22 Basic Electricity Practice Questions & Answers
Every Basic Electricity practice question from the FAA A&P Mechanic (General) Practice Test, with the correct answer and a short explanation.
Start practice test →1. A landing light rated at 100 watts operates on a 28-volt DC bus. What is the operating resistance of the lamp?
- A.28 ohms
- B.3.57 ohms
- C.2,800 ohms
- D.7.84 ohms✓ Answer
Power, voltage and resistance are related by P = E squared divided by R, so R = E squared / P = (28 x 28) / 100 = 784 / 100 = 7.84 ohms. The value 3.57 is the current in amperes (I = P/E), not the resistance, and is the most common wrong pick.
Source: FAA-H-8083-30, Ch. 12 (Fundamentals of Electricity and Electronics) — Ohm's Law / power formulasReport a problem with this question
2. Three resistors of 4 ohms, 8 ohms, and 12 ohms are connected in series across a 24-volt source. What is the voltage drop across the 8-ohm resistor?
- A.3 volts
- B.8 volts✓ Answer
- C.12 volts
- D.24 volts
In a series circuit the same current flows through every element. Total resistance is 4 + 8 + 12 = 24 ohms, so I = 24 V / 24 ohms = 1 ampere, and the drop across the 8-ohm resistor is E = I x R = 1 A x 8 ohms = 8 volts. Kirchhoff's voltage law is satisfied because the drops 4 + 8 + 12 add up to the 24-volt source.
Source: FAA-H-8083-30, Ch. 12 — series DC circuits and Kirchhoff's voltage lawReport a problem with this question
3. Three lamps are wired in series across a 28-volt DC source. One lamp filament opens. With the circuit switched on, a voltmeter placed across the terminals of the failed lamp will indicate approximately what value?
- A.9.3 volts
- B.14 volts
- C.28 volts✓ Answer
- D.0 volts
An open anywhere in a series circuit stops all current flow, so no current passes through the remaining good lamps and they develop no voltage drop. With no drops elsewhere in the loop, the entire source voltage appears across the open point — 28 volts. This is why a voltmeter is the fastest tool for locating an open in a series string.
Source: FAA-H-8083-30, Ch. 12 — troubleshooting opens in series circuitsReport a problem with this question
4. Five 10-ohm resistors are connected in parallel across a 28-volt battery. What is the voltage across the third resistor?
- A.14 volts
- B.2.8 volts
- C.5.6 volts
- D.28 volts✓ Answer
Voltage is common to every branch of a parallel circuit because each branch is connected directly across the source, so all five resistors see the full 28 volts. Only the current divides: 2.8 amperes per branch and 14 amperes total, with a total resistance of 2 ohms. Dividing the source voltage by the number of branches is the classic error on this item.
Source: FAA-H-8083-30, Ch. 12 — parallel DC circuits (voltage common to all branches)Report a problem with this question
5. A 20-volt source supplies three resistors connected in parallel: 10 ohms, 20 ohms, and 20 ohms. What total current does the source deliver?
- A.4 amperes✓ Answer
- B.0.4 ampere
- C.2 amperes
- D.10 amperes
Each branch has the full 20 volts across it, so the branch currents are 20/10 = 2 A, 20/20 = 1 A, and 20/20 = 1 A. Kirchhoff's current law states that the current leaving the source equals the sum of the branch currents, 2 + 1 + 1 = 4 amperes, which corresponds to a total resistance of 5 ohms — always less than the smallest branch resistor.
Source: FAA-H-8083-30, Ch. 12 — parallel circuits and Kirchhoff's current lawReport a problem with this question
6. A 24-volt source delivers 144 watts to four resistors of equal value connected in parallel. What is the resistance of each individual resistor?
- A.16 ohms✓ Answer
- B.64 ohms
- C.6 ohms
- D.4 ohms
Total current is I = P/E = 144 / 24 = 6 amperes, so total circuit resistance is 24 / 6 = 4 ohms. Because the four branches are equal, each carries 6 / 4 = 1.5 amperes and each resistor equals 24 V / 1.5 A = 16 ohms. The 4-ohm answer is the total resistance of the network, not the value of one resistor — the standard trap in this two-step problem.
Source: FAA-H-8083-30, Ch. 12 — Watt's Law combined with parallel circuit analysisReport a problem with this question
7. R1 is a 12-ohm resistor connected in series with a parallel group of three 6-ohm resistors, and the combination is connected across a 28-volt source. What is the total resistance of the circuit?
- A.2 ohms
- B.18 ohms
- C.14 ohms✓ Answer
- D.30 ohms
A series-parallel circuit is solved by reduction: three equal 6-ohm resistors in parallel equal 6 / 3 = 2 ohms. That equivalent resistance is in series with R1, so the total is 12 + 2 = 14 ohms, giving a total current of 28 / 14 = 2 amperes. Adding all four values as though everything were in series produces the incorrect 30 ohms.
Source: FAA-H-8083-30, Ch. 12 — series-parallel (combination) circuit analysisReport a problem with this question
8. In a circuit fed by a 30-volt source, R1 (10 ohms) is in series with a parallel combination of R2 (20 ohms) and R3 (20 ohms). How much current flows through R2?
- A.0.75 ampere✓ Answer
- B.1.5 amperes
- C.0.5 ampere
- D.3 amperes
R2 and R3 in parallel equal 10 ohms, so total resistance is 10 + 10 = 20 ohms and total current is 30 / 20 = 1.5 amperes. That current produces a 15-volt drop across R1, leaving 15 volts across the parallel group, so R2 carries 15 V / 20 ohms = 0.75 ampere — exactly half the total current, as expected with two equal branches.
Source: FAA-H-8083-30, Ch. 12 — series-parallel circuits, branch current determinationReport a problem with this question
9. A bus feeds four independent loads wired in parallel. If the wire to one load breaks open, what happens to the rest of the circuit?
- A.The other three loads keep operating, but at reduced voltage
- B.All four loads stop operating because the circuit is broken
- C.Total resistance decreases and total current from the source increases
- D.The other three loads keep operating at the same voltage, and total current from the source decreases✓ Answer
Each parallel branch is an independent path connected across the source, so losing one branch does not interrupt the others and they still receive the full bus voltage. Removing a current path raises the total resistance of the network, and by Ohm's law a higher total resistance at the same applied voltage means the source delivers less total current.
Source: FAA-H-8083-30, Ch. 12 — effect of an open in a parallel circuitReport a problem with this question
10. A coil in an AC circuit opposes current flow because the changing current generates a counter-emf. What is this opposition called, and what happens to it if the frequency of the applied AC is increased?
- A.Resistance, and it remains unchanged
- B.Inductive reactance, and it increases✓ Answer
- C.Inductive reactance, and it decreases
- D.Capacitive reactance, and it increases
Opposition to alternating current produced by the magnetic field of a coil and its self-induced counter-emf is inductive reactance, XL = 2 x pi x f x L, measured in ohms. Because frequency appears in the numerator, raising the frequency raises XL. Capacitive reactance, Xc = 1 / (2 x pi x f x C), behaves in the opposite way and falls as frequency rises.
Source: FAA-H-8083-30, Ch. 12 — inductance and inductive reactanceReport a problem with this question
11. Which change would decrease the resistance of a given length of copper aircraft wire?
- A.Raising its temperature
- B.Replacing it with aluminum wire of the same size
- C.Increasing its cross-sectional area✓ Answer
- D.Increasing its length
The four factors that determine a conductor's resistance are the material, the length, the cross-sectional area, and the temperature. Resistance varies inversely with cross-sectional area, so a larger conductor (a lower AWG number) has less resistance, while greater length, higher temperature, or a less conductive material such as aluminum all increase it.
Source: FAA-H-8083-30, Ch. 12 — factors affecting the resistance of a conductorReport a problem with this question
12. Why is the allowable current-carrying capacity of an aircraft wire lower when it is installed in a bundle or conduit than when it is run singly in free air?
- A.Bundled wire always has a greater voltage drop per foot than the same wire in free air
- B.Heat generated in the wire cannot dissipate as easily, so the conductor and insulation run hotter for the same current✓ Answer
- C.The magnetic fields of adjacent wires add resistance to the conductor
- D.Bundling raises the inductive reactance of the circuit
Current-carrying capacity is limited by conductor temperature rise. A wire inside a bundle or conduit is surrounded by other heat sources and has far less exposed surface for convection and radiation, so the same current drives the insulation temperature higher and the wire must be derated. Wire size must then also be checked against the maximum allowable voltage drop for the length of the run, and the larger of the two required sizes is used.
Source: FAA-H-8083-30, Ch. 12 — wire size selection (current-carrying capacity, bundle derating, allowable voltage drop); AC 43.13-1, Ch. 11Report a problem with this question
13. A circuit breaker or fuse installed in an aircraft electrical circuit is selected primarily to protect what?
- A.The battery and its contactor
- B.The equipment or component the circuit feeds
- C.The switch that controls the circuit
- D.The wiring of the circuit✓ Answer
Protective devices are rated and located so they open before the circuit's conductors can overheat, so what they protect is the wiring, not the equipment; protection of the equipment itself, when required, is provided separately inside the unit. That is why the rating of the fuse or breaker is derived from the wire gauge and its routing rather than from the load's own current rating.
Source: FAA-H-8083-30, Ch. 12 — circuit protection devices; AC 43.13-1, Ch. 11Report a problem with this question
14. Which statement about circuit breakers used in aircraft electrical systems is correct?
- A.A breaker may be reset as many times as necessary as long as it holds for a few seconds
- B.Holding the breaker control in the closed position will safely keep an overloaded circuit energized
- C.Automatic-reset breakers are preferred because they restore essential circuits without crew action
- D.They must be trip-free, and a breaker that trips repeatedly must be troubleshot rather than reset again✓ Answer
Aircraft circuit breakers must be trip-free, meaning the contacts open on an overload even if the operating control is held in the ON position, and they are the resettable alternative to a fuse. A repeated trip indicates a genuine fault in the wiring, and resetting into that fault risks overheating the conductor and starting a fire, so the circuit must be troubleshot; automatic-reset breakers are not acceptable for aircraft use because they would re-energize a faulted circuit unattended.
Source: FAA-H-8083-30, Ch. 12 — circuit breakers (trip-free requirement, automatic-reset prohibited)Report a problem with this question
15. Which statement correctly describes how electrical meters are connected and how their ranges are extended?
- A.An ammeter must have high internal resistance so that it does not short the source
- B.An ammeter is connected in parallel with the load and uses a multiplier resistor to extend its range
- C.An ammeter is connected in series with the load and uses a low-resistance shunt in parallel with the movement to extend its range✓ Answer
- D.A voltmeter is connected in series with the load and must have very low internal resistance
Because an ammeter must carry the circuit current, it is inserted in series and built with very low internal resistance; a shunt of precisely known low resistance is placed in parallel with the meter movement so most of the current bypasses the movement, which extends the range. A voltmeter is the opposite: it is connected across (in parallel with) the component, has high internal resistance, and uses a series multiplier resistor to extend its range.
Source: FAA-H-8083-30, Ch. 12 — measuring instruments (ammeter shunts, voltmeter multipliers)Report a problem with this question
16. Why must an ohmmeter never be connected to a circuit that is energized?
- A.The ohmmeter's high internal resistance will overload the circuit
- B.Ohmmeters read accurately only on circuits above 28 volts
- C.The circuit's current will recharge the ohmmeter battery and throw off the scale
- D.The ohmmeter supplies its own test voltage from an internal battery, so any external voltage gives a false reading and can damage the meter✓ Answer
An ohmmeter measures resistance by sending a small current from its own internal battery through the component and reading the resulting deflection. Any external source voltage adds to or opposes that current, producing a meaningless reading and possibly burning out the movement, so the circuit must be de-energized and the component isolated from parallel paths first. Insulation resistance is measured instead with a megohmmeter (megger), which applies its own high test voltage.
Source: FAA-H-8083-30, Ch. 12 — ohmmeter and megohmmeter useReport a problem with this question
17. As a lead-acid aircraft battery discharges, what happens to the specific gravity of the electrolyte, and when should distilled water be added?
- A.Specific gravity decreases as the battery discharges, and water should be added when the battery is fully charged✓ Answer
- B.Specific gravity increases as the battery discharges, and water should be added just before charging
- C.Specific gravity decreases, and sulfuric acid rather than water should be added to restore it
- D.Specific gravity is unaffected by state of charge, and water may be added at any time
During discharge the sulfuric acid in the electrolyte combines with the plate material and the solution becomes more like water, so its specific gravity falls (roughly 1.275-1.300 fully charged down to about 1.150 discharged) and a hydrometer therefore indicates state of charge. Water is added only when the battery is fully charged, because the electrolyte level is highest then and topping off earlier would cause overflow and loss of acid; only distilled water is added, never acid.
Source: FAA-H-8083-30, Ch. 12 — lead-acid battery servicing and hydrometer readingsReport a problem with this question
18. Why can the state of charge of a nickel-cadmium battery not be determined with a hydrometer?
- A.The specific gravity of the potassium hydroxide electrolyte does not change appreciably with state of charge✓ Answer
- B.The plates consume the electrolyte during discharge, leaving nothing to test
- C.The electrolyte is too thick to be drawn into a hydrometer
- D.The specific gravity changes so rapidly that no stable reading can be taken
In a nickel-cadmium cell the potassium hydroxide electrolyte acts only as a conductor for ion transfer and is not chemically consumed the way sulfuric acid is in a lead-acid cell, so its specific gravity stays essentially constant and reveals nothing about charge; state of charge is determined instead by a measured discharge (capacity) check. Note also that the electrolyte level is lowest at discharge and highest at full charge, so water is added only when the battery is fully charged, and Ni-Cd and lead-acid batteries must never be serviced in the same area or with the same tools.
Source: FAA-H-8083-30, Ch. 12 — nickel-cadmium battery servicing (KOH electrolyte, capacity check, segregation from lead-acid)Report a problem with this question
19. Which type of DC motor develops the greatest starting torque and is therefore used for engine starters and landing gear actuators?
- A.Shunt-wound motor
- B.Compound-wound motor
- C.Split-phase motor
- D.Series-wound motor✓ Answer
In a series-wound motor the field winding carries the full armature current, so at the instant of starting the very large inrush current produces both a strong field and a strong armature current, and torque — which is proportional to their product — is at its maximum. The trade-off is poor speed regulation: with no load its counter-emf cannot limit speed and the motor can run away, so a series motor must always be connected to a load.
Source: FAA-H-8083-30, Ch. 12 — DC motors (series, shunt and compound characteristics, counter-emf)Report a problem with this question
20. An AC voltmeter connected to an aircraft AC bus indicates 115 volts. What is the peak value of that voltage?
- A.162.6 volts✓ Answer
- B.230 volts
- C.73.3 volts
- D.81.3 volts
Unless stated otherwise, AC voltage and current values — and the readings of AC meters — are effective (RMS) values. Peak equals RMS x 1.414, so 115 x 1.414 = 162.6 volts. Multiplying by 0.707 gives 81.3 volts, which converts peak to RMS (the reverse of what was asked), and 0.637 x peak gives the average value.
Source: FAA-H-8083-30, Ch. 12 — values of alternating current (peak, effective/RMS, average)Report a problem with this question
21. A transformer has 200 turns on the primary and 50 turns on the secondary. If 120 volts AC is applied to the primary, what is the secondary voltage, and does the secondary frequency change?
- A.480 volts, and the frequency is unchanged
- B.60 volts, and the frequency is unchanged
- C.30 volts, and the frequency is unchanged✓ Answer
- D.30 volts, and the frequency is reduced to one-fourth
Secondary voltage follows the turns ratio, Ep/Es = Np/Ns, so Es = 120 x (50 / 200) = 30 volts — a step-down transformer, which lowers voltage while raising the available current. A transformer operates by mutual induction between windings and simply reproduces the primary waveform in the secondary, so it cannot change the frequency of the applied AC.
Source: FAA-H-8083-30, Ch. 12 — transformers and the turns ratioReport a problem with this question
22. A silicon diode can be used as a rectifier because it does what?
- A.Stores AC energy in an electrostatic field and releases it as DC
- B.Increases in resistance as the applied voltage increases, smoothing the output
- C.Conducts when forward biased and blocks current when reverse biased, so it passes only one half of the AC waveform✓ Answer
- D.Conducts equally well in both directions but limits current to a safe value
In a PN junction diode, forward bias (positive applied to the P material) collapses the depletion region once about 0.7 volt is reached in silicon and the diode conducts; reverse bias widens the depletion region so essentially no current flows. Applying AC therefore lets only one polarity through, converting alternating current into pulsating direct current, which is the definition of rectification.
Source: FAA-H-8083-30, Ch. 12 — semiconductors, PN junction diodes and rectifiersReport a problem with this question
Practice questions based on 14 CFR Parts 43 and 65 and the FAA Aviation Maintenance Technician Handbook—General (FAA-H-8083-30). This site is not affiliated with or endorsed by the Federal Aviation Administration. This bank covers the GENERAL written test only; the Airframe and Powerplant written tests and the oral and practical exams are separate. Torque values, servicing quantities, corrosion rework limits, and inspection criteria always come from the manufacturer's current maintenance data and the applicable airworthiness directives — never from a practice test. Confirm current eligibility and testing requirements with the FAA before you test. About the A&P mechanic certificate →