Motors, Generators and Transformers Lesson Test

Answer: C

Correct answer: A motor. Ask what goes in and what comes out. Electrical energy from the battery goes in and a turning fan comes out, so the device converts electrical energy to mechanical energy, which is the job of a motor.

• A: A generator does the reverse job, taking in motion and putting out electricity.
• B: A transformer only changes AC voltage levels and has no moving parts.
• C (correct): It takes in electrical energy from the battery and delivers mechanical energy as a spinning fan.
• D: A rectifier converts AC to DC with diodes; it does not produce motion.

Answer: A

Correct answer: A generator. Ask what goes in and what comes out. The engine supplies mechanical energy to the shaft, and the machine delivers electrical energy, so it is a generator.

• A (correct): Mechanical energy from the engine goes in and electrical energy comes out, which is the job of a generator.
• B: A transformer has no moving parts and only changes the level of an AC voltage that is already present.
• C: A rectifier only converts AC to DC; it cannot turn shaft motion into electricity.
• D: A motor does the reverse job: it takes in electrical energy and produces shaft motion.

Answer: B

Correct answer: The primary winding. Transformer windings are named by role. The input coil, connected to the source, is the primary; the output coil, connected to the load, is the secondary.

• A: The secondary is the output winding, the one connected to the load.
• B (correct): The winding that receives power from the source is named the primary, whatever its number of turns.
• C: Stator is a term for the stationary part of a motor or generator, not a transformer winding.
• D: Cathode is a term used for diodes and batteries, not for transformer coils.

Answer: C

Correct answer: 60 Hz. The secondary voltage is induced by core flux that changes at exactly the rate set by the primary. A transformer changes voltage level, not frequency, so a 60 Hz input gives a 60 Hz output.

• A: An output of 0 Hz would be DC, and a transformer cannot turn AC into DC.
• B: Stepping the voltage down does not divide the frequency; the secondary follows the primary flux exactly.
• C (correct): The core flux rises and falls at the primary's rate, so the induced secondary voltage alternates at the same 60 Hz.
• D: A transformer does not double frequency; changing frequency needs other equipment.

Answer: C

Correct answer: Two continuous slip rings with brushes. Slip rings are two separate, unbroken rings, each tied to one end of the coil. Because the connection never flips, the alternating voltage from the coil reaches the load as AC, which is how an alternator works.

• A: A laminated core reduces eddy currents in transformers; it is not a sliding connection to a rotating coil.
• B: A split-ring commutator reverses the connections each half turn, which turns the output into one-direction pulsating DC.
• C (correct): Each slip ring stays joined to one end of the coil, so the external circuit sees the coil's alternating output unchanged.
• D: A rectifier converts AC to DC, which is the opposite of what is wanted here.

Answer: B

Correct answer: To reverse the rotor coil current every half turn so the torque keeps one direction. Fed with steady DC, a rotor coil would turn until its sides faced the poles and then be pushed back. The segmented commutator, with brushes sliding across it, reverses the coil current every half turn so the force keeps turning the rotor the same way.

• A: Raising voltage is a transformer's job, and a transformer needs AC; the commutator does not change voltage.
• B (correct): Switching the coil connections at the right moment keeps the coil from rocking back, so the rotor keeps turning.
• C: That is the job of the bearings, not the commutator.
• D: Running current is limited by back emf, not the commutator; at start-up there is little back emf to limit it.

Answer: A

Correct answer: Its secondary has more turns than its primary. In an ideal transformer, voltage is proportional to turns. A step-up transformer has more secondary turns than primary turns, so its output voltage is higher, while its output current is lower.

• A (correct): Voltage follows the turns ratio, so more secondary turns give a higher output voltage.
• B: Fewer secondary turns give a lower output voltage, which describes a step-down transformer.
• C: Power out cannot exceed power in, so when voltage rises the current must fall.
• D: A transformer's output frequency always equals its input frequency.

Answer: B

Correct answer: 600 Wh. Energy is power multiplied by time in hours.

That is 0.6 kWh.

• A: This adds watts and hours, which are different kinds of quantities and cannot be added.
• B (correct): Energy equals power times time in hours: 150 W times 4 h is 600 Wh.
• C: This divides the power by the time; energy is power multiplied by time.
• D: The number is right but the unit is not; watts measure power, while energy over time is in watt-hours.

Answer: D

Correct answer: 216 W. Electrical input power is

The shaft output will be somewhat less because of heating and friction.

• A: This adds volts and amperes, which cannot be combined by addition.
• B: This slips a decimal place, as if the current were 45 A.
• C: This divides voltage by current instead of multiplying.
• D (correct): Multiplying 48 V by 4.5 A gives 216 W.

Answer: D

Correct answer: The motor turns in the opposite direction. Rotation depends on the direction of the rotor current and the direction of the magnetic field. Swapping the leads reverses the current while the permanent magnets keep the same field, so the force on each conductor reverses and the motor runs backward.

• A: Swapping the leads still lets current flow through the rotor, so the motor still develops torque.
• B: That happens only if both the current and the field are reversed; here only the current changes.
• C: Lead polarity sets the direction of the force, not the speed; the supply voltage is unchanged.
• D (correct): Reversing the rotor current while the magnet field stays the same reverses the force on each wire, so rotation reverses.

Answer: C

Correct answer: 0.03 N·m. Torque is force times lever arm,

• A: This divides the force by the distance instead of multiplying.
• B: The forces point in opposite directions, but on opposite sides of the shaft, so they twist the same way and do not cancel.
• C (correct): Each side gives 0.5 × 0.03 = 0.015 N·m, and the two equal twists add to 0.03 N·m.
• D: This is the torque from only one side of the coil.

Answer: A

Correct answer: Speed drops slightly, back emf falls, and current rises. The extra load slows the motor. Back emf depends on speed, so it falls, and the current

rises. The larger current produces the extra torque the cutting requires.

• A (correct): Lower speed means less back emf opposing the supply, so more current flows and more torque is produced.
• B: Adding a load slows a motor; it does not speed it up.
• C: Back emf grows with speed, so a slower motor has less back emf, not more.
• D: A motor responds to added load; the current must rise to provide the extra torque.

Answer: B

Correct answer: 4.6 kW. For a generator, input is mechanical and output is electrical. To find output, multiply input by efficiency as a decimal: 5 kW × 0.92 = 4.6 kW. The remaining 0.4 kW becomes heat.

• A: This divides by the efficiency, which would make output larger than input.
• B (correct): Multiplying the 5 kW input by 0.92 gives 4.6 kW of output.
• C: This assumes a perfect machine; a 92 percent efficient generator loses some input.
• D: This is the power lost as heat and friction, not the electrical output.

Answer: B

Correct answer: 20 V. Use secondary over primary on both sides.

The secondary has one-sixth the turns, so it has one-sixth the voltage.

• A: Six is the turns ratio itself, not a voltage; it still has to be applied to the 120 V input.
• B (correct): The secondary voltage is 120 V times 200/1200, which is 20 V.
• C: This multiplies by 1200/200, putting the ratio upside down; fewer secondary turns must give a lower voltage.
• D: Voltage is not equal to the number of turns; it scales with the ratio of turns.

Answer: D

Correct answer: 50 Hz. Use

• A: This uses 2 in place of 4 poles; the formula takes the full number of poles, not pole pairs.
• B: This divides by 60 instead of 120, which double counts the poles.
• C: This multiplies poles by rpm but never divides by 120 to convert to cycles per second.
• D (correct): Frequency is poles times rpm divided by 120: 4 times 1500 divided by 120 is 50 Hz.

Answer: A

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

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

Answer: D

Correct answer: 8 A. Back emf subtracts from the supply, and the remainder drives current through the winding resistance.

• A: This multiplies the 4 V difference by the resistance instead of dividing by it.
• B: This adds the back emf to the supply, but back emf opposes the supply and must be subtracted.
• C: This is supply voltage over resistance, the stall current, which ignores the back emf of a running motor.
• D (correct): The leftover 4 V across 0.5 ohm drives 4/0.5 = 8 A.

Answer: D

Correct answer: About 471 W. Convert the speed to rad/s, then multiply by torque.

• A: This converts rpm to revolutions per second but forgets the 2 pi radians in each revolution.
• B: This uses pi instead of 2 pi radians per revolution, giving half the true power.
• C: This multiplies torque by rpm directly; the speed must first be converted to radians per second.
• D (correct): 1800 rpm is 2 pi times 1800/60, about 188.5 rad/s, and 2.5 times 188.5 is about 471 W.

Answer: C

Correct answer: 0.5 A. For an ideal transformer, power in equals power out. Output power is

• A: This divides 240 V by the 120 W power, which inverts the correct division of power by voltage.
• B: This assumes the current is the same on both sides, but an ideal transformer keeps power, not current, equal.
• C (correct): Output power is 30 times 4 = 120 W, and 120 W divided by 240 V is 0.5 A.
• D: This multiplies 4 A by 240/30; a step-down transformer draws less current on its higher-voltage primary, not more.

Answer: C

Correct answer: An electronic controller switching stator coils. With the coils on the stator, no sliding contact is needed, but current still has to be switched. An electronic controller uses transistors, guided by sensors or coil voltages, to energize the stator coils in sequence; this is electronic commutation.

• A: Slip rings give a fixed sliding connection that never reverses, and a brushless motor has no sliding contacts at all.
• B: A transformer only changes AC voltage level and cannot time current switching to the rotor position.
• C (correct): Transistors in a controller switch current between stator coils in sequence, so the field steps around and the rotor follows.
• D: Permanent magnets keep a fixed polarity; they do not switch on their own.

Answer: B

Correct answer: 97 percent. Efficiency is useful output divided by total input.

The other 24 W becomes heat in the windings and core.

• A: Twenty-four is the loss in watts, not a percentage of the input.
• B (correct): Output over input is 776/800 = 0.97, which is 97 percent.
• C: This divides input by output; an efficiency above 100 percent means the fraction was written upside down.
• D: Three percent is the share of input lost as heat, not the share delivered as useful output.

Answer: D

Correct answer: After the first instant, the secondary has no ongoing output voltage. A transformer works only with changing flux. With steady DC, the flux is constant after the moment of connection, so no voltage is induced in the secondary. The primary current is then limited only by the small wire resistance, so the winding can also overheat.

• A: A transformer cannot create alternation; turning DC into AC is the job of an inverter.
• B: A transformer cannot generate AC from a steady source, so no alternating output appears.
• C: Steady DC gives a constant flux, and a constant flux induces no ongoing voltage in the secondary.
• D (correct): Once the DC current settles, the core flux stops changing, so nothing continues to be induced in the secondary.

Answer: C

Correct answer: Current in the coil creates a magnetic force opposing the rotation. This is Lenz's law. With a load connected, current flows in the coil, and that current in the magnetic field produces a force opposing the rotation. The extra effort is the mechanical energy being turned into the electrical energy the lamp uses.

• A: The lamp only receives energy; the opposition comes from the coil's own current in the magnetic field.
• B: Bearing friction does not depend on whether current flows; it is present with or without the lamp.
• C (correct): Once current flows, the coil feels a force that opposes its motion, so mechanical work must be supplied to keep it turning.
• D: The lamp is wired to the output terminals, not attached to the rotor, so it adds no mechanical mass.

Answer: A

Correct answer: 30 A. A stalled motor produces no back emf, so only the winding resistance limits current.

That is six times the 5 A running current, which is why a jammed motor overheats quickly.

• A (correct): With no rotation there is no back emf, so the current is 12 V divided by 0.4 ohm, which is 30 A.
• B: A stopped motor still has its supply connected; only the back emf drops to zero, so current rises.
• C: This multiplies the voltage by the resistance instead of dividing by it.
• D: This is the normal running current, (12 - 10)/0.4; a jammed motor no longer produces back emf.

Answer: A

Correct answer: To reduce eddy-current losses in the core. Changing flux induces small circulating currents, called eddy currents, inside solid iron, which waste energy as heat. Thin sheets insulated from one another block those current loops, while still giving the flux an easy path between the windings.

• A (correct): Insulated laminations break up the loops in which eddy currents would circulate, cutting core heating.
• B: No core design lets a transformer work on DC, because DC produces no changing flux.
• C: The core is meant to link the two windings with shared flux, not to keep their flux apart.
• D: A transformer's output frequency always equals its input frequency, whatever the core is made of.

Answer: B

Correct answer: About 78 percent. Find the electrical input, then divide the output by it, with both in watts.

The efficiency is about 78 percent.

• A: This is the fraction of input lost as heat and friction, not the useful fraction.
• B (correct): Input is 120 times 8 = 960 W, and 746/960 is about 0.78, or 78 percent.
• C: This divides input by output; efficiency above 100 percent is a sign the fraction is upside down.
• D: This treats 1 hp as 1000 W; it is about 746 W, and no motor can exceed 100 percent.

Answer: D

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

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

Answer: B

Correct answer: 150 turns. Rearrange the turns ratio:

• A: This divides the primary turns by the 24 V target, which mixes turns with volts.
• B (correct): Turns scale with voltage: 1500 times 24/240 gives 150 turns.
• C: Equal turns would give equal voltages, 240 V out, with no step-down.
• D: This multiplies by 240/24, which would make a step-up transformer giving 2400 V.

Answer: D

Correct answer: 2 A. A VA rating is rated voltage times rated current, so current is the rating divided by voltage.

• A: This divides voltage by the VA rating, which is upside down.
• B: This adds the VA rating and the voltage, which are different quantities.
• C: This multiplies the VA rating by the voltage instead of dividing.
• D (correct): Dividing the 48 VA rating by the 24 V secondary voltage gives 2 A.

Answer: B

Correct answer: Pulsating DC in one direction. Every rotating coil makes alternating voltage. A split-ring commutator reverses the coil's connections every half turn, so each half cycle reaches the circuit in the same direction. The result is pulsating DC, not smooth DC.

• A: Reversing the connections every half turn removes the reversals rather than adding more of them.
• B (correct): The commutator flips the coil connections each half turn, so the output always points one way while its size still pulses.
• C: Passing the coil output through unchanged is what slip rings do; a commutator reverses the connections.
• D: The commutator keeps the direction the same, but the size still rises and falls each half turn, so it is not smooth.

Answer: A

Correct answer: About 13.7 kW. In a chain of devices, multiply the efficiencies.

• A (correct): Overall efficiency is 0.90 times 0.95 times 0.80 = 0.684, and 0.684 times 20 kW is about 13.7 kW.
• B: This averages the three efficiencies; losses in a chain multiply, they do not average.
• C: This is the total power lost along the chain, not the power delivered at the motor shaft.
• D: This applies only the motor's 80 percent and ignores the losses in the generator and transformer.

Answer: B

Correct answer: 2 A. Output power is

To find the input, divide by the efficiency.

• A: This is the ideal-transformer answer, 216 W over 120 V; with losses the input must be larger than the output.
• B (correct): Output is 216 W, input is 216/0.90 = 240 W, and 240 W over 120 V gives 2 A.
• C: This scales 9 A by 120/24, which inverts the current ratio; the higher-voltage side carries less current.
• D: This multiplies the output by the efficiency, which makes input smaller than output; it should be divided.

Answer: D

Correct answer: 10 times. When the shaft locks, back emf disappears and only the winding resistance limits current.

Comparing with the 3 A running current gives

times. (While running, back emf is

)

• A: This divides the supply voltage by the running current, which is not a current ratio.
• B: This compares the supply with the 54 V back emf; it is a voltage ratio, not the current ratio.
• C: Thirty is the stall current in amperes, not how many times it exceeds the running current.
• D (correct): Stall current is 60/2 = 30 A, and 30 A divided by the 3 A running current is 10.

Answer: D

Correct answer: 24 poles. Rearrange

to solve for the poles.

Many poles let a slow water turbine still make 60 Hz.

• A: This divides the rpm by 60; pole counts are always even and come from p = 120f/n.
• B: Twelve is the number of pole pairs; the formula uses the total number of poles.
• C: This doubles the correct count, as if each pole had to be counted twice.
• D (correct): Rearranging f = pn/120 gives p = 120 times 60 / 300 = 24 poles.

Answer: A

Correct answer: 200 W. For fixed delivered power,

so doubling V halves I. Line loss is

so it becomes

as large: 800

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

Answer: C

Correct answer: 12.5 W. Find the current, then use

That is one-sixteenth of the 200 W lost at 2500 V.

• A: This is the loss at the original 2500 V, where the current is 20 A.
• B: This divides the 200 W loss at 2500 V by 4, but loss falls with the square of the voltage ratio, 16.
• C (correct): At 10,000 V the current is 50,000/10,000 = 5 A, and 5 squared times 0.5 is 12.5 W.
• D: This multiplies current by resistance, which gives a voltage drop, not the I squared R power loss.

Answer: A

Correct answer: 8 W. First find the running current from the back emf.

Winding heat is

• A (correct): The current is (24 − 22) ÷ 0.5 = 4 A, and the winding heat is 4² × 0.5 = 8 W.
• B: This is the total electrical input, 24 V × 4 A, not just the winding heat.
• C: This is back emf times current, the power converted toward mechanical work, not the heat.
• D: This uses the full 24 V across the winding as if the motor were stalled.

Answer: C

Correct answer: Frequency drops to 50 Hz, and voltage drops. Frequency follows

Slower rotation also means the flux changes more slowly, so the induced voltage falls roughly in proportion to speed, to about five-sixths of normal.

• A: With the pole count fixed, frequency is tied to speed, so it cannot stay at 60 Hz when the shaft slows.
• B: Rpm is not frequency; a two-pole machine gives one cycle per revolution, which is 50 per second here.
• C (correct): Both frequency and induced voltage are roughly proportional to speed, so both fall by one-sixth.
• D: A slower coil changes flux less quickly, so by Faraday's law the induced voltage falls, not rises.

Answer: C

Correct answer: 0.15 A. Work from the secondary back to the primary.

• A: This divides the 3 A by the turns ratio twice; the current ratio uses the turns ratio only once.
• B: This multiplies the 3 A secondary current by the 20:1 ratio, inverting the current relationship.
• C (correct): Secondary voltage is 12 V, so 3 A flows in the load; 36 W divided by 240 V gives 0.15 A.
• D: This is the secondary current; the primary is on the high-voltage side and carries less current.

Answer: A

Correct answer: About 79 percent. Input:

Speed:

rad/s.

The efficiency is about 79 percent.

• A (correct): Converting 1200 rpm to about 125.7 rad/s gives 377 W of output, and 377 ÷ 480 is about 0.79.
• B: This divides input by output; efficiency can never exceed 100 percent.
• C: This is the fraction of input lost, not the fraction delivered as useful output.
• D: This multiplies torque by rpm without converting to radians per second.

Answer: C

Correct answer: The stator. A motor has two main parts: the stator, which stays still, and the rotor, which spins and carries the shaft. The names come from "static" and "rotate."

• A: The rotor is the rotating part that carries the shaft, the opposite of what the question asks.
• B: Primary winding is a transformer term for the coil connected to the source.
• C (correct): The stator is the stationary part of a motor, surrounding the turning rotor.
• D: Slip rings are mounted on the turning shaft of an AC generator to connect the rotating coil to the outside circuit.

Answer: B

Correct answer: 1.25 A. Secondary voltage:

Load current:

so output power is

Input power is output divided by efficiency:

Primary current is

• A: This treats the transformer as ideal and ignores the 96 percent efficiency, so the input power is too small.
• B (correct): Secondary 24 V and 6 A give 144 W; input is 144/0.96 = 150 W, and 150/120 = 1.25 A.
• C: This multiplies the output by the efficiency instead of dividing; the input must be larger than the output.
• D: This uses the turns ratio the wrong way, multiplying the secondary current by 5 instead of dividing.

Answer: B

Correct answer: 220 W. First find the current:

Input power is

and winding heat is

With no friction, the rest becomes mechanical power:

W (the same as

).

• A: This is the total electrical input, $48\times 5$; part of it is lost as heat in the windings.
• B (correct): Current is (48 − 44)/0.8 = 5 A; input 240 W minus 20 W of winding heat leaves 220 W.
• C: This is the heat lost in the winding resistance, $5^2\times 0.8$, not the power converted to motion.
• D: This divides the square of the back emf by the winding resistance, but the back emf is not the voltage across the resistance.

Answer: A

Correct answer: A diode rectifier. Every rotating coil naturally makes AC. A modern alternator lets the machine produce AC and then passes it through a set of diodes, called a rectifier, which allows current in only one direction and delivers DC to the battery.

• A (correct): Diodes let current pass in only one direction, turning the alternator's AC into DC for the battery.
• B: A transformer changes AC voltage levels but its output is still AC; it cannot make DC.
• C: An inverter does the opposite job, turning DC into AC.
• D: Slip rings pass the coil's output to the circuit unchanged, so the output stays AC.

Answer: D

Correct answer: 900 rpm. Rearrange

to solve for speed.

Check: 8 poles give 4 cycles per turn, and 900 rpm is 15 turns per second, so

• A: This uses 60 instead of 120 in the formula, counting each pole rather than each pole pair as one cycle.
• B: 3,600 rpm gives 60 Hz only for a two-pole machine; more poles let the shaft turn more slowly.
• C: 60 Hz divided by 4 pole pairs is 15 revolutions per second, not per minute; times 60 that is 900 rpm.
• D (correct): $n=120f/p=120\times 60/8=900$ rpm.

Answer: A

Correct answer: About 377 W. Convert speed first:

rad/s.

Then

• A (correct): $\omega =2\pi \times 900/60\approx 94.2$ rad/s, and $4\times 94.2\approx 377$ W.
• B: This multiplies torque by rpm directly, skipping the conversion to radians per second.
• C: This converts rpm to revolutions per second but forgets the $2\pi$ radians in each revolution.
• D: This uses only $\pi$ instead of $2\pi$ radians per revolution, giving half the true power.

Answer: C

Correct answer: 200 W. For fixed power,

so tripling V makes the current one third as large. Line loss is $I^{2}R$, so it becomes

of its old value:

• A: This reduces the loss in step with the current, but loss depends on current squared.
• B: Higher voltage lowers the current for the same delivered power, so loss falls rather than rises.
• C (correct): Tripling the voltage cuts current to one third, and $I^{2}R$ loss falls to one ninth: 1800/9 = 200 W.
• D: The line resistance is unchanged, but the current is not; lower current means less heat.

Answer: C

Correct answer: 5 A. In an ideal transformer, power in equals power out:

The side with the higher voltage has the lower current.

• A: This scales the current the same way as voltage; in a step-up transformer the low-voltage side carries the higher current.
• B: The current changes by the inverse of the voltage ratio; it does not stay the same on both sides.
• C (correct): Output power is $240\times 0.5=120$ W, and $120/24=5$ A on the primary.
• D: This multiplies by the ratio twice; a 10-to-1 voltage ratio changes current by a factor of 10, not 100.

Answer: D

Correct answer: 4.0 kW. Efficiency is output divided by input, so input = output ÷ efficiency.

The 0.6 kW difference is lost, mostly as heat.

• A: This multiplies by the efficiency, which would make the input smaller than the output.
• B: This adds 15 percent of the output; the losses are 15 percent of the input, not of the output.
• C: This is only the power lost as heat, not the total mechanical input.
• D (correct): Input equals output divided by efficiency: 3.4/0.85 = 4.0 kW.

Answer: B

Correct answer: It stays the same. The force on a rotor wire depends on both the current direction and the field direction. Reversing one reverses the force; reversing both reverses it twice, leaving it unchanged. This is why a universal motor can run on AC, where both reverse together every half cycle.

• A: Rotation reverses only when one of the two, field or rotor current, is reversed by itself.
• B (correct): Reversing the field flips the force, and reversing the current flips it back, so the two changes cancel.
• C: The force on each wire is still present at full strength; only its two causes have both flipped.
• D: Reversing directions does not change the size of the field or current, so the force and speed are unchanged.