1True or false
In an AC generator, the EMF is zero at the instant when the flux linkage through the coil is greatest.
Show answer
Answer: True
At that instant the flux is momentarily not changing, so the induced EMF is zero.
!Common mistakeExpecting the largest EMF with the largest flux overlooks Faraday's law: EMF depends on the rate of change of flux.
2True or false · ★ Challenge
Eddy-current braking becomes weaker as the vehicle slows down.
Show answer
Answer: True
The induced EMF, and so the eddy currents and the braking force, are proportional to the speed of the disc; at very low speed the force is small.
!Common mistakeThinking the braking force is constant ignores that the eddy currents depend on how fast the disc cuts the field lines.
3True or false
In an electric guitar, the vibrating steel string changes the magnetic flux through a pickup coil, inducing a small EMF.
Show answer
Answer: True
The pickup's magnet magnetises the string; as it vibrates, the flux through the coil changes and an EMF of the same frequency is induced.
!Common mistakeThinking the pickup is a tiny microphone is wrong: it responds to the moving steel string, not to sound in the air.
4True or false · ★ Challenge
When a magnet is pushed into a coil whose ends are not connected to anything, an EMF is induced but no current flows.
Show answer
Answer: True
The changing flux still induces an EMF across the ends, but with an open circuit there is no path for a current.
!Common mistakeThinking nothing is induced in an open coil mixes up EMF and current: the EMF exists whether or not a current can flow.
5True or false
Pulling a coil out of a magnetic field more quickly makes more charge flow round its circuit.
Show answer
Answer: False
Q = NΔΦ/R does not depend on time: faster removal gives a larger current for a shorter time.
!Common mistakeLinking 'faster' with 'more charge' mixes up current (charge per second) with the total charge.
6True or false
The flux through a coil is greatest when the plane of the coil is parallel to the magnetic field.
Show answer
Answer: False
With the plane parallel to the field, no field lines pass through the coil, so Φ = 0; the flux is greatest when the plane is perpendicular to the field.
!Common mistakeConfusing the plane of the coil with its normal reverses the answer: Φ = BA cos θ with θ measured from the normal.
7Fill in the blank · ★ Challenge
To double the energy stored in an inductor, the current through it must be multiplied by ______.
Show answer
Answer: √2 (about 1.41)
E = ½LI², so E ∝ I²; doubling E needs I² to double, so I × √2.
!Common mistakeAnswering 2 forgets the square: doubling the current would make the energy four times as large.
8True or false
A coil's self-inductance opposes changes in the current through it but does not oppose a steady current.
Show answer
Answer: True
Self-induced EMF = −L dI/dt is zero when the current is steady; only the coil's resistance then limits the current.
!Common mistakeThinking an inductor always reduces the current confuses inductance with resistance; with steady DC only R matters.
9Multiple choice · ★ Challenge
The north pole of a magnet is pulled AWAY from one end of a coil connected to a lamp. What magnetic pole is induced at that end of the coil?
- AA north pole, repelling the magnet
- BA south pole, attracting the magnet
- CA north pole, attracting the magnet
- DNo pole, since the magnet is moving away
Show answer
Answer: B. A south pole, attracting the magnet
By Lenz's law the induced current opposes the change (the magnet leaving), so that end becomes a south pole that attracts the retreating north pole.
!Common mistakeAnswering 'north pole' copies the case of a magnet being pushed IN; when it is pulled away, the coil tries to hold it back by attracting it.
10True or false
For the same power, sending electricity at a higher voltage means a smaller current in the transmission lines.
Show answer
Answer: True
P = VI, so for a fixed P a larger V needs a smaller I.
!Common mistakeThinking a higher voltage pushes a larger current applies Ohm's law to the wrong voltage; the power, not a fixed resistance, sets the line current.
11Fill in the blank · ★ Challenge
A coil of 200 turns, each of area 5.0 cm², lies with its plane perpendicular to a 0.30 T field. Its flux linkage is ______ Wb-turns.
Show answer
Answer: 0.030
NΦ = NBA = 200 × 0.30 × 5.0 × 10⁻⁴ = 0.030 Wb-turns.
!Common mistakeUsing A = 5.0 m² forgets to change cm² to m² (× 10⁻⁴), which makes the answer 10 000 times too large.
12Fill in the blank
The coil of a transformer that is connected to the AC supply is called the ______ coil.
Show answer
Answer: primary
The supply feeds the primary; the load is connected to the secondary.
!Common mistakeWriting 'secondary' swaps the coils: the secondary is the output coil, where the EMF is induced.
13Multiple choice · ★ Challenge
The same power is sent along the same line, first at 30 kV and then at 110 kV. By about what factor does the power wasted in the line fall?
- A3.7
- B13
- C0.27
- D80
Show answer
Answer: B. 13
For fixed power I ∝ 1/V, and loss = I²R ∝ 1/V². Factor = (110 ÷ 30)² = 3.67² ≈ 13.
!Common mistake3.7 forgets to square the voltage ratio; the loss depends on I², not on I.
14True or false
Pushing a magnet into a coil quickly induces a larger EMF than pushing it in slowly, even though the total change of flux is the same.
Show answer
Answer: True
EMF = NΔΦ/Δt, so the same ΔΦ in a shorter time gives a larger EMF.
!Common mistakeThinking only the size of the magnet matters forgets that Faraday's law depends on how FAST the flux changes.
15Fill in the blank
The EMF ε = BLv induced in a moving rod is largest when the rod moves at ______ to the magnetic field.
Show answer
Answer: right angles
Only the component of velocity perpendicular to B cuts field lines; moving parallel to the field induces no EMF.
!Common mistakeWriting 'parallel' is wrong: a rod sliding along the field lines cuts none of them, so no EMF is induced.
16Multiple choice · ★ Challenge
A generator has a peak EMF of 20 V. What is the EMF at the instant when the coil has turned 30° from the position where its plane is perpendicular to the field?
- A17 V
- B10 V
- C20 V
- D0 V
Show answer
Answer: B. 10 V
The EMF is zero when the plane is perpendicular (flux maximum) and ε = ε₀ sin θ from there: ε = 20 × sin 30° = 10 V.
!Common mistake17 V uses cos 30°; measured from the perpendicular position the EMF starts at zero, so it follows sin θ.
17Fill in the blank
The EMF induced in the turning coil of an electric motor, which opposes the supply voltage, is called the ______ EMF.
Show answer
Answer: back
The rotating coil acts as a generator; by Lenz's law its EMF opposes the applied voltage.
!Common mistakeWriting 'forward' gets the direction wrong: the induced EMF opposes the supply, so it is called back EMF.
18Short answer · ★ Challenge
In electronic circuits, a diode is often connected in reverse across the coil of a relay. Explain why.
Show answer
Model answer: When the transistor or switch turns the relay off, the current in the coil falls very quickly, so a large self-induced EMF (−L dI/dt) appears. This could damage the transistor. The diode gives the current a safe path to die away gradually, so the stored energy ½LI² is used up in the coil's resistance instead of as a high-voltage spike.
!Common mistakeSaying the diode stops the relay working backwards misses the point: it protects the circuit from the large back EMF when the current is switched off.
19Multiple choice
What is the job of the split-ring commutator in a simple DC generator?
- ATo make the coil turn at a steady speed
- BTo reverse the coil's connections every half turn so the output current does not change direction
- CTo increase the peak EMF of the coil by adding the EMFs of both halves of each turn
- DTo stop sparks forming at the brushes
Show answer
Answer: B. To reverse the coil's connections every half turn so the output current does not change direction
Each half turn the coil's EMF reverses; the commutator swaps the connections at the same moment, so the output is always the same way round (varying DC).
!Common mistakeThinking it increases the EMF confuses it with adding turns or a stronger magnet; the commutator only changes the direction of the output.
20Multiple choice · ★ Challenge
A metal rod 0.50 m long spins at 40 rad/s about one end in a plane perpendicular to a 0.20 T field. What EMF is induced between its ends?
- A2.0 V
- B4.0 V
- C1.0 V
- D0.50 V
Show answer
Answer: C. 1.0 V
Different parts move at different speeds; the average speed is ωL/2, so ε = ½BωL² = ½ × 0.20 × 40 × 0.50² = 1.0 V.
!Common mistake2.0 V uses the speed of the tip (ωL) for the whole rod; the end at the axis is at rest, so the average speed is half of ωL.
21Multiple choice
In a moving-coil microphone, how is sound turned into an electrical signal?
- ASound squeezes carbon granules, changing their resistance and so the current through them
- BSound charges a capacitor
- CSound makes a coil vibrate in a magnetic field, inducing an EMF
- DSound waves heat a wire
Show answer
Answer: C. Sound makes a coil vibrate in a magnetic field, inducing an EMF
The diaphragm moves a coil back and forth in a magnet's field; the changing flux induces an EMF that varies like the sound wave.
!Common mistakeChoosing the carbon granule describes an older carbon microphone; the MOVING-COIL microphone works by electromagnetic induction.
22Multiple choice
A phone lies on a wireless charging pad. How does energy reach the phone's battery?
- AA steady current in the pad magnetises the phone's battery
- BRadio waves from the pad are absorbed as heat
- CAn alternating current in the pad's coil induces an EMF in a coil inside the phone (mutual induction)
- DStatic charge builds up on the pad and jumps across the small gap into the phone's charging socket and battery
Show answer
Answer: C. An alternating current in the pad's coil induces an EMF in a coil inside the phone (mutual induction)
The pad's coil carries AC, giving a changing flux that links the phone's coil; the induced EMF is rectified to charge the battery.
!Common mistakeChoosing 'steady current' misses the key point: only a changing current in the pad induces an EMF in the phone's coil.
23Short answer · ★ Challenge
A motor driving a maize mill becomes jammed and stops turning while still switched on. Explain why it may overheat and burn out.
Show answer
Model answer: When the coil turns, it cuts field lines and produces a back EMF that opposes the supply and keeps the current small. When the motor is jammed, there is no back EMF, so the current is limited only by the small coil resistance. The current becomes very large and the I²R heating in the coil can melt its insulation.
!Common mistakeSaying the motor overheats because it 'works harder' is wrong: a stalled motor does no mechanical work; the heat comes from the large current with no back EMF.
24Multiple choice
A lamp is connected in series with a large coil on an iron core and a DC supply. What is seen when the switch is closed?
- AThe lamp lights gradually, reaching full brightness after a short time
- BThe lamp never lights, because the coil blocks the direct current completely
- CThe lamp lights at once at full brightness
- DThe lamp flashes brightly, then goes out
Show answer
Answer: A. The lamp lights gradually, reaching full brightness after a short time
As the current starts to grow, the coil's self-induced EMF opposes the rise, so the current (and brightness) builds up gradually to the final value V/R.
!Common mistakeChoosing 'lights at once' ignores self-induction; the coil opposes any CHANGE in current, including switching on.
25Multiple choice · ★ Challenge
A small water-pump motor has a coil resistance of 4.0 Ω and runs on 24 V. At full speed its back EMF is 20 V. What current does it take at full speed?
- A6.0 A
- B11 A
- C5.0 A
- D1.0 A
Show answer
Answer: D. 1.0 A
I = (V − εback)/R = (24 − 20) ÷ 4.0 = 1.0 A.
!Common mistake6.0 A (24 ÷ 4.0) is the starting current, before the coil turns; once running, the back EMF opposes the supply.
26Multiple choice
Some heavy trucks have electromagnetic brakes: a metal disc on the axle turns between the poles of an electromagnet. How do these brakes slow the truck?
- AThe magnet attracts the steel disc and grips it like a brake pad
- BEddy currents induced in the turning disc produce a force that opposes its motion
- CThe magnet removes the kinetic energy by magnetising the disc more and more strongly as it turns
- DThe electromagnet stops current flowing in the truck's engine
Show answer
Answer: B. Eddy currents induced in the turning disc produce a force that opposes its motion
The disc cuts the field, so eddy currents flow in it; by Lenz's law their magnetic force opposes the rotation and the kinetic energy becomes heat.
!Common mistakeChoosing 'grips it like a pad' assumes contact friction; eddy-current brakes work without touching the disc.
27Multiple choice · ★ Challenge
An air-cored solenoid is 0.20 m long with 500 turns, each of area 4.0 cm². Using L = μ₀N²A/l with μ₀ = 4π × 10⁻⁷ H/m, what is its inductance?
- A0.63 mH
- B6.3 H
- C1.3 μH
- D2.5 mH
Show answer
Answer: A. 0.63 mH
L = 4π × 10⁻⁷ × 500² × 4.0 × 10⁻⁴ ÷ 0.20 = 6.3 × 10⁻⁴ H ≈ 0.63 mH.
!Common mistake1.3 μH forgets to square N; 6.3 H forgets to change cm² to m².
28Fill in the blank
When the flux linkage of a coil circuit changes by NΔΦ, the total charge that flows is Q = NΔΦ/______, where the missing quantity is the circuit's total resistance.
Show answer
Answer: R
Q = IΔt = (NΔΦ/Δt ÷ R) × Δt = NΔΦ/R.
!Common mistakeWriting Δt here is wrong: time cancels out of the charge, leaving only the resistance.
29Multiple choice · ★ Challenge
An ideal transformer steps 230 V down to 12 V to supply ten 12 V, 6.0 W lamps in parallel at full brightness. What is the current in the primary coil?
- A0.026 A
- B0.50 A
- C5.0 A
- D0.26 A
Show answer
Answer: D. 0.26 A
Power out = 10 × 6.0 = 60 W = power in (ideal), so Ip = 60 ÷ 230 ≈ 0.26 A.
!Common mistake5.0 A is the secondary current (60 ÷ 12); the primary is at 230 V, so for the same power its current is much smaller.
30Short answer
Give three ways of increasing the mutual inductance between two coils.
Show answer
Model answer: Wind both coils on the same soft-iron core (so the flux of one passes through the other); put the coils closer together, coaxially (one around or next to the other); use more turns on the coils.
!Common mistakeSuggesting a bigger current in the first coil is wrong: that increases the induced EMF but not the mutual inductance, which depends only on the coils and their arrangement.
31Multiple choice
A rod is pushed at a steady speed along rails in a magnetic field, driving a current through a resistor. If the resistor is replaced by one of half the resistance and the speed stays the same, the force needed:
- AHalves
- BDoubles
- CStays the same
- DBecomes four times as large
Show answer
Answer: B. Doubles
The EMF BLv is unchanged, so the current I = ε/R doubles; the opposing force BIL doubles too.
!Common mistakeChoosing 'stays the same' assumes the force depends only on speed; the opposing force depends on the current, which rises when R falls.
32Multiple choice · ★ Challenge
A relay coil has L = 2.0 H and R = 10 Ω. When it is switched onto a 12 V battery, what steady current does it finally reach, and what is the time constant of the rise?
- A1.2 A; 0.20 s
- B1.2 A; 5.0 s
- C0.60 A; 0.20 s
- D24 A; 20 s
Show answer
Answer: A. 1.2 A; 0.20 s
Final current I = V/R = 12 ÷ 10 = 1.2 A (no back EMF once steady). Time constant τ = L/R = 2.0 ÷ 10 = 0.20 s.
!Common mistake5.0 s inverts the time constant (R/L); the larger the inductance, the LONGER the current takes to grow, so τ = L/R.
33Multiple choice
In Fleming's right-hand rule for a generator, what do the thumb, first finger and second finger represent?
- AMotion, field, induced current
- BField, motion, induced current
- CMotion, induced current, field
- DInduced current, field, motion
Show answer
Answer: A. Motion, field, induced current
Thumb = motion (thrust) of the conductor, first finger = field, second finger = induced current.
!Common mistakeMixing up the right-hand (generator) rule with the left-hand (motor) rule swaps the roles of current and motion.
34Short answer · ★ Challenge
Describe how the flux linkage through a coil rotating at constant speed in a uniform field varies with time, and how the induced EMF varies. State the phase difference between them.
Show answer
Model answer: Both vary sinusoidally with the same frequency. The flux linkage NΦ = NBA cos ωt is greatest when the coil's plane is perpendicular to the field; the EMF ε = NBAω sin ωt is the negative rate of change of flux linkage, so it is zero when the flux is greatest and greatest when the flux is zero. They are 90° (a quarter of a cycle) out of phase.
!Common mistakeSaying the EMF is largest when the flux is largest confuses the value of the flux with its rate of change, which is what gives the EMF.
35Multiple choice
Why is electricity generated and distributed as AC rather than DC in most countries, including Rwanda?
- AAC is safer to touch than DC
- BAC travels faster along the cables
- CAC voltages can easily be stepped up and down with transformers
- DAC cannot be stored in batteries, so none of it is wasted between the power station and the homes
Show answer
Answer: C. AC voltages can easily be stepped up and down with transformers
Transformers need a changing current; with AC the voltage can be raised for transmission (low current, low I²R loss) and lowered for homes.
!Common mistakeChoosing 'AC travels faster' is wrong: the speed of the signal is not the reason; the advantage is efficient voltage change with transformers.
36Short answer · ★ Challenge
Explain why the total charge that flows when a coil is pulled out of a magnetic field does not depend on how quickly it is pulled out.
Show answer
Model answer: The EMF is NΔΦ/Δt, so the current is I = NΔΦ/(RΔt). The charge is Q = IΔt = NΔΦ/R: the Δt cancels. Pulling faster gives a larger current for a shorter time, so the charge depends only on the flux change and the resistance.
!Common mistakeConfusing charge with current leads to 'faster gives more'; the current is bigger but flows for less time.
37Multiple choice
Two coils have a mutual inductance of 0.15 H. The current in the first coil changes at 40 A/s. What EMF is induced in the second coil?
- A0.0038 V
- B6.0 V
- C270 V
- D40 V
Show answer
Answer: B. 6.0 V
ε = M dI/dt = 0.15 × 40 = 6.0 V.
!Common mistake270 V divides 40 by 0.15; the EMF is the mutual inductance multiplied by the rate of change of current.
38Multiple choice · ★ Challenge
A learner holds a flat 100-turn coil (area 0.010 m²) face-on to a 0.40 T field, then turns it over completely in 0.10 s so that the field passes through it the opposite way. What average EMF is induced?
- A4.0 V
- B8.0 V
- C0 V
- D0.080 V
Show answer
Answer: B. 8.0 V
The flux through each turn goes from +BA to −BA, a change of 2BA = 2 × 0.40 × 0.010 = 0.0080 Wb; ε = NΔΦ/Δt = 100 × 0.0080 ÷ 0.10 = 8.0 V.
!Common mistake4.0 V treats the change as BA; turning the coil over reverses the flux, so it changes by twice BA.
39Multiple choice
A learner connects a 12 V car battery to the primary of a transformer. A voltmeter on the secondary flicks only at the moments when the battery is connected and disconnected. Why?
- AThe secondary coil has too few turns
- BThe battery's voltage is too small to push a current all the way across into the secondary coil
- COnly a changing current gives a changing flux, which is needed to induce an EMF
- DThe iron core blocks direct current
Show answer
Answer: C. Only a changing current gives a changing flux, which is needed to induce an EMF
A steady current produces a steady flux, so no EMF is induced; at switching on and off the current changes briefly, giving a brief EMF.
!Common mistakeChoosing 'the core blocks DC' is wrong: the core carries flux, not current; a steady flux simply induces nothing.
40Short answer
Explain how a walk-through metal detector at Kigali International Airport can detect a metal object.
Show answer
Model answer: A coil in the detector carries an alternating (or pulsed) current, producing a changing magnetic field. This induces eddy currents in any metal object passing through. The eddy currents produce their own changing magnetic field, which induces an extra EMF in the detector coil; the electronics sense this change and sound an alarm.
!Common mistakeSaying the detector attracts metal like a magnet is wrong: it works for non-magnetic metals such as aluminium, through induced eddy currents.
41Multiple choice · ★ Challenge
A 50-turn coil of area 4.0 cm² is pulled out of a 0.25 T field that was perpendicular to it. The circuit has a total resistance of 5.0 Ω. How much charge flows round the circuit?
- A2.0 × 10⁻⁵ C
- B5.0 × 10⁻³ C
- C1.0 × 10⁻³ C
- DIt depends on how fast the coil is removed
Show answer
Answer: C. 1.0 × 10⁻³ C
Q = NΔΦ/R = 50 × (0.25 × 4.0 × 10⁻⁴) ÷ 5.0 = 50 × 1.0 × 10⁻⁴ ÷ 5.0 = 1.0 × 10⁻³ C.
!Common mistakeChoosing 'depends on how fast' is the classic error: faster removal gives a bigger current for a shorter time, and the charge stays the same.
42Multiple choice
The EMF of a generator is given by ε = 340 sin(314t), with ε in volts and t in seconds. What is the frequency of the output?
- A314 Hz
- B340 Hz
- C100 Hz
- D50.0 Hz
Show answer
Answer: D. 50.0 Hz
ω = 2πf = 314 rad/s, so f = 314 ÷ 2π = 50 Hz.
!Common mistake314 Hz takes ω as the frequency; ω is the angular frequency in rad/s and must be divided by 2π.
43Short answer · ★ Challenge
Plan an experiment to test whether the peak EMF induced as a bar magnet falls through a coil is proportional to the speed of the magnet. Name the apparatus and the measurements.
Show answer
Model answer: Hold a coil vertically, connect it to a data logger (or oscilloscope) and drop a bar magnet through it from several measured heights h. Record the peak EMF for each drop and calculate the speed at the coil from v = √(2gh). Repeat each height and average, then plot peak EMF against v: a straight line through the origin shows proportionality. Keep the same magnet, coil and starting orientation.
!Common mistakeUsing an ordinary moving-coil voltmeter is a poor choice: the EMF lasts only a fraction of a second, so the needle cannot show the peak value.
44Short answer
Use the idea of energy to explain why the induced current must oppose the motion of a magnet pushed into a coil.
Show answer
Model answer: If the induced current attracted the magnet instead, the magnet would speed up on its own and the current would grow, producing electrical energy from nothing, which breaks conservation of energy. Because the current repels the magnet, work must be done to push it in, and this work becomes the electrical energy.
!Common mistakeSaying the current opposes the magnet 'because like poles repel' describes what happens but not why; the reason is conservation of energy.
45Short answer · ★ Challenge
A battery charger's transformer is tested: primary 230 V and 0.40 A; secondary 11.5 V and 7.2 A. Calculate its efficiency, and name two energy losses with one way of reducing each.
Show answer
Model answer: Input power = 230 × 0.40 = 92 W; output power = 11.5 × 7.2 = 82.8 W; efficiency = 82.8 ÷ 92 × 100 = 90 %. Losses: heating of the copper windings (use thicker, low-resistance wire); eddy currents in the core (laminate the core); hysteresis in the core (use soft iron); flux leakage (wind coils on top of each other).
!Common mistakeDividing input by output gives 111 %, which is impossible; efficiency is useful output ÷ input and must be below 100 %.
46Multiple choice
The flux through a coil rises steadily from 0 to 4 mWb between t = 0 and t = 2 s, stays at 4 mWb until t = 5 s, then falls steadily to 0 at t = 6 s. During which interval is the induced EMF largest?
- A0 to 2 s
- B2 to 5 s
- C5 to 6 s
- DThe EMF is the same throughout
Show answer
Answer: C. 5 to 6 s
EMF = rate of change of flux: 0–2 s, 2 mWb/s; 2–5 s, zero; 5–6 s, 4 mWb/s. The steepest part of the graph gives the largest EMF.
!Common mistakeChoosing 2 to 5 s confuses the largest FLUX with the largest EMF; a constant flux, however big, induces nothing.
47Short answer · ★ Challenge
A horizontal metal rod of mass 0.010 kg and length 0.20 m slides down two smooth vertical rails joined at the top through a 0.10 Ω resistor. A horizontal 0.50 T field acts at right angles to the rails. Find the steady (terminal) speed of the rod (g = 10 N/kg).
Show answer
Model answer: At terminal speed the magnetic force balances the weight: BIL = mg with I = BLv/R, so B²L²v/R = mg. v = mgR/(B²L²) = 0.010 × 10 × 0.10 ÷ (0.50² × 0.20²) = 0.010 ÷ 0.010 = 1.0 m/s.
!Common mistakeUsing v = √(2gh) treats the rod as falling freely; the induced current gives an upward force that grows with speed until it equals the weight.
48Multiple choice
A flat loop of area 0.20 m² is in a uniform 0.50 T field. The field makes an angle of 60° with the normal to the loop. What is the flux through the loop?
- A0.087 Wb
- B0.10 Wb
- C0.050 Wb
- D2.5 Wb
Show answer
Answer: C. 0.050 Wb
Φ = BA cos θ = 0.50 × 0.20 × cos 60° = 0.10 × 0.5 = 0.050 Wb.
!Common mistake0.087 Wb uses sin 60°; θ is measured from the NORMAL, so cos θ gives the component of B through the loop.
49Multiple choice
An electromagnet of inductance 0.50 H carries a current of 4.0 A. How much energy is stored in its magnetic field?
- A1.0 J
- B4.0 J
- C8.0 J
- D2.0 J
Show answer
Answer: B. 4.0 J
E = ½LI² = ½ × 0.50 × 4.0² = 4.0 J.
!Common mistake8.0 J forgets the ½; 1.0 J forgets to square the current.
50Short answer · ★ Challenge
Compare the cable loss when a village mini-hydro scheme delivers 20 kW at 400 V with the loss when transformers let it deliver the same power at 11 kV. The cables total 2.5 Ω.
Show answer
Model answer: At 400 V: I = 20 000 ÷ 400 = 50 A; loss = I²R = 50² × 2.5 = 6250 W (about 31 % of the power). At 11 kV: I = 20 000 ÷ 11 000 ≈ 1.8 A; loss = 1.82² × 2.5 ≈ 8.3 W (about 0.04 %). The high voltage cuts the loss enormously.
!Common mistakeUsing P = V²/R with the transmission voltage (400² ÷ 2.5) is wrong: the voltage across the CABLES is not the supply voltage; use I²R with the line current.