| English | 意味 |
|---|---|
| internal resistance/ɪnˈtɜːnl rɪˈzɪstəns/ | 内部抵抗 |
| terminal potential difference/ˈtɜːmɪnl pəˈtenʃl ˈdɪfrəns/ | 端子間電位差 |
| lost volts/lɒst vəʊlts/ | 内部損失電圧 |
| short circuit/ʃɔːt ˈsɜːkɪt/ | 短絡 |
| parallel/ˈpærəlel/ | 並列 |
| ammeters/ˈæmiːtəz/ | アンメータ |
| voltmeters/ˈvəʊltmiːtəz/ | ボルトメータ |
| series/ˈsɪəriːz/ | 直列 |
Why the headlights dim
- Start a car and the headlights dim for a moment.
- The starter motor draws a huge current, and the battery's own voltage sags.
- Every real source has some internal resistance 内阻.
e.m.f. and p.d.
- e.m.f. $\varepsilon$ — energy each coulomb is given by the source.
- p.d. — energy each coulomb gives up to a component. Both in volts.
A real circuit built on a breadboard, being measured with a multimeter
Internal resistance
V = ε − I·r
Terminal p.d. falls with current: it starts at the e.m.f. ε and drops by I·r.
The e.m.f. is the energy given to each unit of charge by the ____.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The source (battery, cell, generator) supplies energy to the charge; a component takes it back out (the p.d.).
Internal resistance
- A real source has internal resistance $r$, so some energy is lost inside it.
- The terminal potential difference 端电压 (terminal p.d.) delivered outside is $V = \varepsilon - Ir$.
- The $Ir$ is called the lost volts 内阻压降 — energy per coulomb turned to heat inside the source.
A cell of e.m.f. $12\ \text{V}$ and internal resistance $0.50\ \Omega$ drives a current of $2.0\ \text{A}$. What is the terminal p.d.?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$V = \varepsilon - Ir = 12 - 2.0 \times 0.50 = 11\ \text{V}$.
What happens as current changes
- No current ($I = 0$): terminal p.d. equals the e.m.f.
- More current: the terminal p.d. drops (that $Ir$ loss grows).
- Short circuit 短路 ($R_{\text{ext}} \to 0$): $I = \dfrac{\varepsilon}{r}$ — a large current.
A cell of e.m.f. $6.0\ \text{V}$ and internal resistance $0.50\ \Omega$ is short-circuited. What current flows?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
With $R_{\text{ext}} \to 0$, $I = \dfrac{\varepsilon}{r} = \dfrac{6.0}{0.50} = 12\ \text{A}$.
Worked example: a battery and two lamps
A battery of e.m.f. $6.0\ \text{V}$ and internal resistance $0.80\ \Omega$ drives a current of $1.5\ \text{A}$ through two lamps X and Y in parallel. X has resistance $4.0\ \Omega$.
- Terminal p.d.: $V = \varepsilon - Ir = 6.0 - 1.5 \times 0.80 = 4.8\ \text{V}$.
- Current in X: the parallel lamps share the terminal p.d., so $I_{\text{X}} = \dfrac{4.8}{4.0} = 1.2\ \text{A}$.
- Current in Y: Kirchhoff's first law at the junction: $I_{\text{Y}} = 1.5 - 1.2 = 0.30\ \text{A}$.
- Power in Y: $P = VI = 4.8 \times 0.30 = 1.4\ \text{W}$.
- Check: $1.2\ \text{V}$ is lost inside the battery, so only $4.8\ \text{V}$ of the $6.0\ \text{V}$ reaches the lamps. Both lamps see the same p.d. because they are in parallel.
A battery of e.m.f. $9.0\ \text{V}$ and internal resistance $1.0\ \Omega$ delivers $2.0\ \text{A}$ to two resistors in parallel. One of them, of resistance $10\ \Omega$, takes part of the current. What current, in A, flows in the other resistor?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Terminal p.d. $= 9.0 - 2.0 \times 1.0 = 7.0\ \text{V}$; the $10\ \Omega$ resistor takes $\dfrac{7.0}{10} = 0.70\ \text{A}$; the other takes $2.0 - 0.70 = 1.3\ \text{A}$.
Measuring the internal resistance
- Vary the external resistor and plot terminal p.d. $V$ against current $I$.
- The line's intercept is $\varepsilon$ and its gradient is $-r$.
On a terminal-p.d.-against-current graph, match each feature.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
From $V = \varepsilon - Ir$: at $I = 0$, $V = \varepsilon$ (intercept); the slope is $-r$.
Worked example: reading the graph
A $V$–$I$ graph for a cell is a straight line from $(0,\ 1.5\ \text{V})$ to $(3.0\ \text{A},\ 0)$. The external resistor is then made smaller.
- e.m.f.: the intercept, $\varepsilon = 1.5\ \text{V}$.
- Internal resistance: gradient $= \dfrac{0 - 1.5}{3.0 - 0} = -0.50\ \dfrac{\text{V}}{\text{A}}$, so $r = 0.50\ \Omega$.
- Short-circuit current: where the line meets the $I$ axis, $\dfrac{\varepsilon}{r} = 3.0\ \text{A}$.
- Smaller external resistor: the current increases, so the lost volts $Ir$ increase and the terminal p.d. falls — you move down the line to the right.
- Check: the gradient is negative but a resistance is positive; write $r = 0.50\ \Omega$, not $-0.50\ \Omega$.
The external resistance connected to a real cell is decreased. What happens to the terminal p.d., and why?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Smaller external resistance → larger current → larger lost volts $Ir$ → smaller terminal p.d. $\varepsilon - Ir$. The e.m.f. is fixed, but the terminal p.d. is not.
The e.m.f. is not "the voltage of the battery" read by any voltmeter. A voltmeter across the terminals reads the terminal p.d., which equals $\varepsilon$ only when no current flows — an ideal voltmeter alone across the cell. Once a load is connected the reading drops by $Ir$. And the gradient of the $V$–$I$ line is $-r$: quote the resistance as a positive number.
When does a voltmeter across a cell's terminals read the e.m.f.? Select all that apply.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The reading is $\varepsilon - Ir$. It equals $\varepsilon$ if $r = 0$ or if $I = 0$. With a lamp connected, current flows and the reading drops; on a short circuit it falls to zero.
Ammeters 电流表, voltmeters 电压表 and the diagram
- An ideal ammeter has zero resistance and goes in series 串联.
- An ideal voltmeter has infinite resistance and goes in parallel 并联.
- When asked to draw the circuit, use the standard symbols and show the internal resistance as a resistor next to the cell.
The standard circuit symbols you need to recognise and draw
An ideal ammeter has:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
An ammeter goes in series and should add no resistance, so an ideal one has zero resistance.
An ideal voltmeter has infinite resistance and is connected in parallel.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Yes — in parallel, with infinite resistance so it draws no current from the component it measures.
You've got it
- terminal p.d. $V = \varepsilon - Ir$ (drops as current rises); $Ir$ is the lost volts
- a $V$–$I$ graph gives $\varepsilon$ (intercept) and $-r$ (gradient); short-circuit current $= \dfrac{\varepsilon}{r}$
- ideal ammeter: 0 Ω in series; ideal voltmeter: ∞ Ω in parallel — it reads $\varepsilon$ only with no current