Answer key · Unit 30

In the real world: power and signals.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 6

Lesson 1 of 6 · About 25 minutes

Solar cells

A silicon cell turns light into electricity with no fuel and no moving parts. What a photon does at the junction, why the current follows the light and the voltage hardly does, how much of the light becomes electricity, and why panels are tilted towards the Sun.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
Light in, current out
Practice · Choose
Question
The panel faces the Sun, and Zoom to the junction is at Junction. What happens when a photon of sunlight is absorbed in the silicon near the junction?

Correct answer
- Correct It frees an electron and leaves a hole behind, and the junction’s field pushes the electron to the n side and the hole to the p side
Setup, choices and feedback
The Scene at this step
Sunlight
1000 W/m²
Angle to the Sun
0 °
Zoom to the junction
Junction
Choices
Shown in a random order.
- Correct It frees an electron and leaves a hole behind, and the junction’s field pushes the electron to the n side and the hole to the p side
- Wrong It heats the silicon, and the heat drives the electrons round the circuit
  Feedback
  Heat does not drive the current in a solar cell; a hot cell even gives a lower voltage. The energy passes straight from one photon to one electron.
- Wrong It brings electric charge with it, which flows on through the wires
  Feedback
  Light carries no charge. The electrons that flow round the circuit were in the silicon and the wires all along; the photon only gives one of them the energy to break free.
- Wrong It frees many electrons, more the brighter the light
  Feedback
  Each absorbed photon frees at most one pair. Brighter light is more photons every second, so more pairs, not more pairs from each photon.
Feedback for a right answer
An electron–hole pair, and the junction pulls them apart. The electron’s only way back is round the circuit, and that is the current.

Worked solution
1. A photon of visible light has enough energy to tear an electron out of its bond in the silicon. Where the electron was, a hole is left.
2. Near the junction there is a built-in electric field: it sweeps the electron up into the n layer and the hole down into the p base.
3. The electron can get back to the hole only the long way round: through the silver fingers, the wires and the bulb. That journey is the current.
Without the junction the pair would soon meet again, and the energy would turn into heat.
Step 2 of 7
Open this step
Half the light
Practice · Predict, then look
Question
The cell is in full sun. Sunlight is turned down to half, 500 W/m². What will happen to the cell’s current and voltage?

Correct answer
- Correct The current will halve, and the voltage will hardly change
Setup, choices and feedback
The Scene at this step
Sunlight
1000 W/m²
Angle to the Sun
0 °fixed
Zoom to the junction
Panel
Choices
- Correct The current will halve, and the voltage will hardly change
- Wrong Both the current and the voltage will halve
  Misconception: Thinks a solar cell's voltage falls in proportion to the light, as its current does.
  Feedback from the catalogue
  A solar cell’s current follows the light: each photon frees at most one electron, so half the light gives half the current. The voltage behaves differently. It is set by how strongly the two sides of the junction are charged, and that grows only with the logarithm of the light: half the light lowers it by only about 20 mV, out of about half a volt. So a panel gives almost its full voltage even on a cloudy day, and the power falls mainly because of the current.
- Wrong The voltage will halve, and the current will stay the same
  Feedback
  The other way round: the number of photons each second sets how many electrons are freed each second, which is the current.
- Wrong Nothing: as long as there is light, the cell gives the same
  Feedback
  Each photon frees at most one electron, so less light is fewer electrons each second, and less current.
Once the prediction is locked in, the Scene changes
Sunlight
500 W/m²
The Scene then shows: 0.5292 V · Voltage across one cell/nodes/panel/outputs/solarVoltage
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The current halves, and the voltage falls only from about 0.55 to about 0.53 V. So the power falls almost as fast as the light.

Worked solution
1. The current follows the number of photons: half the light, half the current, from about 8.8 A to about 4.4 A.
2. The voltage is set by how strongly the two sides of the junction are charged, and that grows only with the logarithm of the light. Half the light lowers it by only about 20 mV, from about 0.55 V to about 0.53 V.
3. So the power, $P = VI$ ⁠, falls almost as fast as the light.
That is why a panel gives almost its full voltage on a cloudy day, but much less power.
Step 3 of 7
Open this step
How much of the light becomes electricity
Practice · Calculate
Question
One cell is a square 156 mm across, about 0.0244 m², and Sunlight gives 1000 W/m² on a panel facing it. Read the cell’s voltage and current off the meter. What percentage of the light falling on the cell becomes electrical power?

Correct answer
19.77 ± 0.6
Setup, choices and feedback
The Scene at this step
Sunlight
1000 W/m²fixed
Angle to the Sun
0 °fixed
Zoom to the junction
Panel
Accepted within
± 0.6
Computed from the Scene at this step’s setting
Share of the striking light turned into electrical power, 0..1/nodes/panel/outputs/efficiency
Feedback for a right answer
About $20\%$ ⁠: about a fifth of the light becomes electricity, as in good silicon cells.

Hint
The light power on the cell is $1000 \cdot 0.0244$ ⁠ watts, and the electrical power is $P = VI$ ⁠.

Worked solution
1. The light power falling on the cell: $1000\ \mathrm{W/m^2} \cdot 0.0244\ \mathrm{m^2} = 24.4\ \mathrm{W}$ ⁠.
2. The electrical power from the meter: $P = VI \approx 0.550\ \mathrm{V} \cdot 8.78\ \mathrm{A} \approx 4.82\ \mathrm{W}$ ⁠.
3. The efficiency: $\eta = \dfrac{4.82}{24.4} \approx 0.198$ ⁠, about $20\%$ ⁠.
The meter shows the power too, so you can read it directly instead of working out $VI$ ⁠.
Step 4 of 7
Open this step
Where the rest goes
Practice · Choose
Question
About four fifths of the light energy falling on the cell does not become electricity. Where does most of it go?

Correct answer
- Correct Photons with too little energy pass through the silicon without freeing anything, and photons with too much give up the excess as heat
Setup, choices and feedback
The Scene at this step
Sunlight
1000 W/m²
Angle to the Sun
0 °
Zoom to the junction
Junction
Choices
Shown in a random order.
- Correct Photons with too little energy pass through the silicon without freeing anything, and photons with too much give up the excess as heat
- Wrong It is stored in the cell, which gives it back at night
  Feedback
  A solar cell stores nothing: at night it gives nothing. Storing energy takes a separate battery.
- Wrong Almost all of it is reflected off the glass
  Feedback
  An anti-reflective coating leaves only a few per cent reflected, and this model ignores even those. Most of the energy gets into the silicon and is lost there.
- Wrong It is burnt up in the wires on the way to the bulb
  Feedback
  Losses in the wires are small, and this model has none at all.
Feedback for a right answer
Photons that are too weak are wasted entirely, and those that are too strong waste their excess. That is most of the loss.

Worked solution
1. Each absorbed photon frees at most one pair, and each pair gives the circuit roughly a fixed amount of energy, set by the silicon.
2. Deep red and infrared photons are too weak to free a pair, and they pass through the cell.
3. Blue photons free a pair, but all the energy beyond what the pair needs turns into heat.
These two losses alone take more than half the light, so a single-layer silicon cell cannot get much past about $30\%$ ⁠.
Step 5 of 7
Open this step
Tilted away from the Sun
Practice · Calculate
Question
Angle to the Sun is 60°: the panel is tilted away from the rays. How much light power now falls on one cell, 0.0244 m² in area?

Correct answer
12.2 W ± 2%
Setup, choices and feedback
The Scene at this step
Sunlight
1000 W/m²fixed
Angle to the Sun
60 °fixed
Zoom to the junction
Panel
Accepted within
± 2%
Computed from the Scene at this step’s setting
Sunlight striking one cell, by the cosine law/nodes/panel/outputs/lightPower
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 8.13 W ± 3%
  Misconception: Thinks the light a tilted panel catches falls in a straight line with the angle, to nothing at 90°.
  Feedback from the catalogue
  A beam of sunlight has a fixed width. When the panel is tilted, the same beam spreads over a larger area, and each square metre gets a share $\cos\theta$ ⁠ of the light. The cosine falls slowly at first and fast at the end: at 30° the panel still catches $87\%$ ⁠, at 60° exactly half, and only at 90° nothing.
Feedback for a right answer
$24.4 \cos 60^\circ = 12.2\ \mathrm{W}$ ⁠: half the light, because the same beam spreads over twice the area.

Hint
The light on each square metre of the panel is in proportion to $\cos\theta$ ⁠, where $\theta$ ⁠ is the angle between the way the panel faces and the direction of the Sun.

Worked solution
1. Facing the Sun, $1000 \cdot 0.0244 = 24.4\ \mathrm{W}$ ⁠ falls on the cell.
2. Tilted 60°, the same beam spreads over a larger area, and each square metre gets $\cos 60^\circ = 0.5$ ⁠ of the light.
3. $24.4 \cdot 0.5 = 12.2\ \mathrm{W}$ ⁠.
Answering about 8.1 W assumes the light falls in a straight line with the angle, two thirds of the way to 90°.
Step 6 of 7
Open this step
Half the power
Practice · Reach the target
Question
Facing the Sun, the cell gives about 4.8 W. Turn Angle to the Sun until it gives half that, about 2.4 W on the meter, and press Check.

Goal
- 2.41 W ± 0.06 W
  At the start of the step: 4.824 W
Setup, choices and feedback
The Scene at this step
Sunlight
1000 W/m²fixed
Angle to the Sun
0 °
Zoom to the junction
Panel
Read from the Scene at
Electrical power one cell delivers/nodes/panel/outputs/solarPower
One setting that reaches it
Angle to the Sun
58 °
Feedback for a right answer
About 58–59°, not 60°: in weaker light the voltage drops a little too, so it takes a little more than half the light.

Feedback for a wrong answer
Not 2.4 W yet. Move Angle to the Sun one degree at a time and read the power on the meter.

Hint
At 60°, half the light falls. Does half the light give exactly half the power?

Worked solution
1. At 60° exactly half the light falls, and yet the meter shows only about 2.32 W, a little under half.
2. In weaker light the current halves, but the voltage drops a little too, so the efficiency falls from about $19.8\%$ ⁠ to about $19.0\%$ ⁠.
3. Half the power takes a little more than half the light, so a slightly smaller angle: about 58–59°.
The difference is small, but it shows that a cell’s power is not exactly in proportion to the light.
Step 7 of 7
Open this step
The photocurrent
Practice · Calculate
Question
Facing the Sun, with the cell’s two terminals joined to each other, a short-circuit current of 9.27 A flows. The Angle to the Sun is now locked at 60°. What is the short-circuit current?

Correct answer
4.636 A ± 3%
Setup, choices and feedback
The Scene at this step
Sunlight
1000 W/m²fixed
Angle to the Sun
60 °fixed
Zoom to the junction
Panel
Accepted within
± 3%
Computed from the Scene at this step’s setting
One cell's current with its terminals joined: the photocurrent/nodes/panel/outputs/shortCircuitCurrent
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3.09 A ± 3%
  Misconception: Thinks the light a tilted panel catches falls in a straight line with the angle, to nothing at 90°.
  Feedback from the catalogue
  A beam of sunlight has a fixed width. When the panel is tilted, the same beam spreads over a larger area, and each square metre gets a share $\cos\theta$ ⁠ of the light. The cosine falls slowly at first and fast at the end: at 30° the panel still catches $87\%$ ⁠, at 60° exactly half, and only at 90° nothing.
Feedback for a right answer
About 4.64 A, half: half the light, half the photons, half the current.

Hint
The short-circuit current is the photocurrent: how many electron–hole pairs are made each second. It goes with the light falling on the cell.

Worked solution
1. At an angle $\theta$ ⁠ the cell catches $\cos\theta$ ⁠ of the light it would catch facing the Sun.
2. $\cos 60^\circ = 0.5$ ⁠.
3. $9.27 \times 0.5 \approx 4.64$ ⁠ amperes.
If you answered about 3.1 A, you assumed the light falls in a straight line with the angle, a third every 30°. It falls slowly at first, and fast near 90°.

Lesson 2 of 6 · About 35 minutes

The transistor as a switch

A switch with no moving parts, turned on by a small voltage. The gate that draws no current, the threshold, the square law, why a half-on transistor heats up, and why computers work in only two states.

Open the lesson

The 9 steps and their answers

Step 1 of 9
Open this step
Three legs
Practice · Choose
Question
Gate voltage is 5 V, and the bulb is lit: almost half an ampere flows through it. How much current flows into the gate itself?

Correct answer
- Correct Practically none: the oxide layer insulates the gate from the silicon
Setup, choices and feedback
The Scene at this step
Gate voltage
5 V
Zoom to the channel
Chip
Choices
Shown in a random order.
- Correct Practically none: the oxide layer insulates the gate from the silicon
- Wrong The same current as through the bulb, since the gate is what feeds it
  Misconception: Thinks the current into a transistor's gate is what feeds or drives the bulb.
  Feedback from the catalogue
  A MOSFET’s gate is separated from the silicon by an insulating oxide layer, so no current flows through it, except for a brief moment when its voltage changes. It acts through its electric field: it draws electrons to the surface and makes a channel. All the current in the bulb, and all its energy, comes from the battery.
- Wrong A small current, which the transistor amplifies a hundredfold on the way to the bulb
  Misconception: Thinks the current into a transistor's gate is what feeds or drives the bulb.
  Feedback from the catalogue
  A MOSFET’s gate is separated from the silicon by an insulating oxide layer, so no current flows through it, except for a brief moment when its voltage changes. It acts through its electric field: it draws electrons to the surface and makes a channel. All the current in the bulb, and all its energy, comes from the battery.
- Wrong Half the current in the bulb: the gate and the battery share the work
  Misconception: Thinks the current into a transistor's gate is what feeds or drives the bulb.
  Feedback from the catalogue
  A MOSFET’s gate is separated from the silicon by an insulating oxide layer, so no current flows through it, except for a brief moment when its voltage changes. It acts through its electric field: it draws electrons to the surface and makes a channel. All the current in the bulb, and all its energy, comes from the battery.
Feedback for a right answer
The gate only “presses” with its electric field. The battery supplies the energy for the bulb.

Worked solution
1. The gate is separated from the silicon by a thin layer of silicon oxide, which is an insulator. No charge can cross it.
2. The gate acts from a distance, through its electric field: the positive charge on it draws electrons to the surface and makes the channel.
3. The current in the bulb comes entirely from the 9 V battery, through the source, the channel and the drain. The gate supply’s meter shows 0.000 A.
A tiny current flows only for a moment, while the gate voltage changes and the gate charges up.
Step 2 of 9
Open this step
The threshold
Practice · Predict, then look
Question
Gate voltage is at zero, and the bulb is dark. It is raised to 1 V, two thirds of the 1.5 V threshold. What will the bulb do?

Correct answer
- Correct Stay dark: below the threshold there is no channel
Setup, choices and feedback
The Scene at this step
Gate voltage
0 V
Zoom to the channel
Chip
Choices
- Correct Stay dark: below the threshold there is no channel
- Wrong Light dimly, at about two thirds of its brightness near the threshold
  Feedback
  The current does not grow gradually from zero. Below the threshold the gate only pushes holes away from the surface, and there is not yet a sheet of electrons joining source to drain.
- Wrong Light fully: any voltage on the gate opens the switch
  Feedback
  Then a little electrical noise would switch it on. The threshold is exactly what lets a transistor be truly off.
Once the prediction is locked in, the Scene changes
Gate voltage
1 V
Zoom to the channel
Channel
The Scene then shows: 0 mA · Current through the transistor and the bulb/nodes/switch/outputs/drainCurrent
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Zero current. Below the threshold the transistor is fully off, and that is what makes it a good switch.

Worked solution
1. Between source and drain lies p-type silicon, with almost no free electrons.
2. A positive gate voltage first pushes the holes away from the surface. Only past the threshold do enough electrons gather to make an n-type channel.
3. At 1 V, below the 1.5 V threshold, there is no channel, and the current is zero.
A real transistor leaks a tiny current below the threshold too; the model ignores it.
Step 3 of 9
Open this step
The square law
Practice · Predict, then look
Question
Gate voltage is right at the threshold, 1.5 V. With $V_{GS}$ ⁠ above the threshold but the transistor still limiting the current, $I = \tfrac{k}{2}\left(V_{GS} - V_T\right)^2$ ⁠ with $k = 0.5\ \mathrm{A/V^2}$ ⁠. What will the current be at 2 V, in milliamperes?

Correct answer
62.5 mA ± 5%
A number prediction counts as right within ± 5%
Setup, choices and feedback
The Scene at this step
Gate voltage
1.5 V
Zoom to the channel
Channel
Once the prediction is locked in, the Scene changes
Gate voltage
2 V
The Scene then shows: 62.5 mA · Current through the transistor and the bulb/nodes/switch/outputs/drainCurrent
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$0.25 \cdot 0.5^2 = 0.0625\ \mathrm{A}$ ⁠: about $63\ \mathrm{mA}$ ⁠, and the bulb barely glows.

Hint
The voltage above the threshold is $2 - 1.5 = 0.5\ \mathrm{V}$ ⁠.

Worked solution
1. $V_{GS} - V_T = 2 - 1.5 = 0.5\ \mathrm{V}$ ⁠.
2. $I = \tfrac{0.5}{2} \cdot 0.5^2 = 0.25 \cdot 0.25 = 0.0625\ \mathrm{A}$ ⁠, about $63\ \mathrm{mA}$ ⁠.
3. That is far less than the 0.5 A the bulb would take: the channel is still thin, and it, not the bulb, limits the current.
Working out $\dfrac{9}{18} = 0.5\ \mathrm{A}$ ⁠ treats the transistor as a closed switch already at 2 V.
Step 4 of 9
Open this step
Halfway
Practice · Calculate
Question
Gate voltage is 2.5 V, and the meter shows a current of 0.25 A. The bulb is 18 Ω and the battery 9 V. How much power heats the transistor itself?

Correct answer
1.125 W ± 3%
Setup, choices and feedback
The Scene at this step
Gate voltage
2.5 Vfixed
Zoom to the channel
Channel
Accepted within
± 3%
Computed from the Scene at this step’s setting
Power heating the transistor, I·V_DS/nodes/switch/outputs/transistorPower
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 W ± 0.05 W
  Misconception: Thinks a transistor takes no power whatever its state, even half on.
  Feedback from the catalogue
  The power heating a transistor is $P = I V_{DS}$ ⁠, the current through it times the voltage across it. Off, $I = 0$ ⁠; fully on, $V_{DS}$ ⁠ is very small. Either way $P$ ⁠ is almost zero. But halfway, both the current and the voltage are large, and the transistor heats up like a resistor. That is why switches, and chips, work only at the two extremes.
Feedback for a right answer
$0.25 \cdot 4.5 \approx 1.1\ \mathrm{W}$ ⁠: half the battery’s power heats the transistor instead of the bulb.

Hint
The voltage across the bulb is $IR$ ⁠, and the rest of the $9\ \mathrm{V}$ ⁠ falls across the transistor.

Worked solution
1. The voltage across the bulb: $V = IR = 0.25 \cdot 18 = 4.5\ \mathrm{V}$ ⁠.
2. The bulb and the transistor are in series, so the rest falls across the transistor: $V_{DS} = 9 - 4.5 = 4.5\ \mathrm{V}$ ⁠.
3. The power heating it: $P = I V_{DS} = 0.25 \cdot 4.5 = 1.125\ \mathrm{W}$ ⁠, exactly as much as the bulb gets.
Answering zero assumes a switch never wastes power. That is true only when it is fully on or fully off.
Step 5 of 9
Open this step
Fully on
Practice · Predict, then look
Question
Gate voltage is 3 V, and the current is about 0.45 A. It is raised to 5 V, so the voltage above the threshold grows from 1.5 to 3.5 V. How much will the current grow?

Correct answer
- Correct Hardly at all: now the bulb limits the current
Setup, choices and feedback
The Scene at this step
Gate voltage
3 V
Zoom to the channel
Channel
Choices
- Correct Hardly at all: now the bulb limits the current
- Wrong By $\left(\tfrac{3.5}{1.5}\right)^2 \approx 5.4$ ⁠ times, by the square law
  Feedback
  The square law holds only while the transistor is what limits the current. 5.4 times would give about 2.4 A, but the bulb and the battery alone allow at most $\dfrac{9}{18} = 0.5\ \mathrm{A}$ ⁠.
- Wrong About twice, like the voltage above the threshold
  Feedback
  The current is already close to the bulb’s limit, $\dfrac{9}{18} = 0.5\ \mathrm{A}$ ⁠, and cannot double.
Once the prediction is locked in, the Scene changes
Gate voltage
5 V
The Scene then shows: 0.484 A · Current through the transistor and the bulb/nodes/switch/outputs/drainCurrent
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about 0.45 to about 0.48 A. The transistor is already a closed switch, and the bulb sets the current.

Worked solution
1. With no transistor at all, the bulb and the battery give $I = \dfrac{9}{18} = 0.5\ \mathrm{A}$ ⁠. That is the limit.
2. At 3 V the current is already about 0.45 A: the channel is thick, and less than a volt falls across the transistor.
3. At 5 V the channel thickens further, and the current creeps towards the limit: about 0.48 A.
This is the transistor as a closed switch: past a certain point, more gate voltage hardly changes anything.
Step 6 of 9
Open this step
Zeros and ones
Practice · Choose
Question
In a processor, every transistor is almost always fully on or fully off, and never stays halfway. Why?

Correct answer
- Correct Fully on or off it wastes almost no power, and a voltage that is slightly off still reads as the same 0 or 1
Setup, choices and feedback
The Scene at this step
Gate voltage
5 V
Zoom to the channel
Chip
Choices
Shown in a random order.
- Correct Fully on or off it wastes almost no power, and a voltage that is slightly off still reads as the same 0 or 1
- Wrong It makes no difference to power, since a transistor never takes any power
  Misconception: Thinks a transistor takes no power whatever its state, even half on.
  Feedback from the catalogue
  The power heating a transistor is $P = I V_{DS}$ ⁠, the current through it times the voltage across it. Off, $I = 0$ ⁠; fully on, $V_{DS}$ ⁠ is very small. Either way $P$ ⁠ is almost zero. But halfway, both the current and the voltage are large, and the transistor heats up like a resistor. That is why switches, and chips, work only at the two extremes.
- Wrong A transistor cannot be halfway: it is only on or off
  Feedback
  It can, and you saw it at 2.5 V: the channel is thin, and the transistor limits the current. Amplifiers work like that on purpose.
- Wrong A half-on transistor is slower, but otherwise there is no difference
  Feedback
  The main difference is heat: a half-on transistor wastes power, and with billions of them the chip would burn.
Feedback for a right answer
Less heat and fewer errors. That is why computers think in only two states, 0 and 1.

Worked solution
1. The power heating the transistor is $P = I V_{DS}$ ⁠.
2. Off: $I = 0$ ⁠, so $P = 0$ ⁠. Fully on: $V_{DS}$ ⁠ is very small, so $P$ ⁠ is small. Halfway, both $I$ ⁠ and $V_{DS}$ ⁠ are large, and so is $P$ ⁠.
3. Also, with the two states far apart, a little noise on the voltage does not turn a 0 into a 1. That is the strength of digital information.
Real chips pair n-channel and p-channel transistors, so current flows almost only at the moment of switching.
Step 7 of 9
Open this step
How bright when it is on
Practice · Calculate
Question
Gate voltage is at 5 V, and the transistor is fully on: only 0.29 V is left across it. The battery is 9 V, and the bulb 18 Ω. What power does the bulb take?

Correct answer
4.216 W ± 3%
Setup, choices and feedback
The Scene at this step
Gate voltage
5 Vfixed
Zoom to the channel
Chip
Accepted within
± 3%
Computed from the Scene at this step’s setting
Power the bulb takes/nodes/switch/outputs/bulbPower
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4.5 W ± 2%
  Misconception: Treats a transistor that is fully on as a perfect wire, with no voltage across it.
  Feedback from the catalogue
  A transistor fully on is a very small resistance, not zero. A small voltage stays across it, a few tenths of a volt, and the bulb gets the battery’s voltage less that. The difference is small, which is why a good switch wastes very little.
Feedback for a right answer
About 4.2 W. Only about 0.14 W heats the transistor: a good switch wastes almost nothing.

Hint
How much voltage is left for the bulb after the transistor? The power in a resistor is $\dfrac{V^2}{R}$ ⁠.

Worked solution
1. The voltage across the bulb: $9 - 0.29 = 8.71$ ⁠ volts.
2. $P = \dfrac{V^2}{R} = \dfrac{8.71^2}{18} \approx \dfrac{75.9}{18} \approx 4.2$ ⁠ watts.
3. The current: $\dfrac{8.71}{18} \approx 0.48$ ⁠ amperes, and only $0.29 \times 0.48 \approx 0.14$ ⁠ watts are wasted in the transistor.
If you answered 4.5 W, you gave the bulb the whole 9 V. A transistor fully on is almost a wire, but not quite.
Step 8 of 9
Open this step
The voltage across the switch
Practice · Calculate
Question
Gate voltage is locked at 3 V, and a current of 0.453 A flows through the bulb, 18 Ω. The battery is 9 V. What is the voltage across the transistor, drain to source?

Correct answer
0.8393 V ± 5%
Setup, choices and feedback
The Scene at this step
Gate voltage
3 Vfixed
Zoom to the channel
Chip
Accepted within
± 5%
Computed from the Scene at this step’s setting
Voltage across the transistor, drain to source/nodes/switch/outputs/drainVoltage
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3 V ± 3%
  Misconception: Takes the gate voltage for the voltage across the transistor’s channel.
  Feedback from the catalogue
  The gate voltage is measured between the gate and the source, and only opens or closes the channel. The voltage across the transistor itself, drain to source, is set in the bulb’s circuit: what is left of the battery after the bulb.
Feedback for a right answer
About 0.84 V: the transistor is almost fully on, and the bulb takes nearly all the voltage.

Hint
The bulb and the transistor are in series: the two voltages together make the battery’s.

Worked solution
1. The voltage across the bulb: $V = IR = 0.453 \times 18 \approx 8.16$ ⁠ volts.
2. Left across the transistor: $9 - 8.16 \approx 0.84$ ⁠ volts.
Here the bulb, not the transistor, sets the current: the transistor is almost fully on. The gate voltage controls the switch, but it is not the voltage across it.
Step 9 of 9
Open this step
A tap, not a switch
Practice · Reach the target
Question
Gate voltage is at 5 V, and the transistor is fully on: the bulb sets the current. Change the gate voltage until the transistor limits the current, only partly on and warming up, and press Check.

Goal
- 1 ± 0.1
  At the start of the step: 2
Setup, choices and feedback
The Scene at this step
Gate voltage
5 V
Zoom to the channel
Channel
Read from the Scene at
0 off, 1 saturation (the transistor limits the current), 2 linear (fully on, the bulb limits it)/nodes/switch/outputs/region
One setting that reaches it
Gate voltage
1.6 V
Feedback for a right answer
The transistor is partly on: it sets the current, and a large voltage left across it heats it.

Hint
Below 1.5 V the transistor is fully off. How far above that is it still not fully on?

Worked solution
1. Below the threshold, 1.5 V, there is no current at all.
2. A little above the threshold the current is $\tfrac{k}{2}(V_{GS} - V_{th})^2$ ⁠: the transistor limits it, and a large voltage is left across it.
3. Above about 2.9 V that current is more than the bulb lets through, and the bulb limits it.
4. So any gate voltage from 1.6 to about 2.8 V will do.
In this range the transistor is a tap, not a switch: it controls the current, but wastes power. Amplifiers use this; processors avoid it.

Lesson 3 of 6 · About 30 minutes

Radio links

How a message is written on a radio wave, and why it weakens on the way. Carriers and modulation, the inverse-square law in decibels, why FM is cleaner than AM and what it costs, and the cliff of digital broadcasting.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
A carrier
Practice · Choose
Question
Decode the signal is at Message. Under Modulation pick FM. In the chart of the transmitted wave, which feature of the carrier changes with the tone?

Correct answer
- Correct Its frequency: the waves crowd together and spread apart
Setup, choices and feedback
The Scene at this step
Modulation
AM
Distance
5 km
Decode the signal
Message
Choices
Shown in a random order.
- Correct Its frequency: the waves crowd together and spread apart
- Wrong Its height, the amplitude
  Feedback
  That is AM. In FM the strength is constant, so noise that changes the wave’s height barely touches the message.
- Wrong The speed at which the wave travels from the mast to the house
  Feedback
  All radio waves travel at the speed of light, whatever the modulation. What changes is the frequency, and with it the wavelength.
- Wrong Its direction, which flips between two symbols
  Feedback
  That is the Scene’s digital modulation: the wave flips each time the bit changes.
Feedback for a right answer
The frequency. An FM wave’s strength is constant, and that is what gives it its edge against noise.

Worked solution
1. AM: the tone changes the carrier’s height.
2. FM: the height is constant, and the tone changes the frequency, how many waves arrive each second.
3. Digital: the wave flips between two fixed symbols, following the bits.
An FM receiver reads only the timing of the waves, so it ignores most of the noise, which changes the height.
Step 2 of 7
Open this step
Weaker every kilometre
Practice · Calculate
Question
At 1 km the signal arrives 40 dB above the receiver’s noise, and its power falls like $\dfrac{1}{r^2}$ ⁠ while the noise stays the same. The house’s Distance is 10 km. What is the signal-to-noise ratio there, in decibels?

Correct answer
20 ± 0.3
Setup, choices and feedback
The Scene at this step
Modulation
AM
Distance
10 kmfixed
Decode the signal
Signal
Accepted within
± 0.3
Computed from the Scene at this step’s setting
Received signal-to-noise ratio, dB/nodes/link/outputs/snr
Wrong answers it recognises
- 30 ± 0.3
  Misconception: Thinks a radio signal's power falls in proportion to the distance, not to its square.
  Feedback from the catalogue
  A radio wave spreads out in every direction, and its power is shared over the area of a sphere, $4\pi r^2$ ⁠. So at twice the distance the power is $4$ ⁠ times smaller, and at $10$ ⁠ times the distance it is $100$ ⁠ times smaller. In decibels: 6 dB less for every doubling of the distance, and 20 dB less for every factor of ten.
Feedback for a right answer
$40 - 20 \log_{10} 10 = 20\ \mathrm{dB}$ ⁠: ten times the distance costs $20\ \mathrm{dB}$ ⁠, and each doubling costs $6\ \mathrm{dB}$ ⁠.

Hint
$10$ ⁠ times the distance is $100$ ⁠ times less power. How many decibels is a factor of $100$ ⁠?

Worked solution
1. The power falls like $\dfrac{1}{r^2}$ ⁠: $10$ ⁠ times as far, the signal is $10^2 = 100$ ⁠ times weaker.
2. In decibels, a power ratio $x$ ⁠ is $10 \log_{10} x$ ⁠, so a factor of $100$ ⁠ is $20\ \mathrm{dB}$ ⁠.
3. Ratios that multiply become decibels that subtract: $40 - 20 = 20\ \mathrm{dB}$ ⁠.
Answering 30 dB assumes the power falls only $10$ ⁠ times, like the distance. Answering $4$ ⁠ divides the decibels instead of subtracting.
Step 3 of 7
Open this step
From decibels to a ratio
Practice · Calculate
Question
At 10 km the signal arrives 20 dB above the noise. How many times bigger is the signal’s power than the noise’s?

Correct answer
100 ± 3%
Setup, choices and feedback
The Scene at this step
Modulation
AM
Distance
10 kmfixed
Decode the signal
Signal
Accepted within
± 3%
Computed from the Scene at this step’s setting
Received signal power over noise power, a plain ratio/nodes/link/outputs/signalToNoise
Wrong answers it recognises
- 20 ± 3%
  Misconception: Reads a number of decibels as the power ratio itself.
  Feedback from the catalogue
  A decibel is a logarithm: $\mathrm{dB} = 10 \log_{10} \dfrac{P_1}{P_2}$ ⁠. Every 10 dB is ten times the power, so 20 dB is $100$ ⁠ times, 30 dB $1000$ ⁠ times, and 3 dB about twice. To get back to the ratio: $\dfrac{P_1}{P_2} = 10^{\mathrm{dB}/10}$ ⁠.
Feedback for a right answer
$100$ ⁠ times: decibels are a logarithm, so a small number of decibels covers a huge range of ratios.

Hint
Every $10\ \mathrm{dB}$ ⁠ is ten times the power.

Worked solution
1. The definition: $\mathrm{dB} = 10 \log_{10} \dfrac{P_s}{P_n}$ ⁠.
2. Inverting: $\dfrac{P_s}{P_n} = 10^{\mathrm{dB}/10} = 10^{20/10} = 10^2$ ⁠.
3. The signal is $100$ ⁠ times stronger than the noise.
Answering $20$ ⁠ reads the decibels as if they were the ratio itself. 40 dB is $10{,}000$ ⁠ times, not $40$ ⁠.
Step 4 of 7
Open this step
AM against FM
Practice · Predict, then look
Question
The same mast, the same power and the same distance, 10 km, with Modulation on AM. It is switched to FM. What will happen to the decoded tone?

Correct answer
- Correct It will be much cleaner
Setup, choices and feedback
The Scene at this step
Modulation
AM
Distance
10 kmfixed
Decode the signal
Message
Choices
- Correct It will be much cleaner
- Wrong The same: the signal and the noise have not changed
  Feedback
  The signal-to-noise ratio coming in has not changed, but an FM receiver reads only the timing of the waves, and most of the noise changes their height.
- Wrong Noisier, because FM takes a wider band and collects more noise
  Feedback
  FM does collect noise from a wider band, but as long as the signal is above the threshold, the large swings in frequency outweigh the noise by far.
Once the prediction is locked in, the Scene changes
Modulation
FM
Decode the signal
Message
The Scene then shows: 27.78 · The decoded tone's SNR, dB (AM, FM); for digital, Eb/N0 in dB/nodes/link/outputs/outputSnr
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 28 dB against 14 dB: the same signal, $25$ ⁠ times less noise in the tone.

Worked solution
1. In AM most of the power goes into the carrier itself, which carries no information. The decoded tone comes out about 14 dB above the noise.
2. In FM, with a frequency swing twice the tone’s bandwidth, $\beta = 2$ ⁠, the receiver gains $\tfrac{3}{2}\beta^2 = 6$ ⁠ times, about 8 dB, over the incoming ratio: about 28 dB.
3. The difference is about 14 dB, $25$ ⁠ times in power.
The price: FM takes a wider slice of the spectrum, and past its threshold it falls apart.
Step 5 of 7
Open this step
The price of FM
Practice · Calculate
Question
The tone takes up a band of frequencies $W$ ⁠ wide. AM needs $2W$ ⁠. The Scene’s FM swings the frequency by up to $\Delta f = 2W$ ⁠ each way, and by Carson’s rule it needs $2\left(\Delta f + W\right)$ ⁠. How many $W$ ⁠ does FM need?

Correct answer
6 ± 0.05
Setup, choices and feedback
The Scene at this step
Modulation
FMfixed
Distance
10 km
Decode the signal
Message
Accepted within
± 0.05
Computed from the Scene at this step’s setting
The band the receiver must pass, in units of the message band W/nodes/link/outputs/bandwidth
Feedback for a right answer
$6W$ ⁠, three times AM. FM buys its quiet with bandwidth.

Worked solution
1. $\Delta f = 2W$ ⁠.
2. Carson’s rule: $2\left(\Delta f + W\right) = 2\left(2W + W\right) = 6W$ ⁠.
3. Three times AM’s $2W$ ⁠.
That is the price of the clean tone: the same strip of spectrum could hold three AM stations. That is why FM stations broadcast at high frequencies, where there is room.
Step 6 of 7
Open this step
The digital cliff
Practice · Predict, then look
Question
Modulation is digital, and the house is at 30 km: about one bit in a million arrives wrong. It is moved out to 60 km, twice as far. What will happen to the bit error rate?

Correct answer
- Correct It will leap thousands of times, to about one in a hundred
Setup, choices and feedback
The Scene at this step
Modulation
Digitalfixed
Distance
30 km
Decode the signal
Message
Choices
- Correct It will leap thousands of times, to about one in a hundred
- Wrong It will grow four times, like the power lost
  Misconception: Thinks a digital signal's errors grow steadily with distance, as AM's hiss does.
  Feedback from the catalogue
  A bit is read wrongly only when the noise is big enough to push it past the decision line. Big noise spikes are very rare: their chance falls off like the tail of the bell curve. So the error rate does not grow in step with the distance; it leaps, from one in a million to one in a hundred across only 6 dB. That is the digital cliff.
- Wrong It will double, like the distance
  Misconception: Thinks a digital signal's errors grow steadily with distance, as AM's hiss does.
  Feedback from the catalogue
  A bit is read wrongly only when the noise is big enough to push it past the decision line. Big noise spikes are very rare: their chance falls off like the tail of the bell curve. So the error rate does not grow in step with the distance; it leaps, from one in a million to one in a hundred across only 6 dB. That is the digital cliff.
- Wrong It will stay one in a million: digital is not affected by noise
  Feedback
  A bit comes out wrong when the noise pushes it past the decision line. As the signal weakens, that happens far more often.
Once the prediction is locked in, the Scene changes
Distance
60 km
Decode the signal
Message
The Scene then shows: 0.009211 · Share of bits read wrongly, ½ erfc(√(Eb/N0)); NaN for AM and FM/nodes/link/outputs/bitErrorRate
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about $10^{-6}$ ⁠ to about $10^{-2}$ ⁠ across only 6 dB. In AM the same distance would add hiss gradually.

Worked solution
1. Twice the distance is 6 dB less: from about 10.5 to about 4.4 dB.
2. A bit is read wrongly only when the noise is big enough to flip it. The chance of such a noise spike falls off like the tail of the bell curve: $\tfrac12 \operatorname{erfc}\sqrt{E_b/N_0}$ ⁠.
3. So a few decibels take the rate from about $10^{-6}$ ⁠ to about $10^{-2}$ ⁠: several thousand times.
That is the digital cliff: perfect, then almost all at once unusable.
Step 7 of 7
Open this step
On the edge of the cliff
Practice · Reach the target
Question
A good digital radio can correct about one wrong bit in a thousand, but no more. Move the house out with Distance until just one bit in a thousand arrives wrong, and press Check.

Goal
- 0.001 ± 25%
  At the start of the step: 1.044 × 10⁻⁴⁵
Setup, choices and feedback
The Scene at this step
Modulation
Digitalfixed
Distance
10 km
Decode the signal
Message
Read from the Scene at
Share of bits read wrongly, ½ erfc(√(Eb/N0)); NaN for AM and FM/nodes/link/outputs/bitErrorRate
One setting that reaches it
Distance
45 km
Feedback for a right answer
About 45–46 km: a little farther, and the radio can no longer correct the errors.

Feedback for a wrong answer
Not yet. Use the chart of decoded bits beside the Scene, and move one kilometre at a time.

Hint
One in a thousand needs about 7 dB above the noise. How far is that from 40 dB at 1 km?

Worked solution
1. For one bit in a thousand, $E_b/N_0$ ⁠ must be about 6.8 dB.
2. From 40 dB at 1 km it falls by $20 \log_{10} r$ ⁠: $20 \log_{10} r = 40 - 6.8 = 33.2$ ⁠.
3. $r = 10^{33.2/20} \approx 46\ \mathrm{km}$ ⁠.
At 30 km the rate was one in a million, and at 60 km one in a hundred. The whole cliff lies within a factor of two in distance.

Lesson 4 of 6 · About 30 minutes

GPS

How a phone knows where it is without transmitting anything. Distance from time, trilateration with spheres, why a microsecond of clock error is worth hundreds of metres, why a fourth satellite is needed, and why without relativity GPS would be kilometres out every day.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
Listening
Practice · Choose
Question
A phone finds where it is with the GPS satellites. What does it transmit to them?

Correct answer
- Correct Nothing: it only listens to the time signals they broadcast
Setup, choices and feedback
The Scene at this step
Satellites used
4
Receiver clock error
0 µs
Show the timing spheres
Off
Choices
Shown in a random order.
- Correct Nothing: it only listens to the time signals they broadcast
- Wrong A signal the satellites send back, like radar, and it times how long it took to return
  Misconception: Thinks a GPS receiver sends signals to the satellites, or that they track it like radar.
  Feedback from the catalogue
  A GPS receiver transmits nothing. Each satellite broadcasts, over and over, the time on its atomic clock and where it is, and the receiver only listens. It measures how long each signal was on the way, turns the time into a distance, $d = c\,t$ ⁠, and works out for itself where it is. The signal makes a one-way trip, from satellite to receiver, so there is no halving as in radar.
- Wrong A request, and the satellites work out where it is and send back the answer
  Misconception: Thinks a GPS receiver sends signals to the satellites, or that they track it like radar.
  Feedback from the catalogue
  A GPS receiver transmits nothing. Each satellite broadcasts, over and over, the time on its atomic clock and where it is, and the receiver only listens. It measures how long each signal was on the way, turns the time into a distance, $d = c\,t$ ⁠, and works out for itself where it is. The signal makes a one-way trip, from satellite to receiver, so there is no halving as in radar.
- Wrong Its own time, so the satellites can correct its clock
  Feedback
  The satellites do not know the receiver exists. The receiver corrects its own clock, with the help of a fourth satellite.
Feedback for a right answer
GPS is listen-only: the satellites broadcast time and place, and the receiver does all the working out.

Worked solution
1. Each satellite broadcasts, over and over, the time on its atomic clock and where it is.
2. The receiver compares the time stamp with the time the signal arrived by its own clock. The difference times the speed of light is the distance to the satellite.
3. From several such distances it works out for itself where it is.
That is why any number of receivers can use GPS at once, and the satellites know where none of them are.
Step 2 of 7
Open this step
Distance from time
Practice · Calculate
Question
A GPS satellite straight overhead is about 20,200 km away. How long does its signal take to reach you, in milliseconds? Then compare with the time for the first satellite in the list beside the Scene, the highest in the sky.

Correct answer
68.09 ms ± 4%
Setup, choices and feedback
The Scene at this step
Satellites used
4
Receiver clock error
0 µs
Show the timing spheres
Off
Accepted within
± 4%
Computed from the Scene at this step’s setting
Time the signal takes from the first satellite used, the highest/nodes/gps/outputs/travelTime
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 67 ms: under a tenth of a second, and still long enough to measure.

Hint
Light travels about 300,000 km in a second.

Worked solution
1. $t = \dfrac{d}{c}$ ⁠.
2. $t = \dfrac{20{,}200\ \mathrm{km}}{300{,}000\ \mathrm{km/s}} \approx 0.067\ \mathrm{s}$ ⁠, about $67\ \mathrm{ms}$ ⁠.
3. The highest satellite here is not quite straight overhead, so it is a little farther: about 68 ms.
A short time, but the receiver has to measure it to a billionth of a second to know the distance to tens of centimetres.
Step 3 of 7
Open this step
Spheres
Practice · Predict, then look
Question
With one satellite the receiver is somewhere on a sphere, and with two on a circle. The receiver’s clock is exact. Where can it be with three satellites?

Correct answer
- Correct At one of two points, one of them far out in space
Setup, choices and feedback
The Scene at this step
Satellites used
2
Receiver clock error
0 µsfixed
Show the timing spheres
On
Choices
- Correct At one of two points, one of them far out in space
- Wrong At exactly one point
  Feedback
  The third sphere cuts the first two’s circle at two points, one on each side of the plane through the three satellites. The receiver throws away the one in space.
- Wrong On a smaller circle
  Feedback
  A sphere and a circle meet at points, not on a circle. Each satellite takes away a dimension: sphere, circle, points.
Once the prediction is locked in, the Scene changes
Satellites used
3
Show the timing spheres
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Two points, and only one of them on the Earth. With a perfect clock, three satellites would be enough.

Worked solution
1. One satellite: every point at the right distance from it, a sphere.
2. Two spheres meet in a circle.
3. A third sphere cuts the circle at two points, mirror images in the plane through the three satellites. One is near the ground, and the other far out in space, and the receiver throws it away.
This is trilateration: position from distances. Surveyors did it with measuring tapes long before GPS.
Step 4 of 7
Open this step
The receiver's clock
Practice · Predict, then look
Question
The Receiver clock error is zero. If the clock runs 250 µs fast, a quarter of a millisecond, by how many kilometres will every range come out too long?

Correct answer
74.95 km ± 5%
A number prediction counts as right within ± 5%
Setup, choices and feedback
The Scene at this step
Satellites used
4fixed
Receiver clock error
0 µs
Show the timing spheres
Off
Once the prediction is locked in, the Scene changes
Receiver clock error
250 µs
The Scene then shows: 74.95 km · How much too long every measured range is: c × the clock error/nodes/gps/outputs/rangeError
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$c\,\Delta t \approx 75\ \mathrm{km}$ ⁠: a quarter of a millisecond is already the size of a whole city.

Hint
A clock that runs fast sees every signal arrive 250 µs too late. How far does light go in that time?

Worked solution
1. Every signal seems to arrive later by $\Delta t = 250 \times 10^{-6}\ \mathrm{s}$ ⁠.
2. The receiver turns time into distance: $\Delta d = c\,\Delta t = 3 \times 10^8 \cdot 2.5 \times 10^{-4} = 7.5 \times 10^4\ \mathrm{m}$ ⁠.
3. Every range is about $75\ \mathrm{km}$ ⁠ too long, all by the same amount.
Light goes about $300\ \mathrm{m}$ ⁠ in a microsecond, so an ordinary quartz clock, which can be off by a millisecond, is useless without a correction.
Step 5 of 7
Open this step
Three satellites and a fast clock
Practice · Predict, then look
Question
Three satellites, and the clock runs 100 µs fast, so every range is about 30 km too long. The receiver trusts its clock. How far from the true place will the position it works out be?

Correct answer
- Correct More than 30 km
Setup, choices and feedback
The Scene at this step
Satellites used
3fixed
Receiver clock error
100 µsfixed
Show the timing spheres
Off
Choices
- Correct More than 30 km
- Wrong About 30 km, the same as the range error
  Feedback
  All the spheres grow together, and the point where they meet moves. With the satellites close together in the sky, it moves more than the error in each range.
- Wrong Less than 30 km, because the errors average out
  Feedback
  The errors do not average out: they are all the same way, every range too long.
- Wrong Zero: the same error in every range cancels out
  Feedback
  It would cancel if the receiver solved for the clock too. With three satellites it does not have enough measurements for that.
Once the prediction is locked in, the Scene changes
Show the timing spheres
On
The Scene then shows: 55.58 km · Distance from the computed fix to the receiver; NaN with fewer than three satellites/nodes/gps/outputs/fixError
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 56 km: the geometry of the spheres magnifies the clock error.

Worked solution
1. The three spheres are all 30 km too big.
2. They still meet at a point, but in the wrong place. The satellites are all above the receiver, on the same side of it, so enlarging every sphere together moves the meeting point a long way: here about 28 kilometres down into the ground and sideways too.
3. The computed position is about 56 km off, almost twice the range error.
A small clock error enlarges every range together, and poor geometry magnifies the result further.
Step 6 of 7
Open this step
The fourth satellite
Practice · Predict, then look
Question
The same clock, 100 µs fast. A fourth satellite is added. What will happen to the computed position?

Correct answer
- Correct It will land right on the receiver, and the receiver will find the clock error too
Setup, choices and feedback
The Scene at this step
Satellites used
3
Receiver clock error
100 µsfixed
Show the timing spheres
On
Choices
- Correct It will land right on the receiver, and the receiver will find the clock error too
- Wrong It will get a little closer, because one more measurement averages the error down
  Feedback
  The fourth satellite is not one more reading to average. It gives a fourth equation, so a fourth unknown can be solved: the clock.
- Wrong Nothing: the clock error is in every range, the fourth too
  Feedback
  The error is indeed in every range, and that is exactly why it can be found: it is the same number in all of them.
Once the prediction is locked in, the Scene changes
Satellites used
4
Show the timing spheres
On
The Scene then shows: 0.1 ms · The clock error the receiver solves for; NaN with fewer than four satellites/nodes/gps/outputs/foundClockError
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The position is exact, and the clock error found is 0.1 ms: the fourth satellite solves for the clock.

Worked solution
1. There are four unknowns: three coordinates of the place, and the clock error.
2. Each satellite gives one equation: the measured range equals the true distance plus $c\,\Delta t$ ⁠.
3. Four equations solve four unknowns: the receiver finds the place and the 100 µs at once.
That is why a receiver needs at least four satellites, and why your phone knows the time to a millionth of a second.
Step 7 of 7
Open this step
Relativity in your pocket
Practice · Calculate
Question
By relativity, a clock in a GPS orbit gains about 38.5 µs a day on a clock on the ground. If nobody corrected for it, by how many kilometres would every range drift each day?

Correct answer
11.54 km ± 3%
Setup, choices and feedback
The Scene at this step
Satellites used
4
Receiver clock error
0 µs
Show the timing spheres
On
Accepted within
± 3%
Computed from the Scene at this step’s setting
How far each range would drift each day if that gain were left uncorrected/nodes/gps/outputs/rangeDriftPerDay
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About $11.5\ \mathrm{km}$ ⁠ a day. Without Einstein, GPS would be useless within hours.

Hint
Use $\Delta d = c\,\Delta t$ ⁠ again.

Worked solution
1. $\Delta t = 38.5 \times 10^{-6}\ \mathrm{s}$ ⁠ a day.
2. $\Delta d = c\,\Delta t = 3 \times 10^8 \cdot 3.85 \times 10^{-5} \approx 1.15 \times 10^4\ \mathrm{m}$ ⁠.
3. About $11.5\ \mathrm{km}$ ⁠ every day, and more each day after that.
That is why the clocks are tuned before launch. Answering about 11.5 m divides by a thousand once too often, as if microseconds were nanoseconds.

Lesson 5 of 6 · About 30 minutes

Particle accelerators

How a collider pushes protons, bends them into a ring and records what is made when they collide. Cavities and magnets, the bending radius from p = qBρ, the trade-off between magnets and tunnel, why the protons hardly speed up, and what the detector sees.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Push and bend
Practice · Choose
Question
Protons enter the LHC ring at 0.45 TeV and leave it at 7 TeV. What gives them the energy?

Correct answer
- Correct The electric field in radio-frequency cavities, which they pass through every lap
Setup, choices and feedback
The Scene at this step
Beam energy
7 TeV
Magnet strength
8.33 T
The collision makes
Higgs boson → two photons
Zoom to a collision
Ring
Choices
Shown in a random order.
- Correct The electric field in radio-frequency cavities, which they pass through every lap
- Wrong The bending magnets, which push them round the ring
  Misconception: Thinks a collider's bending magnets are what speed the protons up.
  Feedback from the catalogue
  A magnetic force on a moving charge is always at right angles to its motion, so it does no work: it bends the path without changing the energy. The energy comes from an electric field, in radio-frequency cavities the beam passes through every lap. The magnets strengthen during the acceleration only to hold the protons, whose momentum is growing, on the same circle.
- Wrong The magnets, which get stronger during the acceleration
  Misconception: Thinks a collider's bending magnets are what speed the protons up.
  Feedback from the catalogue
  A magnetic force on a moving charge is always at right angles to its motion, so it does no work: it bends the path without changing the energy. The energy comes from an electric field, in radio-frequency cavities the beam passes through every lap. The magnets strengthen during the acceleration only to hold the protons, whose momentum is growing, on the same circle.
- Wrong The collision with the other beam
  Feedback
  The collision uses up energy; it does not give it. It happens only once the beams have already reached $7\ \mathrm{TeV}$ ⁠.
Feedback for a right answer
An electric field pushes, a magnetic field bends. The ring brings the protons back to the same cavities millions of times.

Worked solution
1. A magnetic force on a moving charge is always at right angles to the motion, so it does no work: it changes direction, not speed or energy.
2. An electric field along the motion does do work, $W = qV$ ⁠. In the cavities the field flips in step with the arriving bunches, so they always get a push forward.
3. Each lap gives only about half a million electronvolts, but more than 11,000 laps a second add up to 7 TeV in about twenty minutes.
The magnets do get stronger during the acceleration, but only to hold the protons, whose momentum is growing, on the same circle.
Step 2 of 6
Open this step
The bending radius
Practice · Calculate
Question
Beam energy 7 TeV and Magnet strength 8.33 T, as in the LHC. For a proton, the energy in TeV is about $0.3 \cdot B \cdot \rho$ ⁠, with $B$ ⁠ in tesla and $\rho$ ⁠ in kilometres. What is the bending radius $\rho$ ⁠?

Correct answer
2.803 km ± 2%
Setup, choices and feedback
The Scene at this step
Beam energy
7 TeVfixed
Magnet strength
8.33 Tfixed
The collision makes
Higgs boson → two photons
Zoom to a collision
Ring
Accepted within
± 2%
Computed from the Scene at this step’s setting
Radius the bending magnets hold the beam on, p/eB/nodes/collider/outputs/bendingRadius
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$\rho \approx 2.8\ \mathrm{km}$ ⁠. With focusing magnets, cavities and detectors in between, the whole ring comes to $26.7\ \mathrm{km}$ ⁠.

Hint
$\rho = \dfrac{E}{0.3\,B}$ ⁠.

Worked solution
1. From $p = qB\rho$ ⁠, for a fast proton: $E\,[\mathrm{TeV}] \approx 0.3 \cdot B\,[\mathrm{T}] \cdot \rho\,[\mathrm{km}]$ ⁠.
2. $\rho = \dfrac{7}{0.3 \cdot 8.33} \approx 2.80\ \mathrm{km}$ ⁠.
3. The bending magnets fill only about two thirds of the ring, so the ring is bigger than $2\pi \cdot 2.8 \approx 17.6\ \mathrm{km}$ ⁠ round: about $26.7\ \mathrm{km}$ ⁠.
Answering about 4.2 km divides the ring’s circumference by $2\pi$ ⁠, as if the whole ring were magnets.
Step 3 of 6
Open this step
More energy
Practice · Predict, then look
Question
Magnet strength is locked at 8.33 T. Beam energy is doubled from 7 to 14 TeV. What will happen to the size of ring needed?

Correct answer
- Correct It will double
Setup, choices and feedback
The Scene at this step
Beam energy
7 TeV
Magnet strength
8.33 Tfixed
The collision makes
Higgs boson → two photons
Zoom to a collision
Ring
Choices
- Correct It will double
- Wrong It will grow four times
  Misconception: Thinks the ring needed grows with the square of the beam energy.
  Feedback from the catalogue
  The magnets have to bend momentum: $p = qB\rho$ ⁠, so $\rho = \dfrac{p}{qB}$ ⁠. For a proton moving at almost the speed of light, $E \approx pc$ ⁠, so the momentum is in proportion to the energy, not to its square root. At the same field, twice the energy needs twice the ring, not four times. Reasoning from $E = \tfrac12 mv^2$ ⁠ does not hold near the speed of light.
- Wrong It will hardly change, because the speed hardly changes
  Feedback
  The speed does hardly change, but what the magnets have to bend is the momentum, and that doubles with the energy.
Once the prediction is locked in, the Scene changes
Beam energy
14 TeV
The Scene then shows: 53.3 km · The ring, with its dipoles filling two-thirds of it/nodes/collider/outputs/circumference
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about 26.7 to about 53 km: the radius is in proportion to the momentum, and the momentum to the energy.

Worked solution
1. $p = qB\rho$ ⁠, so $\rho = \dfrac{p}{qB}$ ⁠: the radius is in proportion to the momentum.
2. For a proton moving at almost the speed of light, $E \approx pc$ ⁠, so the momentum doubles with the energy.
3. At the same field the radius doubles, and so does the circumference: from about $26.7$ ⁠ to about $53\ \mathrm{km}$ ⁠.
Picking four times thinks of $E = \tfrac12 mv^2$ ⁠, which does not hold for a particle moving at almost the speed of light.
Step 4 of 6
Open this step
Magnets or tunnel
Practice · Reach the target
Question
Beam energy is locked at 7 TeV, and Magnet strength is on Iron magnet. With magnets like that, the ring would be more than 100 km round. Find the magnet strength that fits the ring into a tunnel 40 km round, and press Check.

Goal
- 40 km ± 1%
  At the start of the step: 123.3 km
Setup, choices and feedback
The Scene at this step
Beam energy
7 TeVfixed
Magnet strength
1.8 T
The collision makes
Higgs boson → two photons
Zoom to a collision
Ring
Read from the Scene at
The ring, with its dipoles filling two-thirds of it/nodes/collider/outputs/circumference
One setting that reaches it
Magnet strength
5.5 T
The collision makes
Higgs boson → two photons
Feedback for a right answer
About 5.55 T: still superconducting magnets, but well within what niobium–titanium can do.

Feedback for a wrong answer
Not 40 km yet. The stronger the magnets, the smaller the ring.

Hint
At the same energy, $B\rho$ ⁠ is fixed. At 8.33 T the ring is 26.7 km.

Worked solution
1. At the same energy the momentum is fixed, so $B\rho$ ⁠ is fixed: the field and the size are inversely proportional.
2. At 8.33 T the ring is 26.7 km. For a 40 km ring: $B = 8.33 \cdot \dfrac{26.7}{40} \approx 5.55\ \mathrm{T}$ ⁠.
3. That is more than the 2 T iron allows, so these magnets too must be superconducting, niobium–titanium.
Accelerators are always built on this trade-off: a longer tunnel costs money, and stronger magnets need new materials.
Step 5 of 6
Open this step
Almost the speed of light
Practice · Predict, then look
Question
Beam energy is on LHC at injection, 0.45 TeV. It is raised to 7 TeV, 15.5 times as much. What will happen to the protons’ speed?

Correct answer
- Correct It will hardly change: they already move at almost the speed of light
Setup, choices and feedback
The Scene at this step
Beam energy
0.45 TeV
Magnet strength
8.33 Tfixed
The collision makes
Higgs boson → two photons
Zoom to a collision
Ring
Choices
- Correct It will hardly change: they already move at almost the speed of light
- Wrong It will grow almost four times, $\sqrt{15.5}$ ⁠
  Misconception: Uses ½mv² for a proton near light speed, so more energy means proportionally more speed.
  Feedback from the catalogue
  $E = \tfrac12 mv^2$ ⁠ holds only far from the speed of light. By relativity $E = \gamma mc^2$ ⁠, and nothing with mass reaches $c$ ⁠. A 0.45 TeV proton already moves at $99.9998\%$ ⁠ of the speed of light, so extra energy adds almost no speed: it raises $\gamma$ ⁠, and with it the momentum.
- Wrong It will grow $15.5$ ⁠ times, like the energy
  Misconception: Uses ½mv² for a proton near light speed, so more energy means proportionally more speed.
  Feedback from the catalogue
  $E = \tfrac12 mv^2$ ⁠ holds only far from the speed of light. By relativity $E = \gamma mc^2$ ⁠, and nothing with mass reaches $c$ ⁠. A 0.45 TeV proton already moves at $99.9998\%$ ⁠ of the speed of light, so extra energy adds almost no speed: it raises $\gamma$ ⁠, and with it the momentum.
Once the prediction is locked in, the Scene changes
Beam energy
7 TeV
The Scene then shows: 2.693 m/s · How much slower than light the protons go/nodes/collider/outputs/shortfall
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The gap to light falls from about 650 to about 2.7 m/s. The speed barely moves; the momentum grows 15.5 times.

Worked solution
1. At 0.45 TeV the proton carries $\gamma \approx 480$ ⁠ times its rest energy, and it is already only about 650 m/s slower than light.
2. At 7 TeV, $\gamma \approx 7460$ ⁠, and the gap falls to about 2.7 m/s: $c - v \approx \dfrac{c}{2\gamma^2}$ ⁠.
3. The speed rose by less than a millionth of itself. Almost all the extra energy went into momentum.
Picking $\sqrt{15.5}$ ⁠ or $15.5$ ⁠ uses $E = \tfrac12 mv^2$ ⁠, which would take the protons far past the speed of light.
Step 6 of 6
Open this step
What the detector sees
Practice · Choose
Question
Under The collision makes pick Higgs boson → two photons, and move Zoom to a collision to Particles recorded. Two particles leave no track in the silicon tracker, and stop in the crystals. What are they?

Correct answer
- Correct Two photons: they carry no charge, so they leave no track, but the crystals absorb their energy
Setup, choices and feedback
The Scene at this step
Beam energy
7 TeV
Magnet strength
8.33 T
The collision makes
Z boson → two muons
Zoom to a collision
Ring
Choices
Shown in a random order.
- Correct Two photons: they carry no charge, so they leave no track, but the crystals absorb their energy
- Wrong The Higgs boson itself, stopped in the crystals
  Feedback
  A Higgs boson decays within about $10^{-22}$ ⁠ of a second, long before it moves any measurable distance. The detector sees only what it decayed into.
- Wrong Two muons
  Feedback
  Muons are charged, so they leave a track in the silicon, and they cross every layer to the muon chambers in the iron.
- Wrong Two neutrinos
  Feedback
  Neutrinos pass through the whole detector without stopping. They show only as missing push.
Feedback for a right answer
Two photons. Each layer of the detector catches a different kind of particle, and the combination tells what was made.

Worked solution
1. The silicon tracker records only charged particles, so whatever leaves no track has no charge.
2. The calorimeter crystals stop photons and electrons and measure their energy. An electron would leave a track; a photon does not.
3. If the two photons’ energies and directions together give a mass of about $125\ \mathrm{GeV}$ ⁠, they may have come from a Higgs boson.
The Higgs itself is never seen: it is recognised by what is left of it, and only in the statistics of many collisions.

Lesson 6 of 6 · About 30 minutes

Fusion reactors

How to hold a plasma ten times hotter than the centre of the Sun. A cage of field lines, why a plain ring leaks, the twist and the safety factor, Lawson's triple product, why hotter is not always better, and what stronger magnets buy.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A cage of field lines
Practice · Choose
Question
The plasma inside the tokamak is ten times hotter than the centre of the Sun. What keeps it off the walls?

Correct answer
- Correct A magnetic field: the nuclei and electrons are charged, and they circle round the field lines instead of crossing them
Setup, choices and feedback
The Scene at this step
Plasma temperature
150 million K
Field strength
5.3 T
Plasma current
15 MA
Show field and particles
Off
Choices
Shown in a random order.
- Correct A magnetic field: the nuclei and electrons are charged, and they circle round the field lines instead of crossing them
- Wrong Tungsten walls, which can stand even 150 million degrees
  Feedback
  No solid survives contact with such a plasma, and the plasma would cool the moment it touched one.
- Wrong The vacuum round it, which insulates it like a flask
  Feedback
  The vacuum stops air from cooling the plasma, but it does not stop its particles flying to the wall.
- Wrong Nothing: the plasma is too thin to harm the walls
  Feedback
  The plasma is thin, but without a cage its particles would reach the wall in milliseconds, and the plasma would cool.
Feedback for a right answer
A cage of field lines. In a tokamak the lines are bent into a ring so that they have no ends.

Worked solution
1. At 150 million kelvin the atoms are torn apart: bare nuclei and free electrons, all charged.
2. A magnetic force on a moving charge is at right angles to the motion, so it turns the particle in circles round the field line. Along the line it slides freely.
3. Bend the field lines into a ring and they close on themselves, so a particle sliding along them never reaches an end.
That is the tokamak’s idea: a doughnut-shaped cage made of field lines.
Step 2 of 6
Open this step
No twist
Practice · Predict, then look
Question
The field and particles are showing. Plasma current is turned down to zero, no twist: only the coils’ field is left, running round the ring and closing on itself. What will happen to the particles?

Correct answer
- Correct The ions will drift up and the electrons down, into the walls
Setup, choices and feedback
The Scene at this step
Plasma temperature
150 million Kfixed
Field strength
5.3 Tfixed
Plasma current
15 MA
Show field and particles
On
Choices
- Correct The ions will drift up and the electrons down, into the walls
- Wrong They will stay held: the field lines close, so there is nowhere to escape
  Misconception: Thinks a magnetic field running round a closed ring holds a plasma by itself.
  Feedback from the catalogue
  In a ring the coils crowd together on the inside, so the field is stronger on the inside than outside. A particle’s circle is tighter where the field is strong, and each turn does not quite close: ions creep one way and electrons the other. The charges separate, and the electric field this makes pushes the whole plasma into the wall. For the drifts to cancel the field lines must be twisted, in a tokamak by a current in the plasma itself.
- Wrong They will all be thrown outwards together, to the outer wall
  Feedback
  At first ions and electrons drift opposite ways, one up and one down. Only the separated charge this makes then pushes the whole plasma outwards.
Once the prediction is locked in, the Scene changes
Plasma current
0 MA
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Without a twist, the drifts up and down do not cancel, and the plasma reaches the wall within milliseconds.

Worked solution
1. Bend the field into a ring and the coils crowd together on the inside, where the field is stronger.
2. A particle’s circle is tighter where the field is strong, so each turn does not quite close, and the particle creeps sideways: positive ions one way, electrons the other.
3. The charges separate, the electric field this makes pushes the whole plasma into the wall, and within milliseconds it is lost.
The cure is to twist the field lines, so that every particle spends half its time above the plasma and half below, and the drifts cancel.
Step 3 of 6
Open this step
Too much twist
Practice · Predict, then look
Question
ITER’s settings: Field strength 5.3 T and Plasma current 15 MA, with an edge safety factor $q = 3$ ⁠. The field is lowered to JET’s field, 3.45 T, and the current is left alone. What will happen?

Correct answer
- Correct The twist will be too tight, $q$ ⁠ will fall below $2$ ⁠, and the plasma will kink and disrupt
Setup, choices and feedback
The Scene at this step
Plasma temperature
150 million Kfixed
Field strength
5.3 T
Plasma current
15 MAfixed
Show field and particles
On
Choices
- Correct The twist will be too tight, $q$ ⁠ will fall below $2$ ⁠, and the plasma will kink and disrupt
- Wrong It will be held as before: the current, which makes the twist, has not changed
  Feedback
  The twist depends on the ratio of the current’s field to the coils’ field: $q \propto \dfrac{B}{I}$ ⁠. With weaker coils, the same current twists the lines more.
- Wrong The twist will be too weak, and the particles will drift into the wall
  Feedback
  The other way round: a weaker coil field, with the same current, gives more twisted lines, a smaller $q$ ⁠.
Once the prediction is locked in, the Scene changes
Field strength
3.45 T
The Scene then shows: 1.953 · Q95, field-line laps the long way round per lap the short way (999 with no current)/nodes/tokamak/outputs/safetyFactor
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$q \approx 1.95$ ⁠: too much twist for this field. More current is not always better.

Worked solution
1. The safety factor $q$ ⁠ rises with the coils’ field and falls with the current: $q \propto \dfrac{B}{I}$ ⁠.
2. $q = 3 \cdot \dfrac{3.45}{5.3} \cdot \dfrac{15}{15} \approx 1.95$ ⁠.
3. Below $2$ ⁠ the plasma column kinks into a spiral, strikes the wall, and all the heat is dumped within about a millisecond: a disruption.
So a machine with a weaker field must run less current: at 12 MA, $q \approx 2.4$ ⁠, and the plasma is held again.
Step 4 of 6
Open this step
The triple product
Practice · Calculate
Question
At ITER’s settings the plasma holds about $5.1 \times 10^{19}$ ⁠ fuel nuclei in each cubic metre, at 150 million K, and the energy confinement time is about 3.0 s. What is the triple product $nT\tau_E$ ⁠, in keV·s/m³?

Correct answer
1.973 × 10²¹ ± 5%
Setup, choices and feedback
The Scene at this step
Plasma temperature
150 million Kfixed
Field strength
5.3 Tfixed
Plasma current
15 MAfixed
Show field and particles
Off
Accepted within
± 5%
Computed from the Scene at this step’s setting
N·T·τE in keV·s/m³/nodes/tokamak/outputs/tripleProduct
Feedback for a right answer
About $2 \times 10^{21}$ ⁠: two thirds of the way to ignition, and enough for a gain of about $10$ ⁠.

Hint
1 keV is about 11.6 million K.

Worked solution
1. The temperature in keV: $T = \dfrac{150}{11.6} \approx 12.9\ \mathrm{keV}$ ⁠.
2. $nT\tau_E = 5.1 \times 10^{19} \cdot 12.9 \cdot 3.0 \approx 2.0 \times 10^{21}\ \mathrm{keV \cdot s/m^3}$ ⁠.
3. Ignition needs about $3 \times 10^{21}$ ⁠, so ITER is about two thirds of the way there.
The three factors trade off against each other: a plasma twice as dense needs only half the confinement time.
Step 5 of 6
Open this step
Hotter
Practice · Predict, then look
Question
ITER’s settings, with a gain $Q \approx 10$ ⁠. Plasma temperature is doubled from 150 to 300 million K, without changing Field strength. What will happen to $Q$ ⁠?

Correct answer
- Correct It will fall: at the same field, a hotter plasma must be a thinner one
Setup, choices and feedback
The Scene at this step
Plasma temperature
150 million K
Field strength
5.3 Tfixed
Plasma current
15 MAfixed
Show field and particles
Off
Choices
- Correct It will fall: at the same field, a hotter plasma must be a thinner one
- Wrong It will rise: in a hotter plasma the nuclei collide harder and fuse more
  Misconception: Thinks a hotter plasma always gives more fusion power.
  Feedback from the catalogue
  The magnetic field can hold a certain pressure, and the plasma’s pressure is in proportion to $nT$ ⁠. At the same field, a hotter plasma must be a thinner one, and the fusion rate is in proportion to $n^2$ ⁠. Past about 150 million kelvin the chance that each pair of nuclei fuses no longer grows fast enough to make up for it, and the gain falls. So there is a best temperature, not “the hotter the better”.
- Wrong It will not change: fusion has already started
  Feedback
  $Q$ ⁠ depends on all three factors, $n$ ⁠, $T$ ⁠ and $\tau_E$ ⁠, and two of them change when you heat the plasma.
Once the prediction is locked in, the Scene changes
Plasma temperature
300 million K
The Scene then shows: 4.482 · Q, fusion power over heating power (10⁶ at ignition)/nodes/tokamak/outputs/gain
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$Q$ ⁠ falls to about $4.5$ ⁠. Hotter is not always more fusion.

Worked solution
1. The field can hold a certain pressure, and the plasma’s pressure is in proportion to $nT$ ⁠. At the same field, twice the $T$ ⁠ means half the $n$ ⁠.
2. The fusion rate is in proportion to $n^2$ ⁠ times the chance that each pair fuses. Past about 150 million K that chance no longer grows fast enough to make up for $n^2$ ⁠ falling four times.
3. $Q$ ⁠ falls from about $10$ ⁠ to about $4.5$ ⁠.
That is why tokamaks aim for about 150 million K and no hotter: there the triple product needed for ignition is smallest.
Step 6 of 6
Open this step
Stronger magnets
Practice · Reach the target
Question
The idea behind SPARC and other new machines: stronger magnets. Plasma temperature is locked at 150 million K. Change Field strength and Plasma current until the plasma reaches ignition, without kinking, and press Check.

Goal
- 2 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Plasma temperature
150 million Kfixed
Field strength
5.3 T
Plasma current
15 MA
Show field and particles
Off
Read from the Scene at
0 below breakeven, 1 net gain, 2 ignition/nodes/tokamak/outputs/gainBand
One setting that reaches it
Field strength
11.55 T
Plasma current
7 MA
Feedback for a right answer
Ignition: the helium nuclei alone keep the plasma hot. A stronger field buys density, and density buys fusion.

Feedback for a wrong answer
Not ignition yet. Try a stronger field, and watch that the plasma does not kink.

Hint
The field sets how much pressure, and so how much density, the plasma can hold, and the density goes into the triple product.

Worked solution
1. At the same temperature, the density the field holds is in proportion to $B^2$ ⁠, and the confinement time grows a little with the field.
2. So the triple product grows roughly like $B^2$ ⁠. At ITER’s settings it is two thirds of the way, so raising the field $\sqrt{1.5} \approx 1.2$ ⁠ times is enough.
3. At 15 MA, a field of about 6.5 T or more reaches ignition. More current helps too, as long as $q$ ⁠ stays above $2$ ⁠.
That is what new superconducting magnets promise: the same plasma in a much smaller, cheaper machine.

Across the unit · 15 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Solar cells

Light in, current out

Question

Correct answer

The Scene at this step

Choices

Worked solution

Half the light

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

How much of the light becomes electricity

Question

Correct answer

The Scene at this step

Hint

Worked solution

Where the rest goes

Question

Correct answer

The Scene at this step

Choices

Worked solution

Tilted away from the Sun

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Half the power

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The photocurrent

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The transistor as a switch

Three legs

Question

Correct answer

The Scene at this step

Choices

Worked solution

The threshold

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The square law

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Hint

Worked solution

Halfway

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Fully on

Question

Correct answer

The Scene at this step

Choices