Electrical Theory Practice Test
This electrical theory practice test has 40 free, original, unofficial questions with worked answers. These are paper-and-pencil theory problems—not a state-specific licensing exam, an NEC code quiz, or instructions for working on energized equipment.
Practice questions
Ohm's law and power
Question 1 of 40
A 12 Ω heating element is connected to 120 V. How much current flows through it?
- A. 132 A
- B. 1,440 A
- C. 0.1 A
- D. 10 A
Pick one answer for each question, then open the explanation to check your work. Keep a calculator and scratch paper handy.
Show answer and explanation
Answer: D. 10 A
Use Ohm's law solved for current: I = V ÷ R = 120 V ÷ 12 Ω = 10 A. B multiplies voltage by resistance (that math doesn't give current). C divides the wrong way (R ÷ V). A just adds the two numbers.
Question 2 of 40
A water-heater element is rated 4,500 W at 240 V. What is its resistance?
- A. 18.75 Ω
- B. 12.8 Ω
- C. 0.053 Ω
- D. 1,080,000 Ω
Show answer and explanation
Answer: B. 12.8 Ω
When you know power and voltage, use R = V² ÷ P = 240² ÷ 4,500 = 57,600 ÷ 4,500 = 12.8 Ω. A is the element's current in amps (4,500 ÷ 240 = 18.75 A), not its resistance. C divides voltage by power (240 ÷ 4,500 ≈ 0.053), which doesn't give resistance. D multiplies power by voltage (4,500 × 240 = 1,080,000), which doesn't give resistance either.
Source: OpenStax, College Physics 2e, §20.4 Electric Power and Energy
Question 3 of 40
A heater rated 1,500 W at 120 V is mistakenly connected to 240 V. Assuming its resistance stays the same, about how much power does it try to draw?
- A. 1,500 W
- B. 3,000 W
- C. 6,000 W
- D. 750 W
Show answer and explanation
Answer: C. 6,000 W
First find the fixed resistance: R = 120² ÷ 1,500 = 9.6 Ω. At 240 V: P = V² ÷ R = 240² ÷ 9.6 = 6,000 W. Power rises with the square of voltage, so doubling the voltage gives four times the power. B assumes power only doubles. A assumes the rating doesn't change. D halves it. (Real elements heat up and change resistance somewhat; the question tells you to hold it constant.)
Source: OpenStax, College Physics 2e, §20.4 Electric Power and Energy
Question 4 of 40
A 4,500 W element runs steadily for 2 hours 30 minutes. How much energy does it use?
- A. 9 kWh
- B. 11.25 kWh
- C. 11,250 kWh
- D. 1.8 kWh
Show answer and explanation
Answer: B. 11.25 kWh
Energy = power × time. Convert both units first: 4,500 W = 4.5 kW and 2 h 30 min = 2.5 h. So E = 4.5 kW × 2.5 h = 11.25 kWh. A ignores the 30 minutes. C forgets to convert watts to kilowatts (11,250 is watt-hours, not kilowatt-hours). D divides power by time instead of multiplying.
Source: OpenStax, College Physics 2e, §20.4 Electric Power and Energy
Question 5 of 40
10 A flows through a conductor path with a total resistance of 0.2 Ω. How much power is lost as heat in that path?
- A. 2 W
- B. 200 W
- C. 50 W
- D. 20 W
Show answer and explanation
Answer: D. 20 W
Use P = I² × R = 10² × 0.2 = 100 × 0.2 = 20 W. A is the voltage drop across the path (10 A × 0.2 Ω = 2 V), not the power. C divides current by resistance. B squares the current but multiplies by 2 instead of 0.2.
Source: OpenStax, College Physics 2e, §20.4 Electric Power and Energy
Series and parallel circuits
Question 6 of 40
Resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series across 120 V. What is the voltage across the 30 Ω resistor?
- A. 40 V
- B. 60 V
- C. 120 V
- D. 20 V
Show answer and explanation
Answer: B. 60 V
Series resistances add: 10 + 20 + 30 = 60 Ω. Current is I = 120 V ÷ 60 Ω = 2 A, and it's the same through every resistor. Voltage across the 30 Ω resistor is V = 2 A × 30 Ω = 60 V. A splits the voltage equally, which only works if the resistors are equal. C is the full source voltage. D is the drop across the 10 Ω resistor.
Source: OpenStax, College Physics 2e, §21.1 Resistors in Series and Parallel; OpenStax, College Physics 2e, §20.2 Ohm's Law
Question 7 of 40
What is the equivalent resistance of a 6 Ω resistor and a 3 Ω resistor connected in parallel?
- A. 9 Ω
- B. 4.5 Ω
- C. 2 Ω
- D. 0.5 Ω
Show answer and explanation
Answer: C. 2 Ω
For two resistors in parallel, use product over sum: (6 × 3) ÷ (6 + 3) = 18 ÷ 9 = 2 Ω. A adds them like a series circuit. B averages them. D is 1/6 + 1/3 = 0.5, which is the sum of the reciprocals before you flip it back over. Quick check: a parallel total is always smaller than the smallest branch, and 2 Ω is smaller than 3 Ω.
Source: OpenStax, College Physics 2e, §21.1 Resistors in Series and Parallel
Question 8 of 40
A 4 Ω resistor is in series with a parallel pair of 6 Ω and 3 Ω resistors. A 24 V direct-current (DC) source is connected across the whole network. What is the source current?
- A. 4 A
- B. About 1.85 A
- C. 6 A
- D. 12 A
Show answer and explanation
Answer: A. 4 A
Reduce the circuit in steps. The parallel pair is 6 × 3 ÷ 9 = 2 Ω. Add the series resistor: 4 + 2 = 6 Ω total. Source current is 24 V ÷ 6 Ω = 4 A. B treats all three resistors as if they were in series (24 ÷ 13). C uses only the 4 Ω resistor (24 ÷ 4) and leaves out the parallel pair. D uses only the 2 Ω parallel pair (24 ÷ 2) and leaves out the series resistor.
Source: OpenStax, College Physics 2e, §21.1 Resistors in Series and Parallel; OpenStax, College Physics 2e, §20.2 Ohm's Law
Question 9 of 40
Several lamps are wired in series to an ideal 120 V DC source, with no bypass paths. One lamp's filament burns open. What does an ideal voltmeter that draws no current read across the open lamp's terminals?
- A. 0 V
- B. 120 V
- C. 40 V
- D. It can't be measured
Show answer and explanation
Answer: B. 120 V
With the path broken, no current flows. With zero current, the good lamps have zero voltage drop across them (V = I × R = 0). Kirchhoff's voltage law says the drops around the loop still have to add up to the source, so the entire 120 V appears across the open. A is what you'd read across one of the good lamps. C assumes the voltage still divides among the lamps. D is wrong; a voltmeter can read across an open.
Source: OpenStax, College Physics 2e, §21.3 Kirchhoff's Rules; OpenStax, College Physics 2e, §20.2 Ohm's Law; OpenStax, College Physics 2e, §21.4 DC Voltmeters and Ammeters
Question 10 of 40
Another resistive load is connected in parallel with the resistive loads already on a DC circuit. The source voltage stays the same. What happens?
- A. Total current falls
- B. Total resistance and total current both rise
- C. The voltage across each branch drops to share the load
- D. Total resistance falls and total current rises
Show answer and explanation
Answer: D. Total resistance falls and total current rises
Each new parallel path gives current another way through, so the equivalent resistance goes down. With the same voltage and less resistance, total current goes up. Every branch still has the full source voltage across it, which rules out C. A and B get the direction of change backward.
Source: OpenStax, College Physics 2e, §21.1 Resistors in Series and Parallel
Kirchhoff's laws, resistance, and voltage drop
Question 11 of 40
Three resistive loads are in series on a 120 V DC source. Two of them drop 45 V and 30 V. What is the voltage drop across the third load?
- A. 75 V
- B. 45 V
- C. 120 V
- D. 15 V
Show answer and explanation
Answer: B. 45 V
Kirchhoff's voltage law: the drops around a loop add up to the source voltage. 120 − 45 − 30 = 45 V. A adds the two known drops but doesn't subtract them from the source. C is the source itself. D subtracts one drop from the other.
Source: OpenStax, College Physics 2e, §21.3 Kirchhoff's Rules
Question 12 of 40
In a DC circuit, 15 A flows into a junction. Two branches leaving the junction carry 4 A and 6 A. How much current leaves through the third branch?
- A. 25 A
- B. 10 A
- C. 5 A
- D. 11 A
Show answer and explanation
Answer: C. 5 A
Kirchhoff's current law: current in equals current out. 15 = 4 + 6 + x, so x = 5 A. A adds everything together. B is the total of the two known branches. D subtracts only the 4 A branch.
Source: OpenStax, College Physics 2e, §21.3 Kirchhoff's Rules
Question 13 of 40
A conductor is replaced with one made of the same material, at the same temperature, with twice the length and twice the cross-sectional area. How does the new resistance compare?
- A. It doubles
- B. It halves
- C. It quadruples
- D. It stays the same
Show answer and explanation
Answer: D. It stays the same
Resistance follows R = ρL ÷ A. Doubling length doubles R; doubling area halves it. The two changes cancel, so resistance stays the same. A counts only the length. B counts only the area. C multiplies the two changes together instead of letting them cancel. Watch the wording: doubling the area is not the same as doubling the diameter.
Source: OpenStax, College Physics 2e, §20.3 Resistance and Resistivity
Question 14 of 40
A 120 V DC source feeds a load drawing 20 A. The two circuit conductors together have a total resistance of 0.3 Ω. What voltage reaches the load?
- A. 120 V
- B. 117 V
- C. 114 V
- D. 108 V
Show answer and explanation
Answer: C. 114 V
Voltage drop on the conductors is V = I × R = 20 A × 0.3 Ω = 6 V. The load gets 120 − 6 = 114 V. B uses half the resistance (one conductor instead of both). The question already gives the total for both conductors, so don't cut it in half. A ignores the drop. D doubles the drop.
Source: OpenStax, College Physics 2e, §20.2 Ohm's Law; OpenStax, College Physics 2e, §21.3 Kirchhoff's Rules
Question 15 of 40
As a copper conductor gets hotter, what happens to its resistance?
- A. It rises
- B. It falls
- C. It stays constant
- D. It drops to zero
Show answer and explanation
Answer: A. It rises
Metals like copper have a positive temperature coefficient of resistance: as temperature goes up, so does resistance. The hotter metal's atoms vibrate more and get in the way of the moving charges. B describes materials with a negative coefficient, such as some semiconductors, not copper. C ignores copper's temperature dependence; special resistance alloys can have a small temperature coefficient, but that does not make copper's resistance constant. D confuses this with superconductors, which need extreme cold.
Source: OpenStax, College Physics 2e, §20.3 Resistance and Resistivity
Magnetism, inductance, and capacitance
Question 16 of 40
A conductor sits still in a steady, unchanging magnetic field. What voltage is induced in it?
- A. A value equal to the field strength
- B. The maximum possible
- C. It depends on the conductor's metal
- D. Zero
Show answer and explanation
Answer: D. Zero
Faraday's law says induced voltage comes from a change in magnetic flux. If nothing moves and the field doesn't change, the change is zero, so the induced voltage is zero. Generators work because they keep that flux changing. A, B, and C all assume a strong or present field is enough on its own.
Source: OpenStax, College Physics 2e, §23.2 Faraday's Law of Induction: Lenz's Law
Question 17 of 40
Lenz's law says the current from an induced voltage will:
- A. Oppose the change that produced it
- B. Aid the change that produced it
- C. Always be zero in a coil
- D. Depend only on the circuit's resistance
Show answer and explanation
Answer: A. Oppose the change that produced it
Lenz's law: the induced current sets up a magnetic field that opposes the change in flux that caused it. That's why a coil resists sudden changes in current. B is the reverse (it would violate conservation of energy). C and D ignore the change in flux that drives induction.
Source: OpenStax, College Physics 2e, §23.2 Faraday's Law of Induction: Lenz's Law
Question 18 of 40
Two 10 µF capacitors are connected in series. What is the total capacitance?
- A. 20 µF
- B. 10 µF
- C. 5 µF
- D. 100 µF
Show answer and explanation
Answer: C. 5 µF
Capacitors in series combine the way resistors combine in parallel: 1/C = 1/10 + 1/10, so C = 5 µF. For two equal capacitors, it's half of one. A is the parallel answer (capacitors in parallel add). B assumes nothing changes. D multiplies them.
Source: OpenStax, College Physics 2e, §19.6 Capacitors in Series and Parallel
Question 19 of 40
An initially uncharged 10 µF capacitor begins charging through a 100 kΩ resistor from an ideal constant DC supply. What is the time constant, and about how far has the capacitor charged after one time constant?
- A. 0.001 s; about 99% of the supply voltage
- B. 1 s; about 50% of the supply voltage
- C. 1,000 s; about 63% of the supply voltage
- D. 1 s; about 63% of the supply voltage
Show answer and explanation
Answer: D. 1 s; about 63% of the supply voltage
τ = R × C = 100,000 Ω × 0.000010 F = 1 second. After one time constant, the capacitor reaches about 63% of its final voltage. B has the right time but the wrong percentage. C skips the unit conversions (100 × 10 = 1,000). A gets both parts wrong; it takes several time constants, not one, to get near 99%.
Question 20 of 40
Why are transformer and motor cores built from thin, insulated steel laminations instead of one solid block?
- A. To reduce eddy-current loss
- B. To reduce copper (winding) loss
- C. To raise the turns ratio
- D. To reduce windage
Show answer and explanation
Answer: A. To reduce eddy-current loss
A changing magnetic field induces swirling currents, called eddy currents, inside the core metal. Those currents waste power as heat. Thin layers separated by insulation break up the current paths, which cuts eddy-current loss. B is the I²R loss in the copper windings, not the core-current loss that laminations are designed to reduce. C depends on the number of turns, not the core. D is air friction from rotating parts.
Source: OpenStax, College Physics 2e, §23.4 Eddy Currents and Magnetic Damping; Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §10.8 Practical Considerations: Transformers
AC fundamentals
Question 21 of 40
A sinusoidal alternating-current (AC) supply measures 120 V root mean square (RMS). What is its peak voltage, to the nearest volt?
- A. 85 V
- B. 120 V
- C. 170 V
- D. 240 V
Show answer and explanation
Answer: C. 170 V
For a sine wave, peak = RMS × √2 ≈ 120 × 1.414 = about 170 V. A multiplies by 0.707, which converts peak to RMS, the opposite direction. B assumes RMS and peak are the same. D just doubles it (the true peak-to-peak value is about 339 V). This √2 factor applies only to sine waves.
Source: OpenStax, College Physics 2e, §20.5 Alternating Current versus Direct Current
Question 22 of 40
How long does one cycle of a 60 Hz AC waveform last?
- A. About 16.7 ms
- B. 60 ms
- C. About 0.0167 ms
- D. 1 second
Show answer and explanation
Answer: A. About 16.7 ms
Period is the inverse of frequency: T = 1 ÷ f = 1 ÷ (60 Hz) ≈ 0.0167 s = about 16.7 ms. B mistakes the frequency number for a time. C forgets to convert seconds to milliseconds correctly (it's off by 1,000). D is the time for all 60 cycles, not one.
Source: OpenStax, College Physics 2e, §16.2 Period and Frequency in Oscillations
Question 23 of 40
What is the inductive reactance of a 0.2 H coil at 60 Hz?
- A. 12 Ω
- B. About 75.4 Ω
- C. About 0.013 Ω
- D. About 377 Ω
Show answer and explanation
Answer: B. About 75.4 Ω
X_L = 2πfL = 2 × 3.1416 × 60 × 0.2 ≈ 75.4 Ω. A multiplies only f × L and leaves out 2π. C uses the reciprocal, which is the shape of the capacitive reactance formula. D is 2π × 60, which leaves out the inductance.
Source: OpenStax, College Physics 2e, §23.11 Reactance, Inductive and Capacitive
Question 24 of 40
In a purely capacitive circuit in sinusoidal AC steady state, how does current relate to voltage?
- A. Current leads voltage by 180°
- B. Current lags voltage by 90°
- C. Current and voltage are in phase
- D. Current leads voltage by 90°
Show answer and explanation
Answer: D. Current leads voltage by 90°
In a capacitor, current leads voltage by 90°. The capacitor has to take on charge (current) before voltage can build across its plates. B describes an inductor. C describes a pure resistance. A is not the current–voltage relationship of an ideal resistor, inductor, or capacitor. The memory aid "ELI the ICE man" helps: in L (an inductor), E (voltage) comes before I (current); in C (a capacitor), I comes before E.
Source: OpenStax, College Physics 2e, §23.11 Reactance, Inductive and Capacitive
Question 25 of 40
A series circuit in sinusoidal AC steady state has R = 30 Ω, X_L = 60 Ω, and X_C = 20 Ω. What is the impedance magnitude?
- A. 110 Ω
- B. 70 Ω
- C. 50 Ω
- D. 90 Ω
Show answer and explanation
Answer: C. 50 Ω
Inductive and capacitive reactance cancel each other partly: net X = 60 − 20 = 40 Ω. Resistance and reactance combine at right angles, so Z = √(30² + 40²) = √2,500 = 50 Ω. A adds all three. B adds R to the net reactance as if they were in line. D adds R and X_L and ignores X_C.
Source: OpenStax, College Physics 2e, §23.12 RLC Series AC Circuits
AC power and three-phase
Question 26 of 40
A 240 V RMS single-phase AC load draws 20 A RMS at a power factor of 0.8. What is its true (real) power?
- A. 4,800 W
- B. 3,840 W
- C. 2,880 W
- D. 6,000 W
Show answer and explanation
Answer: B. 3,840 W
True power = V × I × power factor = 240 × 20 × 0.8 = 3,840 W. A is V × I, which is the apparent power in volt-amperes (4,800 VA), not watts. C uses 0.6 instead of 0.8. D divides by the power factor (4,800 ÷ 0.8) instead of multiplying by it.
Source: OpenStax, College Physics 2e, §23.12 RLC Series AC Circuits
Question 27 of 40
Reactive power is measured in which unit?
- A. Watts (W)
- B. Volt-amperes (VA)
- C. Kilowatt-hours (kWh)
- D. Volt-amperes reactive (VAR)
Show answer and explanation
Answer: D. Volt-amperes reactive (VAR)
Reactive power is measured in VAR. Watts are for true (real) power. Volt-amperes are for apparent power. Kilowatt-hours measure energy over time, not power at all. Keeping these three power units straight is a common exam trap.
Question 28 of 40
In a balanced sinusoidal three-phase wye (Y) system, the phase-to-neutral voltage is 120 V RMS. What is the line-to-line voltage?
- A. 120 V
- B. About 208 V
- C. 240 V
- D. 360 V
Show answer and explanation
Answer: B. About 208 V
In a wye system, line voltage = phase voltage × √3 = 120 × 1.732 ≈ 208 V. That's the familiar 120/208 V system. A confuses this wye system with the delta rule: in delta, line voltage equals the voltage across a phase winding, not a phase-to-neutral voltage. C simply doubles 120 V, which is the relationship in a single-phase 120/240 V system, not three-phase wye. D multiplies by 3.
Source: Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §7.5 Three-phase Y and Delta Configurations; Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §7.1 Single-Phase Power Systems
Question 29 of 40
A balanced sinusoidal three-phase load has 480 V RMS line-to-line, 20 A RMS line current, and a power factor of 0.9. What is its real power?
- A. About 8.6 kW
- B. About 15.0 kW
- C. About 25.9 kW
- D. About 16.6 kW
Show answer and explanation
Answer: B. About 15.0 kW
Three-phase real power = √3 × V_line × I_line × PF = √3 × 480 × 20 × 0.9 ≈ 14,965 W, or about 15.0 kW. (This comes from three phases × phase power, using the wye or delta line-to-phase relationships.) A uses the single-phase formula with no √3. C multiplies by 3 while still using line values. D leaves out the power factor, which gives apparent power: about 16.6 kVA.
Source: Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §7.5 Three-phase Y and Delta Configurations; OpenStax, College Physics 2e, §23.12 RLC Series AC Circuits
Question 30 of 40
In a balanced sinusoidal three-phase delta load, each phase winding carries 10 A RMS. What is the line current?
- A. 10 A
- B. About 5.8 A
- C. About 17.3 A
- D. 30 A
Show answer and explanation
Answer: C. About 17.3 A
In delta, line current = phase current × √3 = 10 × 1.732 ≈ 17.3 A. A is the wye rule (line current equals phase current in wye). B divides by √3 instead of multiplying. D multiplies by 3.
Transformers
Question 31 of 40
An ideal transformer has 400 primary turns and 100 secondary turns. With 480 V RMS AC applied to the primary, what is the secondary voltage?
- A. 1,920 V
- B. 120 V
- C. 480 V
- D. 380 V
Show answer and explanation
Answer: B. 120 V
Voltage follows the turns ratio: V_s = V_p × (N_s ÷ N_p) = 480 × 100 ÷ 400 = 120 V. That's a 4:1 step-down. A flips the ratio. C assumes no change. D subtracts the number of turns from the voltage.
Question 32 of 40
An ideal AC transformer has a 480 V RMS primary and a 120 V RMS secondary. The secondary supplies 40 A RMS. What is the primary current?
- A. 160 A
- B. 40 A
- C. 10 A
- D. 4 A
Show answer and explanation
Answer: C. 10 A
In an ideal transformer, apparent power in equals apparent power out: 120 V × 40 A = 4,800 VA. Primary current = 4,800 ÷ 480 = 10 A. Current changes opposite to voltage: step the voltage down 4:1 and the current steps up 4:1 on the secondary. A applies the ratio the wrong way. B assumes current doesn't change. D mistakes the 4:1 voltage ratio for a current of 4 A.
Source: OpenStax, College Physics 2e, §23.7 Transformers; Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §11.2 True, Reactive, and Apparent Power
Question 33 of 40
What is the full-load secondary current of a 25 kVA single-phase transformer with a 240 V secondary?
- A. About 60 A
- B. About 104 A
- C. About 208 A
- D. About 10.4 A
Show answer and explanation
Answer: B. About 104 A
For single-phase: I = VA ÷ V = 25,000 ÷ 240 ≈ 104 A. A divides by √3 × 240, which is the three-phase formula wrongly applied to a single-phase transformer. C uses 120 V instead of 240 V. D is a decimal slip.
Source: Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §11.2 True, Reactive, and Apparent Power; OpenStax, College Physics 2e, §20.4 Electric Power and Energy
Question 34 of 40
With winding resistance, supply voltage, and frequency held constant, which transformer loss grows with the square of the load current?
- A. Eddy-current loss
- B. Hysteresis loss
- C. Copper (winding) loss
- D. No loss depends on load current
Show answer and explanation
Answer: C. Copper (winding) loss
Copper loss is the I²R heat in the winding resistance, so it rises with the square of current. Double the load current and copper loss roughly quadruples. A and B are core losses. They come from the changing magnetic flux in the iron and depend mainly on flux density and frequency, rather than following the winding's I²R relationship. D is wrong because copper loss clearly depends on current.
Source: Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §10.8 Practical Considerations: Transformers; OpenStax, College Physics 2e, §20.4 Electric Power and Energy
Question 35 of 40
A conventional current transformer (CT) supplies an ammeter. If its primary conductor carries AC while its secondary circuit is open, what is the main hazard?
- A. The primary current must fall to zero
- B. Dangerously high voltage can develop across the secondary
- C. The supply frequency doubles
- D. The transformer becomes a DC source
Show answer and explanation
Answer: B. Dangerously high voltage can develop across the secondary
An open CT secondary with current in the primary can develop dangerously high voltage. A conventional CT is designed to feed a low-impedance ammeter circuit; an open secondary is not its normal operating condition. A is wrong because opening the secondary does not open the primary conductor. C and D are wrong because opening the secondary does not double the supply frequency or convert AC to DC.
Motors and meters
Question 36 of 40
What is the synchronous speed of a 4-pole induction motor on 60 Hz?
- A. 3,600 rpm
- B. 1,800 rpm
- C. 1,200 rpm
- D. 900 rpm
Show answer and explanation
Answer: B. 1,800 rpm
Synchronous speed N_s = 120 × f ÷ P = 120 × 60 ÷ 4 = 1,800 rpm. A is a 2-pole motor. C is a 6-pole motor. D is an 8-pole motor. More poles means a slower rotating field.
Source: Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §13.7 Tesla Polyphase Induction Motors
Question 37 of 40
A 4-pole, 60 Hz induction motor runs at 1,750 rpm under load. What is its slip?
- A. About 2.8%
- B. About 97.2%
- C. About 2.9%
- D. About 0.028%
Show answer and explanation
Answer: A. About 2.8%
Synchronous speed is 120 × 60 ÷ 4 = 1,800 rpm. Slip = (N_s − N_rotor) ÷ N_s × 100 = (1,800 − 1,750) ÷ 1,800 × 100 ≈ 2.8%. B is rotor speed divided by synchronous speed, the part that isn't slip. C divides by the rotor speed (50 ÷ 1,750) instead of the synchronous speed. D forgets to multiply by 100 to get a percent.
Source: Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §13.7 Tesla Polyphase Induction Motors
Question 38 of 40
Why must a conventional induction motor's rotor run below synchronous speed while supplying steady driving torque?
- A. With no speed difference, no current would be induced in the rotor, so there'd be no torque
- B. Bearing friction always holds it back by a fixed amount
- C. The supply frequency drops under load
- D. The stator windings stop producing a magnetic field at that speed
Show answer and explanation
Answer: A. With no speed difference, no current would be induced in the rotor, so there'd be no torque
An induction motor's rotor current is induced by the rotating stator field cutting across it. If the rotor matched synchronous speed, the field wouldn't be moving relative to the rotor, so no current would be induced and no torque would be produced. While supplying steady driving torque, the rotor runs below synchronous speed. That difference is slip. B is wrong because friction does not impose a fixed speed difference. C is not required: slip occurs with constant supply frequency. D is wrong because the stator field continues rotating at synchronous speed.
Source: Kuphaldt, Lessons in Electric Circuits Vol. II (LibreTexts), §13.7 Tesla Polyphase Induction Motors; OpenStax, College Physics 2e, §23.2 Faraday's Law of Induction: Lenz's Law
Question 39 of 40
In a circuit diagram using a conventional in-line ammeter, how are the ammeter and voltmeter connected to measure a load's current and voltage?
- A. Ammeter in parallel, voltmeter in series
- B. Both in parallel
- C. Both in series
- D. Ammeter in series, voltmeter in parallel
Show answer and explanation
Answer: D. Ammeter in series, voltmeter in parallel
An ammeter goes in series so the load current passes through it; it has very low resistance so it barely changes that current. A voltmeter goes in parallel across the load; it has very high resistance so it draws almost no current. A is backward. It would either short the load through the low-resistance ammeter or block the load current with the high-resistance voltmeter. B and C each misconnect one of the meters.
Source: OpenStax, College Physics 2e, §21.4 DC Voltmeters and Ammeters
Question 40 of 40
A megohmmeter (often called a "megger") is mainly used to measure:
- A. Frequency
- B. Load current
- C. Power factor
- D. Insulation resistance
Show answer and explanation
Answer: D. Insulation resistance
A megohmmeter is a high-voltage ohmmeter used to measure insulation resistance. It's useful for finding high-resistance leakage paths through wet or damaged insulation that an ordinary ohmmeter can miss. Load current (B), power factor (C), and frequency (A) are measured with other instruments: an ammeter, a power-factor meter, and a frequency meter.
Source: Kuphaldt, Lessons in Electric Circuits Vol. I (LibreTexts), §8.7 High Voltage Ohmmeters
Check your score
Count your correct answers in each topic. Every topic has five questions. Award one point per correct answer, with no partial credit; record incorrect and unanswered items separately. Your overall percentage is correct ÷ 40 × 100, not correct ÷ answered.
| Topic | Questions | Your correct |
|---|---|---|
| Ohm's law and power | 1–5 | ___ / 5 |
| Series and parallel circuits | 6–10 | ___ / 5 |
| Kirchhoff's laws, resistance, and voltage drop | 11–15 | ___ / 5 |
| Magnetism, inductance, and capacitance | 16–20 | ___ / 5 |
| AC fundamentals | 21–25 | ___ / 5 |
| AC power and three-phase | 26–30 | ___ / 5 |
| Transformers | 31–35 | ___ / 5 |
| Motors and meters | 36–40 | ___ / 5 |
| Total | 1–40 | ___ / 40 |
Your result shows how you did on these 40 questions. It isn't an official exam score, and it can't tell you whether you'll pass. Five questions per topic is a small sample, so one miss swings a topic a lot. Read a low topic score as "look here first," not as a verdict.
What to do with a miss:
- Rework the question on paper without reading the explanation again. Use the formula sheet below if you need it.
- Say out loud (or write down) why each wrong option is wrong. The explanations are written so you can check yourself.
- Come back to the questions you missed a day or two later. If you only get them right the second time because you remember the letter, that doesn't count. You should be able to explain the steps.
Electrical theory formula sheet
These formulas cover the questions above and closely related calculations. Use consistent units; for the AC power and three-phase formulas, use RMS voltage and current.
| What you're finding | Formula | Use it when |
|---|---|---|
| Voltage, current, resistance (Ohm's law) | V = I × R · I = V ÷ R · R = V ÷ I | Any resistive circuit or part of one (OpenStax §20.2) |
| Power | P = V × I · P = I² × R · P = V² ÷ R | Resistive loads: steady DC, or average AC power using RMS values (OpenStax §20.4) |
| Energy | kWh = kW × hours | Convert watts to kilowatts and minutes to hours first (OpenStax §20.4) |
| Conductor resistance | R = ρ × L ÷ A | Same material and temperature; A is cross-sectional area, not diameter (OpenStax §20.3) |
| Series resistance | R_total = R1 + R2 + R3 … | Same current through every part (OpenStax §21.1) |
| Parallel resistance | 1 ÷ R_total = 1 ÷ R1 + 1 ÷ R2 … (two resistors: R1 × R2 ÷ (R1 + R2)) | Same voltage across every branch (OpenStax §21.1) |
| Kirchhoff's laws | Current in = current out at a junction; the algebraic sum of voltages around a loop is zero | Use signed DC or instantaneous values; do not add out-of-phase AC RMS magnitudes as ordinary numbers (OpenStax §21.3) |
| Capacitors | Series: 1 ÷ C_total = 1 ÷ C1 + 1 ÷ C2 … · Parallel: C_total = C1 + C2 … | The opposite of the resistor rules (OpenStax §19.6) |
| RC time constant | τ = R × C | About 63% of the way to the final value after one τ (Kuphaldt, LibreTexts §16.4); about 99.3% after 5τ (1 − e⁻⁵) |
| Peak and RMS | Peak = RMS × √2 ≈ RMS × 1.414 · RMS = peak ÷ √2 ≈ peak × 0.707 | Sine waves only (OpenStax §20.5) |
| Period | T = 1 ÷ f | f in hertz gives T in seconds (OpenStax §16.2) |
| Reactance | X_L = 2πfL · X_C = 1 ÷ (2πfC) | Sine-wave AC; both in ohms (OpenStax §23.11) |
| Series impedance magnitude | Z = √(R² + (X_L − X_C)²) | Series R, L, C circuits in sinusoidal steady state (OpenStax §23.12) |
| AC power | True power (W) = V × I × PF · Apparent power (VA) = V × I · Reactive power in VAR | Single-phase AC with RMS values (OpenStax §23.12; Kuphaldt §11.2) |
| Wye (Y) | V_line = √3 × V_phase · I_line = I_phase | Balanced sinusoidal three-phase; RMS values (Kuphaldt §7.5) |
| Delta (Δ) | V_line = V_phase · I_line = √3 × I_phase | Balanced sinusoidal three-phase; RMS values (Kuphaldt §7.5) |
| Three-phase power | P = √3 × V_line × I_line × PF | Balanced sinusoidal wye or delta loads; RMS line values; omit PF for apparent power in VA (Kuphaldt §7.5; OpenStax §23.12) |
| Transformer | V_s ÷ V_p = N_s ÷ N_p · I_s ÷ I_p = N_p ÷ N_s | Ideal AC transformer; RMS ratios; power in = power out (OpenStax §23.7) |
| Single-phase full-load current | I = VA ÷ V | Use the rated RMS voltage of the winding you're asking about (Kuphaldt §10.8) |
| Motor synchronous speed | N_s = 120 × f ÷ poles | Rotating-field speed in rpm (Kuphaldt §13.7) |
| Slip (%) | (N_s − N_rotor) ÷ N_s × 100 | Divide by synchronous speed, not rotor speed (Kuphaldt §13.7) |
In these formulas, V is voltage, I is current, R is resistance, f is frequency, C is capacitance, and PF is power factor. L means conductor length in R = ρL/A, but inductance in the reactance formula; ρ is resistivity and A is cross-sectional area. Use henrys for inductance, farads for capacitance, and seconds for time: 1 µF = 10⁻⁶ F, 1 kΩ = 1,000 Ω, and 1 ms = 0.001 s. In transformer ratios, N_p and N_s count primary and secondary turns; in motor formulas, N_s means synchronous speed.
Mistakes these questions are built to catch
Most wrong answers above come from a handful of habits. If you missed a question, check whether one of these was the cause:
- Using V × I as real power on a single-phase AC load. With RMS values, that's apparent power in volt-amperes. Multiply by the power factor to get watts; for a balanced sinusoidal three-phase load using line values, include √3 as well (Question 26, Question 29).
- Forgetting √3 or using it in the wrong place. In wye, √3 goes on the voltage; in delta, it goes on the current. Single-phase gets no √3 at all (Questions 28–30, Question 33).
- Assuming power scales directly with voltage. For a fixed resistance, power scales with voltage squared (Question 3).
- Stopping halfway through a combination circuit. Reduce the parallel group first, then add the series part (Question 8).
- Mixing up the resistor and capacitor rules. Capacitors in series combine like resistors in parallel (Question 18).
- Flipping a ratio. Transformer voltage follows the turns ratio. Current goes the other way (Questions 31–32).
- Skipping unit conversions. Watts to kilowatts, minutes to hours, microfarads to farads (Question 4, Question 19).
How much theory is on a journeyman exam?
It depends on your state, so check the content outline in your own candidate bulletin.
Texas is a useful example because its bulletin breaks it down. The Texas Department of Licensing and Regulation (TDLR) exam, given by PSI, lists these subject areas that name theory:
| Texas exam portion | Scored items | Time | Subject area that names theory | Items in that area |
|---|---|---|---|---|
| Journeyman: NEC Knowledge | 56 (plus 3 unscored) | 130 min | Definitions, Theory, and Plans | 3 |
| Journeyman: Calculations | 24 (plus 2 unscored) | 110 min | Calculations and Theory | 2 |
| Master: NEC Knowledge | 70 (plus 5 unscored) | 150 min | Definitions, Theory, and Plans | 7 |
| Master: Calculations | 30 (plus 3 unscored) | 170 min | Calculations and Theory | 2 |
Source: PSI Texas Electrician Candidate Information Bulletin, content outlines on printed pp. 9–14, checked September 29, 2026.
So in Texas, only a few scored items sit under a theory label. Some of those areas also include code definitions and plan reading. But that count undersells theory's role. The Calculations portions also list subject areas like transformers, motors, generators, and load calculations, and that math draws on the theory you practiced above as well as code-specific calculation rules that this set does not cover. Texas requires 70% on each portion separately, so strong code knowledge can't make up for weak calculations. (PSI content outlines, pp. 9–14)
Two other Texas details: exams are referenced to the 2026 NEC starting September 1, 2026, according to the PSI bulletin's effective-date notice; TDLR's exam reference-materials section also identifies the 2026 edition. You can bring only a silent, battery-operated, non-programmable calculator without a paper tape or an alphabet keyboard (PSI security procedures, p. 4). Other states set their own formats and rules, so confirm yours with your state's board or its testing vendor.
Is this the IBEW apprenticeship aptitude test?
No. The electrical training ALLIANCE aptitude test is a different kind of test. The electrical training ALLIANCE's official sample test page covers Algebra and Functions and Reading Comprehension, not circuit theory. If that's the test you're facing next, practice algebra and reading. Use this set when your course or exam calls for electrical theory.
Sources and how we checked this page
By the Castleport Test Prep Editorial Team. We wrote all 40 questions for this page. We checked every answer and calculation against the cited sections of two open textbooks: OpenStax College Physics 2e and Tony R. Kuphaldt's Lessons in Electric Circuits (LibreTexts edition). The Texas exam figures are drawn from the PSI candidate bulletin, with scored counts calculated by subtracting the unscored items. AI tools assisted with drafting and these source, arithmetic, and consistency checks. These are source and arithmetic checks, not a review by a licensed electrician. You can read more in how we check study content.
Last verified: September 29, 2026 (question principles and calculations; Texas exam counts, time limits, passing requirements, reference edition, and calculator rule; ALLIANCE sample-test subjects).
Teaching sources
- OpenStax, College Physics 2e (2022): §16.2 Period and Frequency · §19.6 Capacitors in Series and Parallel · §20.2 Ohm's Law · §20.3 Resistance and Resistivity · §20.4 Electric Power and Energy · §20.5 AC versus DC · §21.1 Resistors in Series and Parallel · §21.3 Kirchhoff's Rules · §21.4 DC Voltmeters and Ammeters · §23.2 Faraday's Law and Lenz's Law · §23.4 Eddy Currents · §23.7 Transformers · §23.11 Reactance · §23.12 RLC Series AC Circuits
- Tony R. Kuphaldt, Lessons in Electric Circuits (LibreTexts): Vol. I §8.7 High Voltage Ohmmeters · Vol. I §16.4 Time Constant Calculations · Vol. II §7.1 Single-Phase Power Systems · Vol. II §7.5 Three-phase Y and Delta · Vol. II §10.7 Special Transformers · Vol. II §10.8 Practical Considerations: Transformers · Vol. II §11.2 True, Reactive, and Apparent Power · Vol. II §13.7 Polyphase Induction Motors
Exam sources
- PSI Services LLC for TDLR, Texas Electrician Candidate Information Bulletin, accessed September 29, 2026: effective-date notice, cover; security procedures (calculator rule), printed p. 4; content outlines, printed pp. 9–14.
- TDLR, Electrician Exam Information, “Reference Materials,” accessed September 29, 2026.
- electrical training ALLIANCE, Preparing for the Test (sample questions).
Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by PSI, the Texas Department of Licensing and Regulation, the electrical training ALLIANCE, the IBEW, NECA, NFPA, or any state licensing board. Exam, credential, and organization names identify their subjects; trademarks belong to their respective owners. The questions on this page are original practice items, not official exam questions.