---
title: "Electric Circuits and Power"
book: "High School Physics"
subject: physics
language: en
chapter: 12
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power
---

# Chapter 12 — Electric 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](#def-g11-circuits-and-power-voltage), [Ohm’s law](#prop-g11-circuits-and-power-ohm). This chapter reopens it with the account book of [Chapter 9](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#ch-g10-energy-conservation): what the meter charges, why [thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) exchanged per [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) $E = qU$ with it. A voltmeter, connected in parallel, measures it.

**Proposition 12.4 (Circuit laws).**

In a circuit running steadily:

1. *junction law:* the currents arriving at a junction add up to the currents leaving it;
2. *series law:* [voltages](#def-g11-circuits-and-power-voltage) along a path add up: $U_{AC} = U_{AB} + U_{BC}$ ;
3. *parallel law:* two branches joining the same two points carry the same [voltage](#def-g11-circuits-and-power-voltage) .

**Proof.** In a steady state charge does not pile up anywhere, so what flows into a junction flows out. [Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) per [coulomb](#def-g11-circuits-and-power-current) 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](#def-g11-circuits-and-power-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](#def-g11-circuits-and-power-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](#def-g11-circuits-and-power-voltage) add ([Proposition 12.4](#prop-g11-circuits-and-power-laws)), 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.](https://one-course.com/images/onecourse/chapters/physics-2/g11-circuits-and-power/fig-d1c9f5a77cbc.svg)

![Series (left): one current, voltages add. Parallel (right): one voltage, currents add.](https://one-course.com/images/onecourse/chapters/physics-2/g11-circuits-and-power/fig-cbe8d59b292d.svg)

*Series (left): one current, [voltages](#def-g11-circuits-and-power-voltage) add. Parallel (right): one [voltage](#def-g11-circuits-and-power-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](#def-g11-circuits-and-power-voltage) $U$ crossed by a current $I$ receives [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](#def-g11-circuits-and-power-current)) and each [coulomb](#def-g11-circuits-and-power-current) hands over $U$ [joules](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) ([Definition 12.3](#def-g11-circuits-and-power-voltage)): $E = qU = UIt$. ∎

**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](#prop-g11-circuits-and-power-ohm) 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](#def-g11-circuits-and-power-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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) ([Definition 9.17](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh)): $1\,\mathrm{kW}\,\mathrm{h} = 3.6 \times 10^{6}\,\mathrm{J}$. Joules are the physicist’s [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit); [kilowatt-hours](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) 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).](https://one-course.com/images/onecourse/chapters/physics-2/g11-circuits-and-power/fig-0ec8ff08c170.svg)

*Appliance [powers](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-forms) 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](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) 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](#def-g11-circuits-and-power-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.](https://one-course.com/images/onecourse/chapters/physics-2/g11-circuits-and-power/fig-6ba94ca116e3.svg)

*The delivery of [Example 12.15](#ex-g11-circuits-and-power-transmission): 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) $P = UI$, the line loss $R_\ell I^2 = R_\ell
P^2/U^2$ falls as the *square* of the transport [voltage](#def-g11-circuits-and-power-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](#def-g11-circuits-and-power-joule) — 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-forms). Its *electromotive force* (EMF) $\mathcal{E}$, in [volts](#def-g11-circuits-and-power-voltage), is the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) it hands each [coulomb](#def-g11-circuits-and-power-current) crossing it; its *internal resistance* $r$ accounts for the [Joule effect](#def-g11-circuits-and-power-joule) inside its own chemistry. Model: an ideal source $\mathcal{E}$ in series with $r$ (curly $\mathcal{E}$: $E$ is [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](#def-g11-circuits-and-power-emf) eats $rI^2$, the circuit receives $UI$: [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) conservation ([Chapter 9](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#ch-g10-energy-conservation)) 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](#def-g11-circuits-and-power-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.](https://one-course.com/images/onecourse/chapters/physics-2/g11-circuits-and-power/fig-8992ebe78e13.svg)

*Discharge characteristic of the battery of [Example 12.20](#ex-g11-circuits-and-power-discharge): 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) ([Definition 9.12](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency)) 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](#def-g11-circuits-and-power-current)? How many elementary charges is that ($e = 1.6 \times 10^{-19}\,\mathrm{C}$)?

**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](#def-g11-circuits-and-power-resistance)?

**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](#prop-g11-circuits-and-power-networks) of $R_1 = 120\,\Omega$ and $R_2 =
60\,\Omega$ in series, then in parallel. Which is smaller than both, and why?

**Solution of Exercise 12.3.**

Series: $120 + 60 = 180\,\Omega$. Parallel: $\frac{120 \times 60}{180}
= 40\,\Omega$ — smaller than both: same [voltage](#def-g11-circuits-and-power-voltage), but the current gets a second path, so more total current and less [resistance](#def-g11-circuits-and-power-resistance).

**Exercise 12.4 ★.**

A travel iron draws $8.7\,\mathrm{A}$ at $230\,\mathrm{V}$. Compute its [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power), then the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) it uses in $4.0\,\mathrm{min}$, in [joules](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) and in [kilowatt-hours](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh).

**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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) in [kilowatt-hours](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) and [joules](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy), and cost at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$?

**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 $230\,\mathrm{V}$.

1. Compute its [resistance](#def-g11-circuits-and-power-resistance) and the current it draws.
2. Plugged (same [resistance](#def-g11-circuits-and-power-resistance) ) into a $115\,\mathrm{V}$ supply, what [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) does it deliver? Why not half?

**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](#def-g11-circuits-and-power-voltage) across each resistor, and the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) dissipated in each; check the [powers](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) add up to $UI$.

**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](#prop-g11-circuits-and-power-networks) two ways.

**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](#def-g11-circuits-and-power-emf) $r$, then the short-circuit current $\mathcal{E}/r$.

**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](#def-g11-circuits-and-power-joule). A $2.2\,\mathrm{kW}$ kettle and a $1.0\,\mathrm{kW}$ toaster run together: does the [fuse](#def-g11-circuits-and-power-joule) hold? Someone adds a $1.5\,\mathrm{kW}$ heater: show that it blows.

**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](#def-g11-circuits-and-power-joule) 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](#def-g11-circuits-and-power-resistance) $2.0\,\Omega$. Compute the current and the Joule loss if the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) is delivered at $200\,\mathrm{V}$, then at $4.0\,\mathrm{kV}$. Compare the two losses and explain the ratio.

**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](#def-g11-circuits-and-power-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](#prop-g11-circuits-and-power-networks) (series, parallel, mixed); compute each.

**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 $R = 6.0\,\Omega$.

1. Compute $I$ , $U$ , the useful [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) in $R$ , the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) lost in $r$ , and the [efficiency](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) .
2. Show that the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) per degree). Compute the heat required, the [electrical energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-forms) drawn, the heating time, and the cost at $0.25$ euros per $\mathrm{kW}\,\mathrm{h}$.

**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 $60\,\mathrm{A}\,\mathrm{h}$.

1. Convert the capacity to [coulombs](#def-g11-circuits-and-power-current) , and estimate the stored [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) ( $E \approx \mathcal{E} q$ ) in [joules](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) and [kilowatt-hours](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) .
2. The starter draws $150\,\mathrm{A}$ : compute the terminal [voltage](#def-g11-circuits-and-power-voltage) , the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) delivered, and the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) lost inside.
3. Explain why the headlights dim while the engine cranks.

**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](#def-g11-circuits-and-power-voltage) drops from $12.7$ to $9.7\,\mathrm{V}$, so they receive less [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) $0.90$. Through a line of total [resistance](#def-g11-circuits-and-power-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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) $0.99$ each) convert [voltages](#def-g11-circuits-and-power-voltage) 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.**

1. How much potential [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) does the falling water release each second? Deduce the [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) available to the turbines.
2. What [electrical power](#prop-g11-circuits-and-power-power) leaves the alternator?
3. At $20\,\mathrm{kV}$ straight from the alternator, what current flows?
4. Compute the Joule loss $R_\ell I^2$ at that current and compare it with the plant’s output. Conclude.
5. The city’s [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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.**

6. A transformer, treated as lossless for this question, steps the $88\,\mathrm{MW}$ up to $400\,\mathrm{kV}$ : what current flows now?
7. Compute the new line loss, in megawatts.
8. What fraction of the plant’s output is lost in the line?
9. Show that the lost fraction equals $R_\ell P/U^2$ , and check that it explains the factor between questions 4 and 7.
10. Compute the [voltage](#def-g11-circuits-and-power-voltage) drop $R_\ell I$ along the line at $400\,\mathrm{kV}$ . Is it consistent with question 8?

**Part III — In the kitchen.**

11. Convert $1\,\mathrm{kW}\,\mathrm{h}$ to [joules](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) ; what does the meter multiply?
12. Compute the toaster’s current and [resistance](#def-g11-circuits-and-power-resistance) at $230\,\mathrm{V}$ .
13. Three minutes of toasting: the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) in [joules](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) and [kilowatt-hours](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) , and its price.
14. The toaster dissipates $RI^2$ : explain in one sentence why the same formula is the loss on the line, the product in the kitchen.
15. The toaster shares a $16\,\mathrm{A}$ [fuse](#def-g11-circuits-and-power-joule) with a $2.2\,\mathrm{kW}$ kettle: does the [fuse](#def-g11-circuits-and-power-joule) hold? A $2.0\,\mathrm{kW}$ heater joins: and now?

**Part IV — The audit.**

16. The house’s own wiring, [resistance](#def-g11-circuits-and-power-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.
17. Chain the efficiencies (turbine–alternator, step-up, line, step-down, house wiring) into one overall grid [efficiency](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-efficiency) .
18. How many [kilograms](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) (and cubic [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) ) of dam water must fall for *one* [kilowatt-hour](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) to reach your toaster?
19. Evaluate the lost fraction $R_\ell P/U^2$ of question 9 at $U = 20\,\mathrm{kV}$ . What does a value above $1$ mean?
20. The punchline — *the price of distance* : in two sentences, what does one toasted [kilowatt-hour](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) cost the mountain, what fraction dies en route, and which single trick makes it possible?

**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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](#def-g11-circuits-and-power-joule) blows.

**16.** $0.05 \times 13.5^2 = 9.1\,\mathrm{W}$: $9.1/3100 \approx
0.3\%$ of the delivered [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) 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](#def-g11-circuits-and-power-voltage) at all.

**20.** One toasted [kilowatt-hour](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-kwh) costs the mountain about two tonnes — $2.1\,\mathrm{m}^{3}$ — of water falling $200\,\mathrm{m}$, and some $14\%$ of the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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 $400$.