High School Physics · Grades 10–12

12Electric Circuits and Power

Flip a switch: a lamp lights, because a generator many kilometres away pushes charge around a loop of copper through your wall. Earlier years mapped the circuit — current, voltage, Ohm’s law. This chapter reopens it with the account book of Chapter 9: what the meter charges, why thin wires glow, why the grid runs at $400\,\mathrm{kV}$ .

12.1 Charge on the move

Definition 12.1 (Current and the coulomb)

An electric current is an ordered flow of charge (electrons, in a metal). Its intensity $I$ , in amperes ( $\mathrm{A}$ ), is the charge passing through a cross-section of the wire per second: a steady current $I$ flowing for a time $t$ transports the charge

q = I\,t ,

in coulombs ( $\mathrm{C}$ ), with $1\,\mathrm{C} = 1\,\mathrm{A}\,\mathrm{s}$ . The elementary charge is $e = 1.6 \times 10^{-19}\,\mathrm{C}$ . Current is measured by an ammeter, inserted in series.

Example 12.2 (Lightning versus phone)

A lightning stroke carries some $3.0 \times 10^{4}\,\mathrm{A}$ for about $1.0 \times 10^{-4}\,\mathrm{s}$ : $q = 3.0\times10^{4} \times 10^{-4} = 3.0\,\mathrm{C}$ . Your phone, at $2.0\,\mathrm{A}$ , moves the same charge every $1.5$ seconds.

Definition 12.3 (Voltage and the volt)

The voltage $U$ between two points of a circuit, in volts ( $\mathrm{V}$ ), is the energy exchanged per unit of charge travelling between them: $1\,\mathrm{V} = 1\,\mathrm{J}/\mathrm{C}$ , so a charge $q$ crossing a device under voltage $U$ exchanges the energy $E = qU$ with it. A voltmeter, connected in parallel, measures it.

Proposition 12.4 (Circuit laws)

In a circuit running steadily:

junction law: the currents arriving at a junction add up to the currents leaving it;
series law: voltages along a path add up: $U_{AC} = U_{AB} + U_{BC}$ ;
parallel law: two branches joining the same two points carry the same voltage.

Proof. In a steady state charge does not pile up anywhere, so what flows into a junction flows out. Energy per coulomb is additive along a path and depends only on its endpoints — not on the branch taken between them. ∎

12.2 Ohm’s law and resistor networks

Definition 12.5 (Resistance)

The resistance of a conductor is the ratio $R = U/I$ of the voltage across it to the current through it, measured in ohms ( $\Omega$ ): $1\,\Omega = 1\,\mathrm{V}/\mathrm{A}$ .

Proposition 12.6 (Ohm’s law)

For a metallic conductor held at fixed temperature, $R$ is constant: voltage and current are proportional,

U = R\,I .

Proof. Admitted at this level. ∎

Remark 12.7

An experimental law, not a universal one: a diode, a heating lamp filament or your own skin do not obey it. Why metals do is derived in the university volumes.

Proposition 12.8 (Series and parallel resistors)

Two resistors in series are equivalent to one resistor $R_{\text{s}} = R_1 + R_2$ ; two in parallel, to one resistor $R_{\text{p}}$ given by

\frac{1}{R_{\text{p}}} = \frac{1}{R_1} + \frac{1}{R_2} , \qquad\text{i.e.}\quad R_{\text{p}} = \frac{R_1 R_2}{R_1 + R_2} < \min(R_1, R_2) .

Proof. Series: the same $I$ crosses both and voltages add (Proposition 12.4), so $U = (R_1 + R_2)\,I$ . Parallel: both sit under the same $U$ and currents add, so $I = \frac{U}{R_1} + \frac{U}{R_2}$ . ∎

Series (left): one current, voltages add. Parallel (right): one voltage, currents add.

Example 12.9 (Equivalent resistance)

$R_1 = 100\,\Omega$ and $R_2 = 150\,\Omega$ : in series, $250\,\Omega$ ; in parallel, $\frac{100 \times 150}{250} = 60\,\Omega$ — less than either: the current gets a second road.

12.3 Energy and power in a circuit

Proposition 12.10 (Electrical power)

A device under voltage $U$ crossed by a current $I$ receives energy at the rate

P = U\,I , \qquad\text{i.e.}\quad E = U\,I\,t \ \text{in a time } t .

Proof. In a time $t$ the charge $q = It$ crosses the device (Definition 12.1) and each coulomb hands over $U$ joules (Definition 12.3): $E = qU = UIt$ . ∎

Proposition 12.11 (Power in a resistor)

A resistor $R$ carrying a current $I$ dissipates

P = R\,I^2 = \frac{U^2}{R} .

Proof. Substitute Ohm’s law into $P = UI$ : $P = (RI)I = RI^2 = U(U/R) = U^2/R$ . ∎

Example 12.12 (Reading a nameplate)

A kettle marked “ $230\,\mathrm{V}$ – $2200\,\mathrm{W}$ ” draws $I = P/U = 2200/230 = 9.6\,\mathrm{A}$ and has resistance $R = U^2/P = 230^2/2200 = 24\,\Omega$ . Running $150\,\mathrm{s}$ it uses $E = 2200 \times 150 = 3.3 \times 10^{5}\,\mathrm{J} \approx 0.09\,\mathrm{kW}\,\mathrm{h}$ — about two cents.

Remark 12.13 (Joules and kilowatt-hours)

The meter counts kilowatt-hours (Definition 9.17): $1\,\mathrm{kW}\,\mathrm{h} = 3.6 \times 10^{6}\,\mathrm{J}$ . Joules are the physicist’s unit; kilowatt-hours price the bills.

Appliance powers on a logarithmic axis: everything that heats (red) lives a factor of ten above everything that computes or shines (blue).

12.4 The Joule effect: from toaster to pylon

Definition 12.14 (Joule effect)

The dissipation $P = RI^2$ turning electrical energy into heat in every resistor is the Joule effect. Heaters, kettles, toasters and old-style bulbs are resistors by design; a fuse is a thin wire calibrated to melt — opening the circuit — when the current exceeds its rating.

Example 12.15 (Why pylons run high)

A village needs $P = 100\,\mathrm{kW}$ through a line of resistance $R_\ell = 5.0\,\Omega$ . Delivered at $U = 500\,\mathrm{V}$ : $I = P/U = 200\,\mathrm{A}$ and the line wastes $R_\ell I^2 = 200\,\mathrm{kW}$ — twice the delivery: absurd. At $U = 20\,\mathrm{kV}$ : $I = 5.0\,\mathrm{A}$ and $R_\ell I^2 = 125\,\mathrm{W}$ . Forty times less current, $1600$ times less loss.

The delivery of : raising U at fixed P = UI shrinks I, and the loss falls with I2. — The delivery of Example 12.15: raising $U$ at fixed $P = UI$ shrinks $I$ , and the loss falls with $I^2$ .

Remark 12.16 (The grid’s iron rule)

At fixed delivered power $P = UI$ , the line loss $R_\ell I^2 = R_\ell P^2/U^2$ falls as the square of the transport voltage. Hence: transform up ( $400\,\mathrm{kV}$ ) to travel, down ( $230\,\mathrm{V}$ ) to live with.

Remark 12.17 (Household safety)

A $230\,\mathrm{V}$ line fused at $16\,\mathrm{A}$ supplies at most $230 \times 16 \approx 3.7\,\mathrm{kW}$ : two kettles at once melt the fuse — by design the thinnest, weakest link, far cheaper than a fire inside the wall.

12.5 Real batteries: electromotive force

Definition 12.18 (EMF and internal resistance)

A battery converts chemical into electrical energy. Its electromotive force (EMF) $\mathcal{E}$ , in volts, is the energy it hands each coulomb crossing it; its internal resistance $r$ accounts for the Joule effect inside its own chemistry. Model: an ideal source $\mathcal{E}$ in series with $r$ (curly $\mathcal{E}$ : $E$ is energy).

Proposition 12.19 (Terminal voltage)

A battery delivering a current $I$ shows at its terminals

U = \mathcal{E} - r\,I .

Proof. The chemistry supplies $\mathcal{E}I$ , the internal resistance eats $rI^2$ , the circuit receives $UI$ : energy conservation (Chapter 9) gives $UI = \mathcal{E}I - rI^2$ . ∎

Example 12.20 (A battery under load)

A battery ( $\mathcal{E} = 9.0\,\mathrm{V}$ , $r = 1.0\,\Omega$ ) feeds a bulb of resistance $R = 8.0\,\Omega$ . The loop obeys $\mathcal{E} = rI + RI$ , so $I = \mathcal{E}/(R + r) = 1.0\,\mathrm{A}$ and $U = 9.0 - 1.0 \times 1.0 = 8.0\,\mathrm{V}$ : the bulb receives $8.0\,\mathrm{W}$ while the battery wastes $1.0\,\mathrm{W}$ warming itself.

Discharge characteristic of the battery of : measured points sit on the line U = E - rI, of slope -r = -1.0\, V/ A. — Discharge characteristic of the battery of Example 12.20: measured points sit on the line $U = \mathcal{E} - rI$ , of slope $-r = -1.0\,\mathrm{V}/\mathrm{A}$ .

Remark 12.21 (Receivers)

A motor or a charging battery runs the conversion backwards: it absorbs under $U = \mathcal{E}' + r'I$ , converting $\mathcal{E}'I$ into motion or chemistry and losing $r'I^2$ . Sources push, receivers push back.

Example 12.22 (Efficiency of a battery)

The efficiency (Definition 9.12) of the discharge above: useful $UI = 8.0\,\mathrm{W}$ out of $\mathcal{E}I = 9.0\,\mathrm{W}$ , so $\eta = U/\mathcal{E} = 8/9 \approx 0.89$ — dropping as the current climbs: hard-working batteries run hot.

12.6 Exercises

Exercise 12.1 ★

A phone battery charges at a steady $2.0\,\mathrm{A}$ for $90\,\mathrm{min}$ . What charge passes through the cable, in coulombs? How many elementary charges is that ( $e = 1.6 \times 10^{-19}\,\mathrm{C}$ )?

Solution

Solution of Exercise 12.1.

$q = It = 2.0 \times 90 \times 60 = 1.1 \times 10^{4}\,\mathrm{C}$ . Number of elementary charges: $q/e = 1.08\times10^{4}/1.6\times10^{-19} \approx 6.8 \times 10^{22}$ .

Exercise 12.2 ★

A $220\,\Omega$ resistor across $12\,\mathrm{V}$ : what current flows? A heating element draws $0.50\,\mathrm{A}$ at $230\,\mathrm{V}$ : what resistance?

Solution

Solution of Exercise 12.2.

$I = U/R = 12/220 = 0.055\,\mathrm{A} = 55\,\mathrm{mA}$ ; $R = U/I = 230/0.50 = 460\,\Omega$ .

Exercise 12.3 ★

Compute the equivalent resistance of $R_1 = 120\,\Omega$ and $R_2 = 60\,\Omega$ in series, then in parallel. Which is smaller than both, and why?

Solution

Solution of Exercise 12.3.

Series: $120 + 60 = 180\,\Omega$ . Parallel: $\frac{120 \times 60}{180} = 40\,\Omega$ — smaller than both: same voltage, but the current gets a second path, so more total current and less resistance.

Exercise 12.4 ★

A travel iron draws $8.7\,\mathrm{A}$ at $230\,\mathrm{V}$ . Compute its power, then the energy it uses in $4.0\,\mathrm{min}$ , in joules and in kilowatt-hours.

Solution

Solution of Exercise 12.4.

$P = UI = 230 \times 8.7 = 2.0\,\mathrm{kW}$ . In $t = 240\,\mathrm{s}$ : $E = Pt = 2000 \times 240 = 4.8 \times 10^{5}\,\mathrm{J} = 4.8\times10^{5}/3.6 \times10^{6} \approx 0.13\,\mathrm{kW}\,\mathrm{h}$ .

Exercise 12.5 ★

A laptop charger delivers $60\,\mathrm{W}$ for $5.0\,\mathrm{h}$ a day: daily energy in kilowatt-hours and joules, and cost at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ ?

Solution

Solution of Exercise 12.5.

$E = 60 \times 5.0 = 300\,\mathrm{W}\,\mathrm{h} = 0.30\,\mathrm{kW}\,\mathrm{h} = 0.30 \times 3.6\times10^{6} = 1.1 \times 10^{6}\,\mathrm{J}$ . Cost: $0.30 \times 0.25 = 0.075$ euros a day.

Exercise 12.6 ★★

An electric heater is rated $1200\,\mathrm{W}$ at $230\,\mathrm{V}$ .

Compute its resistance and the current it draws.
Plugged (same resistance) into a $115\,\mathrm{V}$ supply, what power does it deliver? Why not half?

Solution

Solution of Exercise 12.6.

1. $R = U^2/P = 230^2/1200 = 44\,\Omega$ ; $I = P/U = 1200/230 = 5.2\,\mathrm{A}$ .

2. $P' = U'^2/R = 115^2/44.1 = 300\,\mathrm{W}$ : a quarter, not half, because halving $U$ also halves $I$ , and $P = UI \propto U^2$ .

Exercise 12.7 ★★

$R_1 = 40\,\Omega$ and $R_2 = 20\,\Omega$ in series across $12\,\mathrm{V}$ . Compute the current, the voltage across each resistor, and the power dissipated in each; check the powers add up to $UI$ .

Solution

Solution of Exercise 12.7.

$R_{\text{s}} = 60\,\Omega$ , so $I = 12/60 = 0.20\,\mathrm{A}$ . $U_1 = R_1 I = 8.0\,\mathrm{V}$ , $U_2 = 4.0\,\mathrm{V}$ (sum $12\,\mathrm{V}$ ); $P_1 = U_1 I = 1.6\,\mathrm{W}$ , $P_2 = 0.8\,\mathrm{W}$ : total $2.4\,\mathrm{W}$ $= UI = 12 \times 0.20$ , as it must.

Exercise 12.8 ★★

$R_1 = 60\,\Omega$ and $R_2 = 30\,\Omega$ in parallel across $12\,\mathrm{V}$ . Compute each branch current, the total current, and the equivalent resistance two ways.

Solution

Solution of Exercise 12.8.

$I_1 = 12/60 = 0.20\,\mathrm{A}$ , $I_2 = 12/30 = 0.40\,\mathrm{A}$ , total $I = 0.60\,\mathrm{A}$ . Equivalent: $R = U/I = 12/0.60 = 20\,\Omega$ , and indeed $\frac{60 \times 30}{60 + 30} = 20\,\Omega$ .

Exercise 12.9 ★★

A battery delivers $U = 5.5\,\mathrm{V}$ at $I = 0.50\,\mathrm{A}$ , and $U = 4.0\,\mathrm{V}$ at $I = 2.0\,\mathrm{A}$ . Find its EMF $\mathcal{E}$ and internal resistance $r$ , then the short-circuit current $\mathcal{E}/r$ .

Solution

Solution of Exercise 12.9.

$\mathcal{E} - 0.50\,r = 5.5$ and $\mathcal{E} - 2.0\,r = 4.0$ ; subtracting, $1.5\,r = 1.5$ , so $r = 1.0\,\Omega$ and $\mathcal{E} = 6.0\,\mathrm{V}$ . Short circuit: $\mathcal{E}/r = 6.0\,\mathrm{A}$ .

Exercise 12.10 ★★

A $230\,\mathrm{V}$ kitchen line is protected by a $16\,\mathrm{A}$ fuse. A $2.2\,\mathrm{kW}$ kettle and a $1.0\,\mathrm{kW}$ toaster run together: does the fuse hold? Someone adds a $1.5\,\mathrm{kW}$ heater: show that it blows.

Solution

Solution of Exercise 12.10.

Kettle $2200/230 = 9.6\,\mathrm{A}$ , toaster $1000/230 = 4.3\,\mathrm{A}$ : total $13.9\,\mathrm{A}$ $< 16\,\mathrm{A}$ , the fuse holds. Heater $1500/230 = 6.5\,\mathrm{A}$ : total $20.4\,\mathrm{A}$ $> 16\,\mathrm{A}$ , it blows (equivalently, $4.7\,\mathrm{kW} > 230 \times 16 \approx 3.7\,\mathrm{kW}$ ).

Exercise 12.11 ★★

A workshop needs $10\,\mathrm{kW}$ through a feeder cable of resistance $2.0\,\Omega$ . Compute the current and the Joule loss if the power is delivered at $200\,\mathrm{V}$ , then at $4.0\,\mathrm{kV}$ . Compare the two losses and explain the ratio.

Solution

Solution of Exercise 12.11.

At $200\,\mathrm{V}$ : $I = 10^4/200 = 50\,\mathrm{A}$ , loss $RI^2 = 2.0 \times 50^2 = 5.0\,\mathrm{kW}$ — half the delivery. At $4.0\,\mathrm{kV}$ : $I = 2.5\,\mathrm{A}$ , loss $2.0 \times 2.5^2 = 12.5\,\mathrm{W}$ . Ratio $400 = 20^2$ : twenty times the voltage means twenty times less current, and the loss goes as $I^2$ .

Exercise 12.12 ★★★

Using all of three identical $60\,\Omega$ resistors, list every distinct equivalent resistance (series, parallel, mixed); compute each.

Solution

Solution of Exercise 12.12.

Four layouts: all in series, $3 \times 60 = 180\,\Omega$ ; all in parallel, $60/3 = 20\,\Omega$ ; one in series with two in parallel, $60 + 30 = 90\,\Omega$ ; one in parallel with two in series, $\frac{60 \times 120}{180} = 40\,\Omega$ .

Exercise 12.13 ★★★

A battery ( $\mathcal{E} = 4.5\,\mathrm{V}$ , $r = 1.5\,\Omega$ ) feeds a resistor $R = 6.0\,\Omega$ .

Compute $I$ , $U$ , the useful power in $R$ , the power lost in $r$ , and the efficiency.
Show that the power received by a variable $R$ , $P = \mathcal{E}^2 R/(R+r)^2$ , is greatest when $R = r$ (hint: $(R+r)^2/R = (R-r)^2/R + 4r$ ), and compute that maximum.

Solution

Solution of Exercise 12.13.

1. $I = \mathcal{E}/(R + r) = 4.5/7.5 = 0.60\,\mathrm{A}$ ; $U = RI = 3.6\,\mathrm{V}$ ; useful $P = UI = 2.2\,\mathrm{W}$ ; lost $rI^2 = 1.5 \times 0.60^2 = 0.54\,\mathrm{W}$ ; $\eta = U/\mathcal{E} = 3.6/4.5 = 0.80$ .

2. By the hint, $P = \mathcal{E}^2 \big/ \big[(R - r)^2/R + 4r\big]$ ; the bracket is least ( $= 4r$ ) exactly when $R = r$ , so $P_{\max} = \mathcal{E}^2/(4r) = 4.5^2/6.0 = 3.4\,\mathrm{W}$ .

Exercise 12.14 ★★★

A $2.2\,\mathrm{kW}$ kettle of efficiency $0.90$ brings $1.0\,\mathrm{kg}$ of water from $20{}^{\circ}\mathrm{C}$ to the boil (heating water takes $4180\,\mathrm{J}$ per kilogram per degree). Compute the heat required, the electrical energy drawn, the heating time, and the cost at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ .

Solution

Solution of Exercise 12.14.

Heat: $Q = 1.0 \times 4180 \times (100 - 20) = 3.3 \times 10^{5}\,\mathrm{J}$ . Electrical: $E = Q/0.90 = 3.7 \times 10^{5}\,\mathrm{J}$ . Time: $t = E/P = 3.7\times10^{5}/2200 \approx 170\,\mathrm{s}$ , under three minutes. Cost: $3.7\times10^{5}/3.6\times10^{6} \approx 0.10\,\mathrm{kW}\,\mathrm{h}$ , about $2.6$ cents.

Exercise 12.15 ★★★

A car battery: $\mathcal{E} = 12.7\,\mathrm{V}$ , $r = 0.020\,\Omega$ , capacity $60\,\mathrm{A}\,\mathrm{h}$ .

Convert the capacity to coulombs, and estimate the stored energy ( $E \approx \mathcal{E} q$ ) in joules and kilowatt-hours.
The starter draws $150\,\mathrm{A}$ : compute the terminal voltage, the power delivered, and the power lost inside.
Explain why the headlights dim while the engine cranks.

Solution

Solution of Exercise 12.15.

1. $q = 60 \times 3600 = 2.2 \times 10^{5}\,\mathrm{C}$ ; $E \approx \mathcal{E}q = 12.7 \times 2.16\times10^{5} = 2.7 \times 10^{6}\,\mathrm{J} \approx 0.76\,\mathrm{kW}\,\mathrm{h}$ .

2. $U = 12.7 - 0.020 \times 150 = 9.7\,\mathrm{V}$ ; delivered $UI = 1.5\,\mathrm{kW}$ ; lost $rI^2 = 0.020 \times 150^2 = 450\,\mathrm{W}$ .

3. The headlights sit across the terminals: while cranking, the terminal voltage drops from $12.7$ to $9.7\,\mathrm{V}$ , so they receive less power and dim.

12.7 Problem: From the dam to the toaster

Problem 12.1

Weekend problem — the thousand-kilometre journey of a kilowatt-hour, from falling water to browning bread, with the full bill of the trip

A mountain dam holds water $200\,\mathrm{m}$ above its turbines, which swallow $50\,\mathrm{m}^{3}$ — $5.0 \times 10^{4}\,\mathrm{kg}$ — per second; turbine and alternator convert with efficiency $0.90$ . Through a line of total resistance $R_\ell = 32\,\Omega$ ( $1000\,\mathrm{km}$ there and back) the plant feeds a city where your $900\,\mathrm{W}$ toaster waits at $230\,\mathrm{V}$ . The alternator outputs $20\,\mathrm{kV}$ ; transformers (efficiency $0.99$ each) convert voltages at will. Electricity sells at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$ ; $g = 9.81\,\mathrm{N}/\mathrm{kg}$ .

Part I — At the dam.

How much potential energy does the falling water release each second? Deduce the power available to the turbines.
What electrical power leaves the alternator?
At $20\,\mathrm{kV}$ straight from the alternator, what current flows?
Compute the Joule loss $R_\ell I^2$ at that current and compare it with the plant’s output. Conclude.
The city’s power $P = UI$ is fixed by demand. What is the only way to shrink $I$ , and which device performs it?

Part II — On the line.

A transformer, treated as lossless for this question, steps the $88\,\mathrm{MW}$ up to $400\,\mathrm{kV}$ : what current flows now?
Compute the new line loss, in megawatts.
What fraction of the plant’s output is lost in the line?
Show that the lost fraction equals $R_\ell P/U^2$ , and check that it explains the factor between questions 4 and 7.
Compute the voltage drop $R_\ell I$ along the line at $400\,\mathrm{kV}$ . Is it consistent with question 8?

Part III — In the kitchen.

Convert $1\,\mathrm{kW}\,\mathrm{h}$ to joules; what does the meter multiply?
Compute the toaster’s current and resistance at $230\,\mathrm{V}$ .
Three minutes of toasting: the energy in joules and kilowatt-hours, and its price.
The toaster dissipates $RI^2$ : explain in one sentence why the same formula is the loss on the line, the product in the kitchen.
The toaster shares a $16\,\mathrm{A}$ fuse with a $2.2\,\mathrm{kW}$ kettle: does the fuse hold? A $2.0\,\mathrm{kW}$ heater joins: and now?

Part IV — The audit.

The house’s own wiring, resistance $0.05\,\Omega$ , carries the $13.5\,\mathrm{A}$ of question 15: compute the loss and its fraction of the roughly $3.1\,\mathrm{kW}$ delivered.
Chain the efficiencies (turbine–alternator, step-up, line, step-down, house wiring) into one overall grid efficiency.
How many kilograms (and cubic metres) of dam water must fall for one kilowatt-hour to reach your toaster?
Evaluate the lost fraction $R_\ell P/U^2$ of question 9 at $U = 20\,\mathrm{kV}$ . What does a value above $1$ mean?
The punchline — the price of distance: in two sentences, what does one toasted kilowatt-hour cost the mountain, what fraction dies en route, and which single trick makes it possible?

Solution

Solution of Problem 12.1.

1. Each second, $E_p = mgh = 5.0\times10^{4} \times 9.81 \times 200 = 9.8 \times 10^{7}\,\mathrm{J}$ : available power $98\,\mathrm{MW}$ .

2. $0.90 \times 98 = 88\,\mathrm{MW}$ .

3. $I = P/U = 8.8\times10^{7}/2.0\times10^{4} = 4.4 \times 10^{3}\,\mathrm{A}$ .

4. $R_\ell I^2 = 32 \times (4.4\times10^{3})^2 \approx 6.2 \times 10^{8}\,\mathrm{W} = 620\,\mathrm{MW}$ — seven times the plant’s output: transmitting at $20\,\mathrm{kV}$ is impossible.

5. With $P = UI$ fixed, only raising $U$ shrinks $I$ : a transformer.

6. $I = 8.8\times10^{7}/4.0\times10^{5} = 220\,\mathrm{A}$ .

7. $R_\ell I^2 = 32 \times 220^2 \approx 1.6\,\mathrm{MW}$ .

8. $1.6/88 \approx 0.018$ : about $1.8\%$ .

9. Fraction $= R_\ell I^2/P = R_\ell (P/U)^2/P = R_\ell P/U^2 \propto 1/U^2$ : from $20\,\mathrm{kV}$ to $400\,\mathrm{kV}$ it falls by $20^2 = 400$ , and indeed $620\,\mathrm{MW}/400 \approx 1.6\,\mathrm{MW}$ .

10. $R_\ell I = 32 \times 220 \approx 7.1\,\mathrm{kV}$ : $1.8\%$ of $400\,\mathrm{kV}$ , the same fraction as question 8, since $R_\ell I/U = R_\ell I^2/UI$ .

11. $1\,\mathrm{kW}\,\mathrm{h} = 3.6 \times 10^{6}\,\mathrm{J}$ : the meter multiplies the power drawn by the time it is drawn, and adds up.

12. $I = 900/230 = 3.9\,\mathrm{A}$ ; $R = U^2/P = 230^2/900 \approx 59\,\Omega$ .

13. $E = 900 \times 180 = 1.6 \times 10^{5}\,\mathrm{J} = 0.045\,\mathrm{kW}\,\mathrm{h}$ : about $1.1$ cents.

14. The physics of $RI^2$ is identical; only the purpose differs — on the line the heat is unwanted, in the toaster the heat is the product.

15. $3.9 + 9.6 = 13.5\,\mathrm{A} < 16\,\mathrm{A}$ : holds. Heater $2000/230 = 8.7\,\mathrm{A}$ : total $22.2\,\mathrm{A} > 16\,\mathrm{A}$ , the fuse blows.

16. $0.05 \times 13.5^2 = 9.1\,\mathrm{W}$ : $9.1/3100 \approx 0.3\%$ of the delivered power — negligible over short wires.

17. $\eta = 0.90 \times 0.99 \times 0.982 \times 0.99 \times 0.997 \approx 0.86$ .

18. One toasted kilowatt-hour needs $3.6\times10^{6}/0.86 \approx 4.2 \times 10^{6}\,\mathrm{J}$ upstream: $m = E/(gh) = 4.2\times10^{6}/(9.81 \times 200) \approx 2.1 \times 10^{3}\,\mathrm{kg}$ , about $2.1\,\mathrm{m}^{3}$ of water.

19. $R_\ell P/U^2 = 32 \times 8.8\times10^{7}/(2.0\times 10^{4})^2 \approx 7$ . A “fraction” above $1$ means the line would dissipate more than it carries: the delivery cannot happen at that voltage at all.

20. One toasted kilowatt-hour costs the mountain about two tonnes — $2.1\,\mathrm{m}^{3}$ — of water falling $200\,\mathrm{m}$ , and some $14\%$ of the energy dies en route, mostly in turbines and transformers (under $2\%$ in $1000\,\mathrm{km}$ of line). The single trick that makes the journey possible is the transformer: stepping up to $400\,\mathrm{kV}$ divides the current by $20$ and the Joule loss by $400$ .