage in reverse breakdown, making it useful as a reference/regulator.
Citation: ASVAB content domain — Electronics Information; underlying principle: Zener breakdown for voltage reference.

Q060. In a low-side NPN transistor switch, the load is typically connected between the supply and which terminal?

Correct Answer: B
Explanation: In a low-side NPN switch, the collector connects to the load and the emitter connects to ground.
Citation: ASVAB content domain — Electronics Information; underlying principle: BJT switch wiring (load at collector).

Q061. A MOSFET used as an electronic switch is controlled primarily by:

Correct Answer: A
Explanation: MOSFETs are voltage-controlled; gate-to-source voltage sets conduction with minimal steady gate current.
Citation: ASVAB content domain — Electronics Information; underlying principle: MOSFET gate-to-source voltage control.

Q062. A digital clock signal has frequency 3053 Hz. Its period is closest to:

Correct Answer: A
Explanation: Period T = 1/f. For f = 3053 Hz, T = 1/3053 s = 0.0003 s = 0.33 ms.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (T = 1/f).

Q063. A current of 1.70 A flows through a 6 Ω resistor. Approximately how much power is dissipated?

Correct Answer: A
Explanation: Power is P = I²R = (1.70)²×6 ≈ 17.34 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = I²R).

Q064. Two resistors 8 Ω and 15 Ω are in series across 24 V. What is the voltage drop across the 8 Ω resistor (closest)?

Correct Answer: A
Explanation: Series current I = V/(R1+R2) = 24/(8+15) ≈ 1.04 A. Then V_R1 = IR1 ≈ 8.35 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: series circuits (voltage division via IR drops).

Q065. Resistors 9 Ω and 9 Ω are connected in parallel across 15 V. What is the total current drawn (closest)?

Correct Answer: A
Explanation: Each branch has 15 V. I1 = 15/9 ≈ 1.67 A and I2 = 15/9 ≈ 1.67 A. Total is 3.33 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel circuits (branch currents add).

Q066. Two capacitors, 16 µF and 30 µF, are connected in series. The equivalent capacitance is closest to:

Correct Answer: A
Explanation: C_eq = (C1·C2)/(C1+C2) = (16·30)/(46) ≈ 10.43 µF.
Citation: ASVAB content domain — Electronics Information; underlying principle: series capacitance (reciprocal sum).

Q067. An inductor of 38 mH is used at 860 Hz. Its inductive reactance X_L is closest to:

Correct Answer: A
Explanation: X_L = 2πfL. L = 38 mH = 0.038 H, so X_L ≈ 205.33 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance (X_L = 2πfL).

Q068. A capacitor of 1.24 µF is used at 775 Hz. Its capacitive reactance X_C is closest to:

Correct Answer: A
Explanation: X_C = 1/(2πfC). With C = 1.24 µF, X_C ≈ 166 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance (X_C = 1/(2πfC)).

Q069. An ideal transformer has 700 primary turns and 50 secondary turns. If primary current is 1.34 A, the secondary current is closest to:

Correct Answer: A
Explanation: I_s/I_p = N_p/N_s, so I_s = 1.34×(700/50) ≈ 18.76 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer current ratio (inverse turns).

Q070. In a rectifier + capacitor power supply, the filter capacitor’s main job is to:

Q071. In a simple regulator circuit, a Zener diode is used primarily to create:

Q072. For an NPN transistor used as a low-side switch, the load is usually placed at the:

Q073. For a MOSFET switch, conduction is mainly set by the:

Q074. A digital clock signal has frequency 4529 Hz. Its period is closest to:

Correct Answer: A
Explanation: Period T = 1/f. For f = 4529 Hz, T = 1/4529 s = 0.0002 s = 0.22 ms.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (T = 1/f).

Q075. A current of 2.30 A flows through a 7 Ω resistor. Approximately how much power is dissipated?

Correct Answer: A
Explanation: Power is P = I²R = (2.30)²×7 ≈ 37.03 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = I²R).

Q076. Two resistors 6 Ω and 18 Ω are in series across 24 V. What is the voltage drop across the 6 Ω resistor (closest)?

Correct Answer: A
Explanation: Series current I = V/(R1+R2) = 24/(6+18) ≈ 1.00 A. Then V_R1 = IR1 ≈ 6.00 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: series circuits (voltage division via IR drops).

Q077. Resistors 5 Ω and 11 Ω are connected in parallel across 15 V. What is the total current drawn (closest)?

Correct Answer: A
Explanation: Each branch has 15 V. I1 = 15/5 ≈ 3.00 A and I2 = 15/11 ≈ 1.36 A. Total is 4.36 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel circuits (branch currents add).

Q078. Two capacitors, 20 µF and 18 µF, are connected in series. The equivalent capacitance is closest to:

Correct Answer: A
Explanation: C_eq = (C1·C2)/(C1+C2) = (20·18)/(38) ≈ 9.47 µF.
Citation: ASVAB content domain — Electronics Information; underlying principle: series capacitance (reciprocal sum).

Q079. An inductor of 62 mH is used at 1340 Hz. Its inductive reactance X_L is closest to:

Correct Answer: A
Explanation: X_L = 2πfL. L = 62 mH = 0.062 H, so X_L ≈ 522.01 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance (X_L = 2πfL).

Q080. A capacitor of 1.96 µF is used at 1195 Hz. Its capacitive reactance X_C is closest to:

Correct Answer: A
Explanation: X_C = 1/(2πfC). With C = 1.96 µF, X_C ≈ 68 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance (X_C = 1/(2πfC)).

Q081. An ideal transformer has 1000 primary turns and 50 secondary turns. If primary current is 1.70 A, the secondary current is closest to:

Correct Answer: A
Explanation: I_s/I_p = N_p/N_s, so I_s = 1.70×(1000/50) ≈ 34.00 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer current ratio (inverse turns).

Q082. A large electrolytic capacitor placed after a bridge rectifier is primarily used to:

Q083. A Zener diode is valuable in electronics because it can act as:

Q084. In a common low-side NPN switching circuit, the load connects to the transistor’s:

Q085. A key reason MOSFETs are popular switches is that they are controlled by:

Q086. A digital clock signal has frequency 6005 Hz. Its period is closest to:

Correct Answer: A
Explanation: Period T = 1/f. For f = 6005 Hz, T = 1/6005 s = 0.0002 s = 0.17 ms.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (T = 1/f).

Q087. A current of 2.90 A flows through a 8 Ω resistor. Approximately how much power is dissipated?

Correct Answer: A
Explanation: Power is P = I²R = (2.90)²×8 ≈ 67.28 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = I²R).

Q088. Two resistors 4 Ω and 12 Ω are in series across 24 V. What is the voltage drop across the 4 Ω resistor (closest)?

Correct Answer: A
Explanation: Series current I = V/(R1+R2) = 24/(4+12) ≈ 1.50 A. Then V_R1 = IR1 ≈ 6.00 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: series circuits (voltage division via IR drops).

Q089. Resistors 9 Ω and 13 Ω are connected in parallel across 15 V. What is the total current drawn (closest)?

Correct Answer: A
Explanation: Each branch has 15 V. I1 = 15/9 ≈ 1.67 A and I2 = 15/13 ≈ 1.15 A. Total is 2.82 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel circuits (branch currents add).

Q090. Two capacitors, 24 µF and 30 µF, are connected in series. The equivalent capacitance is closest to:

Correct Answer: A
Explanation: C_eq = (C1·C2)/(C1+C2) = (24·30)/(54) ≈ 13.33 µF.
Citation: ASVAB content domain — Electronics Information; underlying principle: series capacitance (reciprocal sum).

Q091. An inductor of 86 mH is used at 1820 Hz. Its inductive reactance X_L is closest to:

Correct Answer: A
Explanation: X_L = 2πfL. L = 86 mH = 0.086 H, so X_L ≈ 983.44 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance (X_L = 2πfL).

Q092. A capacitor of 2.68 µF is used at 1615 Hz. Its capacitive reactance X_C is closest to:

Correct Answer: A
Explanation: X_C = 1/(2πfC). With C = 2.68 µF, X_C ≈ 37 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance (X_C = 1/(2πfC)).

Q093. An ideal transformer has 1300 primary turns and 50 secondary turns. If primary current is 2.06 A, the secondary current is closest to:

Correct Answer: A
Explanation: I_s/I_p = N_p/N_s, so I_s = 2.06×(1300/50) ≈ 53.56 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer current ratio (inverse turns).

Q094. In a DC supply made from rectified AC, the smoothing capacitor mainly reduces:

Q095. When reverse-biased into breakdown, a Zener diode is typically used as a:

Q096. When wiring an NPN as a low-side switch, the load is normally tied to the:

Q097. A MOSFET turns on mainly when sufficient ______ is applied.

Correct Answer: A
Explanation: A MOSFET turns on when V_GS exceeds threshold enough to form a conducting channel.
Citation: ASVAB content domain — Electronics Information; underlying principle: MOSFET threshold and V_GS control.

Q098. A digital clock signal has frequency 7481 Hz. Its period is closest to:

Correct Answer: A
Explanation: Period T = 1/f. For f = 7481 Hz, T = 1/7481 s = 0.0001 s = 0.13 ms.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (T = 1/f).

Q099. A current of 3.50 A flows through a 9 Ω resistor. Approximately how much power is dissipated?

Correct Answer: A
Explanation: Power is P = I²R = (3.50)²×9 ≈ 110.25 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = I²R).

Q100. Two resistors 2 Ω and 15 Ω are in series across 24 V. What is the voltage drop across the 2 Ω resistor (closest)?

Correct Answer: A
Explanation: Series current I = V/(R1+R2) = 24/(2+15) ≈ 1.41 A. Then V_R1 = IR1 ≈ 2.82 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: series circuits (voltage division via IR drops).

Q101. A 12 Ω load is connected across 30 V DC. Which pair is closest to the current and power?

Correct Answer: A
Explanation: Use Ohm’s law: I = V/R = 30/12 = 2.50 A. Power is P = VI = 30×2.50 ≈ 75 W (also V²/R).
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law and power (I = V/R, P = VI).

Q102. A 10 Ω resistor is in series with two 20 Ω resistors in parallel. The total resistance is closest to:

Correct Answer: A
Explanation: First combine the parallel pair: R_parallel = (20||20) = 10 Ω. Then add series: R_total = 10 + 10 = 20 Ω (≈ 20 Ω).
Citation: ASVAB content domain — Electronics Information; underlying principle: series/parallel resistance reduction.

Q103. A voltage divider has 2 kΩ on top and 3 kΩ on bottom across 15 V. What is the output voltage across the bottom resistor?

Correct Answer: A
Explanation: V_out = V_in × R_bottom/(R_top+R_bottom) = 15×3000/(2000+3000) = 15×0.6 = 9.0 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: voltage divider (series resistors).

Q104. A sine-wave source is 120 V RMS. The peak voltage is closest to:

Correct Answer: A
Explanation: For a sine wave, V_pk = V_rms·√2 = 120×1.414 ≈ 170 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: RMS-to-peak conversion (sine wave).

Q105. A simple RC low-pass filter uses R = 10 kΩ and C = 0.01 µF. The cutoff frequency is closest to:

Correct Answer: A
Explanation: Cutoff f_c = 1/(2πRC). Here RC = 10,000×0.01×10⁻⁶ = 1.0×10⁻⁴ s, so f_c ≈ 1/(2π×1e-4) ≈ 1592 Hz.
Citation: ASVAB content domain — Electronics Information; underlying principle: RC filter cutoff (f_c = 1/(2πRC)).

Q106. A capacitor placed in series with an audio signal line is mainly used to:

Correct Answer: B
Explanation: At DC (0 Hz), a capacitor’s reactance is extremely high, so it blocks DC. For changing signals, it can pass current depending on frequency.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive coupling (blocking DC).

Q107. An inductor of 50 mH carries 2.0 A. Approximately how much energy is stored in its magnetic field?

Correct Answer: A
Explanation: Inductor energy U = ½LI² = 0.5×0.050×(2.0)² = 0.10 J.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductor energy (½LI²).

Q108. A 2.2 µF capacitor is used at 500 Hz. Its capacitive reactance is closest to:

Correct Answer: A
Explanation: X_C = 1/(2πfC) = 1/(2π×500×2.2×10⁻⁶) ≈ 145 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance (X_C = 1/(2πfC)).

Q109. An inductor of 10 mH is used at 1,000 Hz. Its inductive reactance is closest to:

Correct Answer: A
Explanation: X_L = 2πfL = 2π×1000×0.010 ≈ 62.8 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance (X_L = 2πfL).

Q110. A 12 V supply powers an LED that drops about 2.0 V at 20 mA. The series resistor value is closest to:

Correct Answer: A
Explanation: Resistor must drop 12−2 = 10 V at 0.020 A, so R = V/I = 10/0.020 = 500 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: current limiting with series resistor.

Q111. A bridge rectifier converts AC to pulsating DC using:

Correct Answer: C
Explanation: A full-wave bridge rectifier uses four diodes so both half-cycles produce the same output polarity.
Citation: ASVAB content domain — Electronics Information; underlying principle: full-wave bridge rectification.

Q112. In an AC circuit containing a capacitor, the current through an ideal capacitor leads the capacitor’s voltage by approximately:

Correct Answer: B
Explanation: For an ideal capacitor, current leads voltage by 90° because the current is proportional to the rate of change of voltage.
Citation: ASVAB content domain — Electronics Information; underlying principle: phase relationship in capacitors (I leads V).

Q113. In an AC circuit containing an inductor, the current through an ideal inductor lags the inductor’s voltage by approximately:

Correct Answer: C
Explanation: For an ideal inductor, voltage leads current by 90°, so current lags voltage by 90°.
Citation: ASVAB content domain — Electronics Information; underlying principle: phase relationship in inductors (I lags V).

Q114. A transistor described as having “current gain” is primarily used because a small base current can control:

Correct Answer: B
Explanation: In a BJT, a small base current allows a much larger collector current, providing current amplification or switching control.
Citation: ASVAB content domain — Electronics Information; underlying principle: transistor current gain (base controls collector).

Q115. A digital logic gate outputs HIGH only when both inputs are HIGH. This is a(n):

Correct Answer: B
Explanation: An AND gate requires both inputs to be HIGH for the output to be HIGH.
Citation: ASVAB content domain — Electronics Information; underlying principle: basic logic gates (AND).

Q116. A digital XOR gate outputs HIGH when:

Correct Answer: B
Explanation: XOR is HIGH only when exactly one input is HIGH (inputs differ).
Citation: ASVAB content domain — Electronics Information; underlying principle: basic logic gates (XOR).

Q117. A voltage regulator in an electronic circuit is primarily used to:

Correct Answer: B
Explanation: Regulators control output voltage to stay near a set value even as input voltage or load current varies.
Citation: ASVAB content domain — Electronics Information; underlying principle: voltage regulation concept.

Q118. Two resistors, 6 Ω and 3 Ω, are connected in parallel across 12 V. What is the total current drawn (closest)?

Correct Answer: A
Explanation: Branch currents: I1 = 12/6 = 2.0 A and I2 = 12/3 = 4.0 A. Total is I_total = I1+I2 = 6.0 A.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel circuits (currents add).

Q119. Kirchhoff’s Voltage Law (KVL) states that around any closed loop in a circuit, the sum of voltage rises and drops equals:

Correct Answer: B
Explanation: KVL expresses conservation of energy: the algebraic sum of voltages around a closed loop is zero.
Citation: ASVAB content domain — Electronics Information; underlying principle: Kirchhoff’s Voltage Law (loop rule).

Q120. Kirchhoff’s Current Law (KCL) states that at a circuit node, the sum of currents entering equals:

Correct Answer: B
Explanation: KCL expresses conservation of charge: currents into a node equal currents out of the node.
Citation: ASVAB content domain — Electronics Information; underlying principle: Kirchhoff’s Current Law (node rule).

Q121. A 12.0 V battery has internal resistance 0.2 Ω. If it supplies 30 A, the terminal voltage under load is closest to:

Correct Answer: A
Explanation: Voltage drop inside the battery is I·r = 30×0.2 = 6.0 V. Terminal voltage ≈ 12.0 − 6.0 = 6.0 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: internal resistance and voltage sag (V = E − Ir).

Q122. A circuit normally draws 2.5 A. Which fuse rating is the best match for basic protection without nuisance blowing?

Correct Answer: C
Explanation: A fuse should be rated just above normal operating current; 3 A provides margin over 2.5 A while still offering protection.
Citation: ASVAB content domain — Electronics Information; underlying principle: overcurrent protection (selecting fuse rating).

Q123. Two capacitors, 10 µF and 20 µF, are connected in series. The equivalent capacitance is closest to:

Correct Answer: A
Explanation: For capacitors in series: 1/C_eq = 1/C1 + 1/C2, so C_eq = (C1·C2)/(C1+C2) = (10·20)/(30) ≈ 6.7 µF.
Citation: ASVAB content domain — Electronics Information; underlying principle: series capacitance (reciprocal sum).

Q124. Capacitors 4.7 µF, 2.2 µF, and 1.0 µF are connected in parallel. The equivalent capacitance is closest to:

Correct Answer: A
Explanation: Capacitances in parallel add directly: C_eq = 4.7+2.2+1.0 = 7.9 µF.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel capacitance (direct sum).

Q125. A radio signal has frequency 100 MHz. Its wavelength in free space is closest to:

Correct Answer: A
Explanation: Wavelength λ = c/f. Using c ≈ 3×10⁸ m/s and f = 1×10⁸ Hz gives λ ≈ 3 m.
Citation: ASVAB content domain — Electronics Information; underlying principle: wave relationship (λ = c/f).

Q126. A resistor is labeled 2.6 kΩ ±5%. What resistance range is acceptable?

Correct Answer: A
Explanation: Tolerance is ±5%: ±0.05×2.6 kΩ = ±0.128 kΩ, giving 2.423 to 2.678 kΩ.
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor tolerance (% of nominal).

Q127. A divider has 5 kΩ (top) and 6 kΩ (bottom) across 7.9 V. A 8 kΩ load is placed across the bottom resistor. The loaded output is closest to:

Correct Answer: A
Explanation: The load is in parallel with the bottom resistor, reducing the effective bottom resistance. Use R_bottom_eff = R_bottom||R_load, then V_out = V_in×R_bottom_eff/(R_top+R_bottom_eff).
Citation: ASVAB content domain — Electronics Information; underlying principle: loaded voltage divider (parallel loading effect).

Q128. To minimize measurement error when checking a circuit’s node voltage, a voltmeter should have:

Correct Answer: B
Explanation: A voltmeter is placed in parallel; high input resistance draws negligible current and avoids loading the circuit.
Citation: ASVAB content domain — Electronics Information; underlying principle: instrument loading (high voltmeter input resistance).

Q129. An RC circuit uses R = 11 kΩ and C = 0.37 µF. About how long is one time constant τ?

Correct Answer: A
Explanation: The time constant is τ = RC. After one τ, a charging capacitor reaches about 63% of its final value.
Citation: ASVAB content domain — Electronics Information; underlying principle: RC time constant (τ = RC).

Q130. An amplifier increases power from 2.5 W to 25.0 W. The power gain in decibels is closest to:

Correct Answer: A
Explanation: Power gain in dB is 10·log10(Pout/Pin).
Citation: ASVAB content domain — Electronics Information; underlying principle: decibel power ratio (10·log10(P2/P1)).

Q131. A 120 V AC load draws 5.0 A at power factor 0.70. Real power is closest to:

Correct Answer: A
Explanation: Real power is P = VI·pf; apparent power is S = VI.
Citation: ASVAB content domain — Electronics Information; underlying principle: real vs apparent power (P = VI·pf).

Q132. An LC circuit has L = 2.1 mH and C = 24 nF. The resonant frequency is closest to:

Correct Answer: A
Explanation: Use f0 = 1/(2π√(LC)).
Citation: ASVAB content domain — Electronics Information; underlying principle: LC resonance (f0 = 1/(2π√(LC))).

Q133. A standard silicon diode is reverse-biased well below its breakdown rating. Ideally, it will:

Correct Answer: B
Explanation: In reverse bias (below breakdown), a diode blocks current except for very small leakage.
Citation: ASVAB content domain — Electronics Information; underlying principle: diode reverse bias behavior (blocking current).

Q134. On most through-hole LEDs, the shorter lead typically indicates the:

Correct Answer: B
Explanation: Common convention: shorter lead is the cathode (negative), longer lead is the anode (positive).
Citation: ASVAB content domain — Electronics Information; underlying principle: component identification (LED polarity convention).

Q135. An op-amp configured with high gain and no negative feedback is most likely to operate as a:

Correct Answer: B
Explanation: Without negative feedback, small input differences drive the output to a rail, acting like a comparator.
Citation: ASVAB content domain — Electronics Information; underlying principle: op-amp open-loop gain and saturation.

Q136. A 13-bit ADC uses a 2.4 V reference. The ideal voltage per least significant bit (LSB) is closest to:

Correct Answer: A
Explanation: Ideal step size is approximately Vref/2^N.
Citation: ASVAB content domain — Electronics Information; underlying principle: ADC resolution (LSB ≈ Vref/2^N).

Q137. A pull-up resistor on a digital input pin is used primarily to:

Correct Answer: B
Explanation: A pull-up provides a defined HIGH state when undriven, preventing floating/noise-triggered states.
Citation: ASVAB content domain — Electronics Information; underlying principle: digital inputs (pull-up to prevent floating).

Q138. A 3.2 V PWM signal with duty cycle 58% drives a slow-responding load. The average (DC) voltage is closest to:

Correct Answer: A
Explanation: For a load that averages pulses, Vavg ≈ V_high×duty (assuming 0 V low).
Citation: ASVAB content domain — Electronics Information; underlying principle: PWM averaging (Vavg = V·duty).

Q139. In an ideal series RLC circuit at resonance, the total impedance is closest to:

Correct Answer: C
Explanation: At resonance, X_L and X_C cancel, leaving only resistance as the net impedance.
Citation: ASVAB content domain — Electronics Information; underlying principle: series resonance (X_L cancels X_C).

Q140. A common cause of hum in an audio system with multiple grounded devices is:

Correct Answer: B
Explanation: Multiple ground return paths allow small AC currents to circulate and inject hum into signal references.
Citation: ASVAB content domain — Electronics Information; underlying principle: ground loops and common-mode noise.

Q141. A resistor is labeled 7.8 kΩ ±10%. What resistance range is acceptable?

Correct Answer: A
Explanation: Tolerance is ±10%: ±0.10×7.8 kΩ = ±0.780 kΩ, giving 7.020 to 8.580 kΩ.
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor tolerance (% of nominal).

Q142. A divider has 7 kΩ (top) and 9 kΩ (bottom) across 8.3 V. A 8 kΩ load is placed across the bottom resistor. The loaded output is closest to:

Q143. When measuring voltage across a component, the meter should have what internal characteristic to avoid loading?

Q144. An RC circuit uses R = 26 kΩ and C = 0.27 µF. About how long is one time constant τ?

Q145. An amplifier increases power from 2.0 W to 28.0 W. The power gain in decibels is closest to:

Correct Answer: A
Explanation: Power gain in dB is 10·log10(Pout/Pin).
Citation: ASVAB content domain — Electronics Information; underlying principle: decibel power ratio (10·log10(P2/P1)).

Q146. A 120 V AC load draws 5.0 A at power factor 0.82. Real power is closest to:

Q147. An LC circuit has L = 1.4 mH and C = 21 nF. The resonant frequency is closest to:

Correct Answer: A
Explanation: Use f0 = 1/(2π√(LC)).
Citation: ASVAB content domain — Electronics Information; underlying principle: LC resonance (f0 = 1/(2π√(LC))).

Q148. In normal reverse bias (not in breakdown), a silicon diode will mostly:

Q149. For a typical 5 mm LED, the shorter leg is usually the:

Q150. With no negative feedback, an op-amp will usually:

Q151. A 10-bit ADC uses a 2.2 V reference. The ideal voltage per least significant bit (LSB) is closest to:

Correct Answer: A
Explanation: Ideal step size is approximately Vref/2^N.
Citation: ASVAB content domain — Electronics Information; underlying principle: ADC resolution (LSB ≈ Vref/2^N).

Q152. Why would a designer add a pull-up resistor to a logic input?

Q153. A 3.7 V PWM signal with duty cycle 46% drives a slow-responding load. The average (DC) voltage is closest to:

Q154. At the resonant frequency of a series RLC circuit, the net impedance is approximately:

Q155. In audio systems, a persistent 60 Hz hum is often caused by:

Q156. A resistor is labeled 13.1 kΩ ±5%. What resistance range is acceptable?

Correct Answer: A
Explanation: Tolerance is ±5%: ±0.05×13.1 kΩ = ±0.653 kΩ, giving 12.398 to 13.703 kΩ.
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor tolerance (% of nominal).

Q157. A divider has 9 kΩ (top) and 12 kΩ (bottom) across 8.7 V. A 8 kΩ load is placed across the bottom resistor. The loaded output is closest to:

Q158. A good voltmeter is designed to draw ______ from the circuit being tested.

Q159. An RC circuit uses R = 18 kΩ and C = 1.02 µF. About how long is one time constant τ?

Q160. An amplifier increases power from 1.5 W to 10.5 W. The power gain in decibels is closest to:

Correct Answer: A
Explanation: Power gain in dB is 10·log10(Pout/Pin).
Citation: ASVAB content domain — Electronics Information; underlying principle: decibel power ratio (10·log10(P2/P1)).

Q161. A 120 V AC load draws 5.0 A at power factor 0.94. Real power is closest to:

Q162. An LC circuit has L = 0.6 mH and C = 18 nF. The resonant frequency is closest to:

Correct Answer: A
Explanation: Use f0 = 1/(2π√(LC)).
Citation: ASVAB content domain — Electronics Information; underlying principle: LC resonance (f0 = 1/(2π√(LC))).

Q163. A diode operated in reverse bias below breakdown behaves most like a:

Q164. On a common LED package, lead length most often identifies the:

Q165. An op-amp used open-loop is commonly used as a:

Q166. A 13-bit ADC uses a 2.0 V reference. The ideal voltage per least significant bit (LSB) is closest to:

Correct Answer: A
Explanation: Ideal step size is approximately Vref/2^N.
Citation: ASVAB content domain — Electronics Information; underlying principle: ADC resolution (LSB ≈ Vref/2^N).

Q167. A pull-up resistor ensures a digital input reads what when nothing else drives it?

Q168. A 3.2 V PWM signal with duty cycle 34% drives a slow-responding load. The average (DC) voltage is closest to:

Q169. For a series RLC tuned to resonance, the circuit’s impedance is mainly set by:

Q170. Unwanted hum can appear when two devices share more than one ground path; this is called a:

Q171. A resistor is labeled 18.3 kΩ ±10%. What resistance range is acceptable?

Correct Answer: A
Explanation: Tolerance is ±10%: ±0.10×18.3 kΩ = ±1.830 kΩ, giving 16.470 to 20.130 kΩ.
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor tolerance (% of nominal).

Q172. A divider has 11 kΩ (top) and 15 kΩ (bottom) across 9.1 V. A 8 kΩ load is placed across the bottom resistor. The loaded output is closest to:

Q173. If you want the voltage reading to be as accurate as possible, the voltmeter’s input resistance should be:

Q174. An RC circuit uses R = 10 kΩ and C = 0.92 µF. About how long is one time constant τ?

Q175. An amplifier increases power from 1.0 W to 11.0 W. The power gain in decibels is closest to:

Correct Answer: A
Explanation: Power gain in dB is 10·log10(Pout/Pin).
Citation: ASVAB content domain — Electronics Information; underlying principle: decibel power ratio (10·log10(P2/P1)).

Q176. A 120 V AC load draws 5.0 A at power factor 0.58. Real power is closest to:

Q177. An LC circuit has L = 4.3 mH and C = 15 nF. The resonant frequency is closest to:

Correct Answer: A
Explanation: Use f0 = 1/(2π√(LC)).
Citation: ASVAB content domain — Electronics Information; underlying principle: LC resonance (f0 = 1/(2π√(LC))).

Q178. Below reverse breakdown, the current through a reverse-biased diode is best described as:

Q179. In many standard LEDs, the shorter lead is connected to the:

Q180. If an op-amp has no feedback path, the output tends to:

Q181. A 10-bit ADC uses a 3.4 V reference. The ideal voltage per least significant bit (LSB) is closest to:

Correct Answer: A
Explanation: Ideal step size is approximately Vref/2^N.
Citation: ASVAB content domain — Electronics Information; underlying principle: ADC resolution (LSB ≈ Vref/2^N).

Q182. The main purpose of a pull-up on an input is to prevent:

Q183. A 3.7 V PWM signal with duty cycle 22% drives a slow-responding load. The average (DC) voltage is closest to:

Q184. In a series RLC circuit at resonance, the reactive parts effectively:

Q185. A ground-referenced signal may pick up hum if there is a:

Q186. A resistor is labeled 23.5 kΩ ±5%. What resistance range is acceptable?

Correct Answer: A
Explanation: Tolerance is ±5%: ±0.05×23.5 kΩ = ±1.177 kΩ, giving 22.372 to 24.727 kΩ.
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor tolerance (% of nominal).

Q187. A divider has 13 kΩ (top) and 6 kΩ (bottom) across 9.5 V. A 8 kΩ load is placed across the bottom resistor. The loaded output is closest to:

Q188. To reduce disturbance of the circuit while measuring a voltage, the meter should have:

Q189. An RC circuit uses R = 25 kΩ and C = 0.82 µF. About how long is one time constant τ?

Q190. An amplifier increases power from 0.5 W to 7.5 W. The power gain in decibels is closest to:

Correct Answer: A
Explanation: Power gain in dB is 10·log10(Pout/Pin).
Citation: ASVAB content domain — Electronics Information; underlying principle: decibel power ratio (10·log10(P2/P1)).

Q191. A 120 V AC load draws 5.0 A at power factor 0.70. Real power is closest to: (Pick the best answer.)

Q192. An LC circuit has L = 3.6 mH and C = 12 nF. The resonant frequency is closest to:

Correct Answer: A
Explanation: Use f0 = 1/(2π√(LC)).
Citation: ASVAB content domain — Electronics Information; underlying principle: LC resonance (f0 = 1/(2π√(LC))).

Q193. With reverse polarity applied (within ratings), a diode primarily:

Q194. Lead length on many through-hole LEDs commonly marks the:

Q195. An op-amp without stabilizing feedback behaves most like a:

Q196. A 13-bit ADC uses a 3.2 V reference. The ideal voltage per least significant bit (LSB) is closest to:

Correct Answer: A
Explanation: Ideal step size is approximately Vref/2^N.
Citation: ASVAB content domain — Electronics Information; underlying principle: ADC resolution (LSB ≈ Vref/2^N).

Q197. A pull-up resistor is used to give an undriven input a:

Q198. A 3.2 V PWM signal with duty cycle 10% drives a slow-responding load. The average (DC) voltage is closest to:

Q199. When a series RLC circuit is at resonance, the impedance is dominated by:

Q200. Multiple chassis grounds tied together can create noise due to a:

Q201. A device draws 1.8 A from a 12 V source. Its effective resistance is closest to:

Correct Answer: A
Explanation: Use Ohm’s law: R = V/I. With V=12 V and I=1.8 A, R ≈ 6.67 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law (R = V/I).

Q202. A 10 Ω resistor is connected across 15 V. Approximately how much power does it dissipate?

Correct Answer: A
Explanation: With voltage across a resistor, power is P = V²/R = 15.0²/10.0 ≈ 22.50 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = V²/R).

Q203. Two resistors 12 Ω and 18 Ω are in parallel. Their equivalent resistance is closest to:

Correct Answer: A
Explanation: For two resistors in parallel: 1/Req = 1/R1 + 1/R2. Here Req ≈ 7.20 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel resistance (reciprocal sum).

Q204. A 10 Ω resistor is in series with a parallel pair of 12 Ω and 6 Ω. The total resistance is closest to:

Correct Answer: A
Explanation: Combine the parallel pair first: R|| = (R1·R2)/(R1+R2) ≈ 4.00 Ω, then add the series resistor: R_total ≈ 14.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: series-parallel reduction (parallel then series).

Q205. An RC low-pass has R = 3.3 kΩ and C = 0.047 µF. The cutoff frequency is closest to:

Correct Answer: A
Explanation: Cutoff frequency is f_c = 1/(2πRC). Substituting values gives f_c ≈ 1026.1 Hz.
Citation: ASVAB content domain — Electronics Information; underlying principle: RC cutoff frequency (f_c = 1/(2πRC)).

Q206. A silicon diode is forward-biased and carries 0.1 A. Its forward voltage drop is typically closest to:

Correct Answer: A
Explanation: A conducting silicon diode is commonly approximated as about 0.7 V for quick estimates (actual drop varies with conditions).
Citation: ASVAB content domain — Electronics Information; underlying principle: silicon diode forward drop (≈0.7 V).

Q207. A full-wave bridge rectifier is connected to 60 Hz AC. The ripple frequency at the rectifier output is:

Correct Answer: A
Explanation: Full-wave rectification flips the negative half-cycle, doubling the pulse rate. Ripple frequency ≈ 2×input frequency.
Citation: ASVAB content domain — Electronics Information; underlying principle: full-wave rectification (ripple frequency doubles).

Q208. An ideal transformer has 500 primary turns and 50 secondary turns with 120 V on the primary. Secondary voltage is closest to:

Correct Answer: A
Explanation: For an ideal transformer, V_s/V_p = N_s/N_p, so V_s = 120×(50/500) ≈ 12.00 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer voltage ratio (V ∝ turns).

Q209. An inductor of 20 mH carries 0.50 A. The stored energy is closest to:

Correct Answer: A
Explanation: E = ½LI². With L=0.020 H and I=0.50 A, E ≈ 0.002500 J = 2.50 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductor energy (E = ½LI²).

Q210. At a fixed frequency, if capacitance is doubled, the capacitive reactance X_C will:

Correct Answer: A
Explanation: X_C = 1/(2πfC). Doubling C doubles the denominator, so X_C halves.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance dependence (X_C ∝ 1/C).

Q211. A resistor with color bands brown-black-red (ignore tolerance band) has a resistance of:

Correct Answer: A
Explanation: For 3-band resistors, the first two bands are digits and the third is the multiplier (10^n). Decode digits, then apply the multiplier.
Citation: ASVAB content domain — Electronics Information; underlying principle: resistor color code (digits + multiplier).

Q212. For an AND gate, the output is HIGH only when:

Correct Answer: A
Explanation: AND outputs HIGH only when all inputs are HIGH.
Citation: ASVAB content domain — Electronics Information; underlying principle: logic gate truth tables (AND).

Q213. The binary number 10110₂ equals what decimal value?

Correct Answer: A
Explanation: Sum powers of 2 where bits are 1. 10110₂ equals 22₁₀.
Citation: ASVAB content domain — Electronics Information; underlying principle: binary place value conversion.

Q214. An ideal inverting op-amp has R_in = 2 kΩ and R_f = 20 kΩ. The voltage gain (Vout/Vin) is closest to:

Correct Answer: A
Explanation: In an ideal inverting op-amp, A_v = -R_f/R_in = -20/2 ≈ -10.00.
Citation: ASVAB content domain — Electronics Information; underlying principle: inverting op-amp gain (A_v = -R_f/R_in).

Q215. A BJT used as a switch is considered ‘saturated’ when it is:

Correct Answer: A
Explanation: Saturation means the transistor is driven hard on; V_CE is small and it behaves like a closed switch.
Citation: ASVAB content domain — Electronics Information; underlying principle: BJT switching regions (saturation).

Q216. If a signal voltage increases by a factor of 10 (same impedance), the change in dB is closest to:

Correct Answer: A
Explanation: For equal impedances, voltage gain in dB is 20·log10(V2/V1). Apply the given voltage ratio.
Citation: ASVAB content domain — Electronics Information; underlying principle: decibel voltage ratio (20·log10(V2/V1)).

Q217. A square wave has frequency 2.5 kHz. Its period is closest to:

Correct Answer: A
Explanation: T = 1/f. For 2.5 kHz, T = 0.0004 s = 400 µs.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (T = 1/f).

Q218. A capacitor of 100 µF is charged to 12 V. The stored energy is closest to:

Correct Answer: A
Explanation: E = ½CV². With C=0.000100 F and V=12.0 V, E ≈ 0.007200 J = 7.20 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor energy (E = ½CV²).

Q219. A 9 V source powers two red LEDs (assume 2.0 V each) in series with a resistor for 20 mA. The resistor value is closest to:

Correct Answer: A
Explanation: First find resistor voltage: V_R = V_s − n·V_f = 9.0 − 2×2.0 = 5.0 V. Then R = V_R/I ≈ 5.0/0.020 ≈ 250.0 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: LED current limiting (R = (V_s − ΣV_f)/I).

Q220. Which filter primarily passes low frequencies and attenuates high frequencies?

Correct Answer: A
Explanation: A low-pass passes frequencies below cutoff and attenuates higher frequencies.
Citation: ASVAB content domain — Electronics Information; underlying principle: filter types (low-pass behavior).

Q221. A circuit current is 0.75 A when connected to 9.0 V. The circuit resistance is closest to:

Correct Answer: A
Explanation: Use Ohm’s law: R = V/I. With V=9 V and I=0.75 A, R ≈ 12.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law (R = V/I).

Q222. A 5.6 Ω load is connected across 12 V. Approximately how much power is delivered to the load?

Correct Answer: A
Explanation: With voltage across a resistor, power is P = V²/R = 12.0²/5.6 ≈ 25.71 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = V²/R).

Q223. Resistors 30 Ω and 10 Ω are connected in parallel. The equivalent resistance is closest to:

Correct Answer: A
Explanation: For two resistors in parallel: 1/Req = 1/R1 + 1/R2. Here Req ≈ 7.50 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel resistance (reciprocal sum).

Q224. A circuit has a 4 Ω resistor in series with a parallel pair of 8 Ω and 8 Ω. Total resistance is closest to:

Correct Answer: A
Explanation: Combine the parallel pair first: R|| = (R1·R2)/(R1+R2) ≈ 4.00 Ω, then add the series resistor: R_total ≈ 8.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: series-parallel reduction (parallel then series).

Q225. An RC filter uses R = 10 kΩ and C = 0.01 µF. The cutoff frequency is closest to:

Correct Answer: A
Explanation: Cutoff frequency is f_c = 1/(2πRC). Substituting values gives f_c ≈ 1591.5 Hz.
Citation: ASVAB content domain — Electronics Information; underlying principle: RC cutoff frequency (f_c = 1/(2πRC)).

Q226. In many circuits, a forward-biased silicon diode is approximated as having a voltage drop of about:

Q227. A full-wave rectifier uses a 50 Hz AC input. The main ripple frequency after rectification is:

Q228. An ideal transformer has 200 primary turns and 800 secondary turns. If 12 V is applied to the primary, the secondary voltage is closest to:

Correct Answer: A
Explanation: For an ideal transformer, V_s/V_p = N_s/N_p, so V_s = 12×(800/200) ≈ 48.00 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer voltage ratio (V ∝ turns).

Q229. A 5 mH inductor carries 2.0 A. The energy stored in its magnetic field is closest to:

Correct Answer: A
Explanation: E = ½LI². With L=0.005 H and I=2.00 A, E ≈ 0.010000 J = 10.00 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductor energy (E = ½LI²).

Q230. At a fixed frequency, if inductance is doubled, the inductive reactance X_L will:

Correct Answer: A
Explanation: X_L = 2πfL. Doubling L doubles X_L.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance dependence (X_L ∝ L).

Q231. A resistor marked yellow-violet-orange (ignore tolerance) has a value of:

Q232. For an OR gate, the output is LOW only when:

Correct Answer: A
Explanation: OR outputs LOW only when all inputs are LOW.
Citation: ASVAB content domain — Electronics Information; underlying principle: logic gate truth tables (OR).

Q233. Convert 11001₂ to decimal.

Correct Answer: A
Explanation: Sum powers of 2 where bits are 1. 11001₂ equals 25₁₀.
Citation: ASVAB content domain — Electronics Information; underlying principle: binary place value conversion.

Q234. An ideal inverting amplifier uses R_in = 5 kΩ and R_f = 15 kΩ. Its gain is closest to:

Correct Answer: A
Explanation: In an ideal inverting op-amp, A_v = -R_f/R_in = -15/5 ≈ -3.00.
Citation: ASVAB content domain — Electronics Information; underlying principle: inverting op-amp gain (A_v = -R_f/R_in).

Q235. In BJT switching, saturation means the transistor is:

Correct Answer: A
Explanation: In saturation, extra base drive does not significantly increase collector current; the device is essentially ‘maxed out’ on.
Citation: ASVAB content domain — Electronics Information; underlying principle: BJT switching regions (saturation condition).

Q236. A voltage gain of 2 (same impedance) corresponds to approximately how many dB?

Q237. A clock runs at 8.0 MHz. Its period is closest to:

Correct Answer: A
Explanation: T = 1/f. At 8.0 MHz, T = 125 ns.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (T = 1/f).

Q238. A 10 µF capacitor is charged to 50 V. Stored energy is closest to:

Correct Answer: A
Explanation: E = ½CV². With C=0.000010 F and V=50.0 V, E ≈ 0.012500 J = 12.50 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor energy (E = ½CV²).

Q239. A 12 V supply drives three LEDs (assume 2.1 V each) in series at 15 mA. Series resistor is closest to:

Correct Answer: A
Explanation: First find resistor voltage: V_R = V_s − n·V_f = 12.0 − 3×2.1 = 5.7 V. Then R = V_R/I ≈ 5.7/0.015 ≈ 380.0 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: LED current limiting (R = (V_s − ΣV_f)/I).

Q240. Which filter primarily passes high frequencies and attenuates low frequencies?

Correct Answer: A
Explanation: A high-pass attenuates low frequencies and passes higher frequencies.
Citation: ASVAB content domain — Electronics Information; underlying principle: filter types (high-pass behavior).

Q241. A 24 V supply pushes 3.0 A through a load. The load resistance is closest to:

Correct Answer: A
Explanation: Use Ohm’s law: R = V/I. With V=24 V and I=3 A, R ≈ 8.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law (R = V/I).

Q242. A 33 Ω resistor has 9.0 V across it. Approximately how much power is dissipated?

Correct Answer: A
Explanation: With voltage across a resistor, power is P = V²/R = 9.0²/33.0 ≈ 2.45 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = V²/R).

Q243. Two resistors 8 Ω and 24 Ω are placed in parallel. The equivalent resistance is closest to:

Correct Answer: A
Explanation: For two resistors in parallel: 1/Req = 1/R1 + 1/R2. Here Req ≈ 6.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel resistance (reciprocal sum).

Q244. A 2 Ω resistor is in series with a parallel pair of 3 Ω and 6 Ω. Total resistance is closest to:

Correct Answer: A
Explanation: Combine the parallel pair first: R|| = (R1·R2)/(R1+R2) ≈ 2.00 Ω, then add the series resistor: R_total ≈ 4.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: series-parallel reduction (parallel then series).

Q245. An RC circuit has R = 2.2 kΩ and C = 0.1 µF. Its cutoff frequency is closest to:

Correct Answer: A
Explanation: Cutoff frequency is f_c = 1/(2πRC). Substituting values gives f_c ≈ 723.4 Hz.
Citation: ASVAB content domain — Electronics Information; underlying principle: RC cutoff frequency (f_c = 1/(2πRC)).

Q246. A standard silicon rectifier diode conducting normally is often modeled as dropping approximately:

Q247. With a full-wave rectifier on a 60 Hz line, the rectified waveform peaks occur at what frequency?

Q248. A transformer has a turns ratio of 10:1 (primary:secondary). If the primary is 240 V, the secondary is closest to:

Correct Answer: A
Explanation: For an ideal transformer, V_s/V_p = N_s/N_p, so V_s = 240×(1/10) ≈ 24.00 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer voltage ratio (V ∝ turns).

Q249. An inductor of 40 mH carries 0.20 A. Stored energy is closest to:

Correct Answer: A
Explanation: E = ½LI². With L=0.040 H and I=0.20 A, E ≈ 0.000800 J = 0.80 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductor energy (E = ½LI²).

Q250. At a fixed capacitance, if frequency is doubled, the capacitive reactance X_C will:

Correct Answer: A
Explanation: X_C = 1/(2πfC). Doubling f halves X_C.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance dependence (X_C ∝ 1/f).

Q251. A resistor with bands red-red-brown (ignore tolerance) is:

Q252. A NOT gate (inverter) outputs HIGH when the input is:

Correct Answer: A
Explanation: NOT inverts: input HIGH becomes LOW.
Citation: ASVAB content domain — Electronics Information; underlying principle: logic gate truth tables (NOT).

Q253. The binary value 100111₂ equals what in decimal?

Correct Answer: A
Explanation: Sum powers of 2 where bits are 1. 100111₂ equals 39₁₀.
Citation: ASVAB content domain — Electronics Information; underlying principle: binary place value conversion.

Q254. An inverting op-amp uses R_in = 1 kΩ and R_f = 47 kΩ. Gain is closest to:

Correct Answer: A
Explanation: In an ideal inverting op-amp, A_v = -R_f/R_in = -47/1 ≈ -47.00.
Citation: ASVAB content domain — Electronics Information; underlying principle: inverting op-amp gain (A_v = -R_f/R_in).

Q255. A BJT in cutoff is best described as:

Correct Answer: A
Explanation: Cutoff means base-emitter is not forward biased, so collector current is approximately zero.
Citation: ASVAB content domain — Electronics Information; underlying principle: BJT switching regions (cutoff).

Q256. If voltage is reduced to half (same impedance), the change is closest to:

Q257. A signal has frequency 400 Hz. Its period is closest to:

Correct Answer: A
Explanation: T = 1/f. For 400 Hz, T = 0.0025 s = 2.50 ms.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (T = 1/f).

Q258. A capacitor of 470 µF is charged to 5.0 V. Stored energy is closest to:

Correct Answer: A
Explanation: E = ½CV². With C=0.000470 F and V=5.0 V, E ≈ 0.005875 J = 5.88 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor energy (E = ½CV²).

Q259. A 5 V supply drives one LED (assume 2.0 V) at 10 mA. The resistor is closest to:

Correct Answer: A
Explanation: First find resistor voltage: V_R = V_s − n·V_f = 5.0 − 1×2.0 = 3.0 V. Then R = V_R/I ≈ 3.0/0.010 ≈ 300.0 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: LED current limiting (R = (V_s − ΣV_f)/I).

Q260. A band-pass filter is designed to:

Correct Answer: A
Explanation: Band-pass filters pass a middle band and attenuate frequencies outside the band.
Citation: ASVAB content domain — Electronics Information; underlying principle: filter types (band-pass behavior).

Q261. A resistor has 6.0 V across it and 0.20 A through it. Its resistance is closest to:

Correct Answer: A
Explanation: Use Ohm’s law: R = V/I. With V=6 V and I=0.2 A, R ≈ 30.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law (R = V/I).

Q262. A 4.7 Ω resistor is placed across 6.0 V. Approximately how much power is dissipated?

Correct Answer: A
Explanation: With voltage across a resistor, power is P = V²/R = 6.0²/4.7 ≈ 7.66 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = V²/R).

Q263. Resistors 15 Ω and 60 Ω are in parallel. The equivalent resistance is closest to:

Correct Answer: A
Explanation: For two resistors in parallel: 1/Req = 1/R1 + 1/R2. Here Req ≈ 12.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel resistance (reciprocal sum).

Q264. A 7 Ω resistor is in series with a parallel pair of 14 Ω and 28 Ω. Total resistance is closest to:

Correct Answer: A
Explanation: Combine the parallel pair first: R|| = (R1·R2)/(R1+R2) ≈ 9.33 Ω, then add the series resistor: R_total ≈ 16.33 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: series-parallel reduction (parallel then series).

Q265. A simple RC filter has R = 47 kΩ and C = 0.0022 µF. Cutoff frequency is closest to:

Correct Answer: A
Explanation: Cutoff frequency is f_c = 1/(2πRC). Substituting values gives f_c ≈ 1539.2 Hz.
Citation: ASVAB content domain — Electronics Information; underlying principle: RC cutoff frequency (f_c = 1/(2πRC)).

Q266. For quick estimates, the forward drop of a silicon PN junction diode is taken as about:

Q267. A bridge rectifier on a 50 Hz source produces pulsating DC at approximately:

Q268. An ideal transformer has 150 primary turns and 75 secondary turns with 18 V on the primary. Secondary voltage is closest to:

Correct Answer: A
Explanation: For an ideal transformer, V_s/V_p = N_s/N_p, so V_s = 18×(75/150) ≈ 9.00 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer voltage ratio (V ∝ turns).

Q269. A 12 mH inductor carries 1.5 A. The energy stored is closest to:

Correct Answer: A
Explanation: E = ½LI². With L=0.012 H and I=1.50 A, E ≈ 0.013500 J = 13.50 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductor energy (E = ½LI²).

Q270. At a fixed inductance, if frequency is doubled, the inductive reactance X_L will:

Correct Answer: A
Explanation: X_L = 2πfL. Doubling f doubles X_L.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductive reactance dependence (X_L ∝ f).

Q271. A resistor with bands green-blue-brown (ignore tolerance) has a resistance of:

Q272. For a NAND gate, the output is LOW only when:

Correct Answer: A
Explanation: NAND is the inverse of AND, so it is LOW only when both inputs are HIGH.
Citation: ASVAB content domain — Electronics Information; underlying principle: logic gate truth tables (NAND).

Q273. Convert 111000₂ to decimal.

Correct Answer: A
Explanation: Sum powers of 2 where bits are 1. 111000₂ equals 56₁₀.
Citation: ASVAB content domain — Electronics Information; underlying principle: binary place value conversion.

Q274. An inverting op-amp has R_in = 10 kΩ and R_f = 100 kΩ. Gain is closest to:

Correct Answer: A
Explanation: In an ideal inverting op-amp, A_v = -R_f/R_in = -100/10 ≈ -10.00.
Citation: ASVAB content domain — Electronics Information; underlying principle: inverting op-amp gain (A_v = -R_f/R_in).

Q275. For a BJT switch, ‘cutoff’ means:

Correct Answer: A
Explanation: Cutoff corresponds to the transistor being off due to insufficient base-emitter forward bias.
Citation: ASVAB content domain — Electronics Information; underlying principle: BJT switching regions (cutoff condition).

Q276. A voltage gain of 5 (same impedance) is closest to how many dB?

Q277. A waveform repeats every 0.002 s. Its frequency is closest to:

Correct Answer: A
Explanation: f = 1/T. If T = 0.002 s, f = 500 Hz.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (f = 1/T).

Q278. A 1.0 µF capacitor is charged to 200 V. Stored energy is closest to:

Correct Answer: A
Explanation: E = ½CV². With C=0.000001 F and V=200.0 V, E ≈ 0.020000 J = 20.00 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor energy (E = ½CV²).

Q279. A 24 V supply drives four LEDs (assume 3.0 V each) at 25 mA. The resistor is closest to:

Correct Answer: A
Explanation: First find resistor voltage: V_R = V_s − n·V_f = 24.0 − 4×3.0 = 12.0 V. Then R = V_R/I ≈ 12.0/0.025 ≈ 480.0 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: LED current limiting (R = (V_s − ΣV_f)/I).

Q280. A notch (band-stop) filter is designed to:

Correct Answer: A
Explanation: A notch filter attenuates a selected band while passing others.
Citation: ASVAB content domain — Electronics Information; underlying principle: filter types (band-stop/notch behavior).

Q281. A load draws 2.4 A from a 36 V source. The load resistance is closest to:

Correct Answer: A
Explanation: Use Ohm’s law: R = V/I. With V=36 V and I=2.4 A, R ≈ 15.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: Ohm’s law (R = V/I).

Q282. A 22 Ω resistor is connected across 48 V. Approximately how much power is dissipated?

Correct Answer: A
Explanation: With voltage across a resistor, power is P = V²/R = 48.0²/22.0 ≈ 104.73 W.
Citation: ASVAB content domain — Electronics Information; underlying principle: power in resistors (P = V²/R).

Q283. Two resistors 20 Ω and 5 Ω are in parallel. The equivalent resistance is closest to:

Correct Answer: A
Explanation: For two resistors in parallel: 1/Req = 1/R1 + 1/R2. Here Req ≈ 4.00 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: parallel resistance (reciprocal sum).

Q284. A 5 Ω resistor is in series with a parallel pair of 10 Ω and 20 Ω. Total resistance is closest to:

Correct Answer: A
Explanation: Combine the parallel pair first: R|| = (R1·R2)/(R1+R2) ≈ 6.67 Ω, then add the series resistor: R_total ≈ 11.67 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: series-parallel reduction (parallel then series).

Q285. An RC network has R = 1.0 kΩ and C = 0.22 µF. The cutoff frequency is closest to:

Q286. A silicon diode in forward conduction typically drops roughly:

Q287. A full-wave rectifier fed by 60 Hz produces a pulsating output at:

Q288. A transformer has a 1:4 turns ratio (primary:secondary). If the primary is 6 V, the secondary is closest to:

Correct Answer: A
Explanation: For an ideal transformer, V_s/V_p = N_s/N_p, so V_s = 6×(4/1) ≈ 24.00 V.
Citation: ASVAB content domain — Electronics Information; underlying principle: ideal transformer voltage ratio (V ∝ turns).

Q289. An inductor of 2.0 mH carries 3.0 A. Stored energy is closest to:

Correct Answer: A
Explanation: E = ½LI². With L=0.002 H and I=3.00 A, E ≈ 0.009000 J = 9.00 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: inductor energy (E = ½LI²).

Q290. At a fixed frequency, if capacitance is halved, the capacitive reactance X_C will:

Correct Answer: A
Explanation: X_C = 1/(2πfC). Halving C halves the denominator, so X_C doubles.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitive reactance dependence (X_C ∝ 1/C).

Q291. A resistor with bands orange-orange-brown (ignore tolerance) is:

Q292. For an XOR gate, the output is HIGH when the inputs are:

Correct Answer: A
Explanation: XOR outputs HIGH when the inputs are different.
Citation: ASVAB content domain — Electronics Information; underlying principle: logic gate truth tables (XOR).

Q293. The binary number 010101₂ equals what decimal value?

Correct Answer: A
Explanation: Sum powers of 2 where bits are 1. 010101₂ equals 21₁₀.
Citation: ASVAB content domain — Electronics Information; underlying principle: binary place value conversion.

Q294. An inverting op-amp uses R_in = 3.3 kΩ and R_f = 33 kΩ. Gain is closest to:

Correct Answer: A
Explanation: In an ideal inverting op-amp, A_v = -R_f/R_in = -33/3.3 ≈ -10.00.
Citation: ASVAB content domain — Electronics Information; underlying principle: inverting op-amp gain (A_v = -R_f/R_in).

Q295. A BJT operating in saturation behaves most like a:

Correct Answer: A
Explanation: A saturated BJT behaves like a closed switch with a small collector-emitter voltage.
Citation: ASVAB content domain — Electronics Information; underlying principle: BJT as a switch (saturation ≈ closed switch).

Q296. If voltage increases by a factor of 3 (same impedance), the change is closest to:

Q297. A signal has a period of 20 µs. Its frequency is closest to:

Correct Answer: A
Explanation: f = 1/T. With T = 20 µs, f = 50,000 Hz = 50.0 kHz.
Citation: ASVAB content domain — Electronics Information; underlying principle: frequency–period relationship (f = 1/T).

Q298. A capacitor of 22 µF is charged to 24 V. Stored energy is closest to:

Correct Answer: A
Explanation: E = ½CV². With C=0.000022 F and V=24.0 V, E ≈ 0.006336 J = 6.34 mJ.
Citation: ASVAB content domain — Electronics Information; underlying principle: capacitor energy (E = ½CV²).

Q299. A 6 V source drives two LEDs (assume 1.8 V each) at 30 mA. The resistor is closest to:

Correct Answer: A
Explanation: First find resistor voltage: V_R = V_s − n·V_f = 6.0 − 2×1.8 = 2.4 V. Then R = V_R/I ≈ 2.4/0.030 ≈ 80.0 Ω.
Citation: ASVAB content domain — Electronics Information; underlying principle: LED current limiting (R = (V_s − ΣV_f)/I).

Q300. Compared to a low-pass, a high-pass filter mainly differs in that it:

Correct Answer: A
Explanation: A high-pass is defined by attenuating low frequencies more than high frequencies.
Citation: ASVAB content domain — Electronics Information; underlying principle: filter types (high-pass vs low-pass).