| English | 意味 |
|---|---|
| energy stored/ˈenədʒi stɔːd/ | 蓄積されたエネルギー |
| defibrillator/dɪˈfɪbrɪleɪtə/ | 除細動器 |
Half the energy never reaches the capacitor
- Connect an ideal battery of e.m.f. $V$ to a capacitor through a wire. The battery pushes charge $Q = CV$ through a p.d. of $V$, so it gives out $QV$ of energy.
- The capacitor ends up holding $\tfrac12 QV$. Exactly half the energy has gone, as heat in the wire, and it makes no difference how good the wire is.
- The missing half is not an inefficiency to be engineered away. It is a consequence of the fact that the p.d. across the capacitor starts at zero and only reaches $V$ at the end.
- This lesson is the energy stored 储存的能量, where the factor of a half comes from, and what happens when charge is shared.
The three forms
- Charging a capacitor takes work, because each extra bit of charge must be pushed against the p.d. that is already there. Adding up all those bits gives:
- All three are the same expression with $Q = CV$ substituted differently. Choose the one that uses the two quantities the question gives you, and no rearranging is needed.
コンデンサに蓄えられるエネルギーとは：
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
同等の式：$W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$。
コンデンサ$2.0\ \mu\text{F}$が$1000\ \text{V}$まで充電された。これに蓄えられているエネルギーはいくらか？
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 2.0 \times 10^{-6} \times (1000)^{2} = 1.0\ \text{J}$.
The area under the graph
- A graph of $V$ against $Q$ for a capacitor is a straight line through the origin, of gradient $1/C$.
- The energy stored is the area under that line up to the charge $Q$, which is a triangle of area $\tfrac12 QV$.
- That is where the half comes from: the p.d. grows from zero to $V$ as the capacitor charges, so the average p.d. during charging is $V/2$, not $V$.
A triangle, not a rectangle, and that is the whole reason for the half
コンデンサ内のエネルギー
E = ½C·V²
蓄えられたエネルギーは電圧の二乗に比例して増大する。
なぜ蓄エネルギーの公式に½という係数が含まれているのでしょうか？
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
充電に伴い電位差は0からVまで上昇するため、平均値はV/2 — よって三角形の面積 $\tfrac{1}{2}QV$ となる。
蓄えられるエネルギーはV–Qグラフの ____ に相当する。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
直線下の三角形の面積は $\tfrac{1}{2}QV$ — 蓄えられるエネルギーである。
Worked example: energy from a graph
- A graph of $Q$ against $V$ is a straight line through the origin passing through $(10\ \text{V},\ 1.2 \times 10^{-3}\ \text{C})$. Find the capacitance, the energy at $10\ \text{V}$, and the extra energy needed to reach $12\ \text{V}$.
- The gradient of $Q$ against $V$ is the capacitance: $C = 1.2 \times 10^{-3}/10 = 1.2 \times 10^{-4}\ \text{F} = 120\ \mu\text{F}$.
- Energy at $10\ \text{V}$: the area, $\tfrac12 QV = \tfrac12 (1.2 \times 10^{-3})(10) = 6.0 \times 10^{-3}\ \text{J}$.
- At $12\ \text{V}$: $\tfrac12 CV^2 = \tfrac12 (1.2 \times 10^{-4})(144) = 8.6 \times 10^{-3}\ \text{J}$, so the extra energy is $2.6 \times 10^{-3}\ \text{J}$.
- Note that a 20% rise in voltage costs a 44% rise in energy, because $W \propto V^2$.
コンデンサの電荷対電位差のグラフは原点を通る直線です。その傾きは何を表しますか？
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Q対Vのグラフの傾きはCであり、エネルギーはV対Qの直線下の面積に相当する。傾きと面積を混同しないように注意すること。
Energy goes as the square
- Because $W = \tfrac12 CV^2$, doubling the p.d. quadruples the energy for the same capacitor.
- That is why a camera flash or a defibrillator 除颤器 charges its capacitor to hundreds or thousands of volts rather than using a larger capacitor at a low voltage: voltage is the cheaper way to buy energy.
- It is also why the same capacitor at half the voltage stores only a quarter of the energy, which is worth checking your answers against.
導線を介してコンデンサを充電すると、供給されたエネルギーの約半分が熱として散逸する。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
バッテリーは$QV$のエネルギーを提供するが、コンデンサに蓄えられるのは$\tfrac{1}{2}QV$のみである。残りのエネルギーは導線の抵抗に関わらずすべて熱となる。
120 μFのコンデンサを12 Vまで充電した。蓄えられたエネルギーはいくらか？単位はmJで答えること。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
W = C V^2 / 2 = 1.2e-4 × 144 / 2 = 8.6e-3 J。10 Vの場合は6.0 mJしか蓄えられない。なぜなら、エネルギーは電圧の二乗に比例するためである。
コンデンサ両端の電位差を倍にすると、蓄えられるエネルギーも倍になる。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
WはVの二乗に比例するため、電位差を倍にすればエネルギーは4倍になる。これがフラッシュや除細動器回路が高電圧を使用する理由である。
Worked example: sharing charge
- A charged capacitor $C$ at p.d. $V$ is disconnected and then connected across an uncharged capacitor of capacitance $3C$. Find the final p.d. and the energy lost.
- Charge is conserved. The total charge is still $Q = CV$, now on a parallel pair of total capacitance $4C$, so the common p.d. is $V' = Q/4C = V/4$.
- Energy is not conserved. Before: $\tfrac12 CV^2$. After: $\tfrac12 (4C)(V/4)^2 = \tfrac18 CV^2$. Three-quarters of the energy has gone.
- It is dissipated as heat in the connecting wires, and as a spark or radiation, while the charge moves, however small the resistance. Saying where it went is the mark.
充電されたコンデンサを未充電のコンデンサに並列接続した。論理の順序を正しく並べ替えること。
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
電荷は保存されるが、エネルギーは保存されない。失われたエネルギーの去向を説明できることが最終点の要件となる。
Energy during a discharge falls twice as fast
- During a discharge the charge and the p.d. both decay as $e^{-t/RC}$, but the energy goes as $V^2$, so it decays as $e^{-2t/RC}$.
- After one time constant the p.d. has fallen to $37\%$ of its start, but the energy has fallen to $e^{-2} = 14\%$.
- After two time constants the p.d. is $14\%$ and the energy only $1.8\%$.
- So the energy has an effective half-life of half that of the charge. A question that gives you a time constant and asks for the remaining energy is testing exactly this factor of two.
- The same square is why a capacitor charged to twice the p.d. stores four times the energy.
コンデンサをちょうど1つの時間定数分だけ放電させた。残存しているエネルギーの割合（%）はいくらか？
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
電位差は37%に低下するが、エネルギーは電圧の二乗に比例するため、e^-2 ≈ 13.5%が残る。エネルギーに対して37%と答えることは、この問題の罠であることに注意せよ。
Marks that slip away
- The factor of $\tfrac12$ comes from the average p.d. being $V/2$ during charging, or equivalently from the area of a triangle. Say one of those, not "it is in the formula".
- $W \propto V^2$, so doubling the voltage gives four times the energy, not twice.
- In a charge-sharing question, charge is conserved but energy is not. Name where the lost energy goes.
- Half the battery's energy is lost as heat during charging whatever the wire's resistance. It is not a fault in the circuit.
You've got it
- energy stored $W = \tfrac12 QV = \tfrac12 CV^2 = \dfrac{Q^2}{2C}$; pick the form matching the quantities given
- it is the area under the $V$ against $Q$ line, a triangle, and the $\tfrac12$ is because the average p.d. while charging is $V/2$
- $W \propto V^2$: double the p.d., four times the energy
- when charge is shared, charge is conserved and energy is not: the difference is dissipated as heat in the wires, whatever their resistance