| English | 의미 |
|---|---|
| energy stored/ˈenədʒi stɔːd/ | 저장된 에너지 |
| defibrillator/dɪˈfɪbrɪleɪtə/ | 제세동기 |
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.
某 capacitor에 저장된 에너지는:
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
등가 형태: $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$.
$2.0\ \mu\text{F}$의 축전기에 $1000\ \text{V}$의 전압을 가했을 때, 저장된 에너지는 얼마입니까?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
$W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 2.0 \times 10^{-6} \times (1000)^{2} = 1.0\ \text{J}$.
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
축전기의 에너지(Energy in a capacitor)
E = ½C·V²
저장된 에너지는 전압의 제곱에 비례하여 증가합니다.
저장 에너지 공식에 왜 ½ 인자가 존재합니까?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
충전过程中前위差从0到V까지 상승하므로 평균은 V/2입니다 — 따라서 삼각형 면적 $\tfrac{1}{2}QV$가 됩니다.
저장된 에너지는 V–Q 그래프 아래 ____에 해당합니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
그래프 아래 삼각형의 면적은 $\tfrac{1}{2}QV$ — 이것이 저장된 에너지입니다.
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 $W \propto V^2$.
축전기의 전하-전위차 그래프는 원점을 지나는 직선입니다. 이 기울기는 무엇을 나타냅니까?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Q 대 V 그래프의 기울기는 C이며, 에너지는 V 대 Q 선 아래 면적입니다. 기울기와 면적을 혼동하지 마십시오.
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.
전선을 통해 콘덴서를 충전하면 공급된 에너지의 약 절반이 열로 소모됩니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
배터리는 $QV$를 공급하지만 콘덴서는 오직 $\tfrac{1}{2}QV$만 저장하며, 나머지는 저항에 상관없이 전선에서 열로 변합니다.
용량 120 uF인 콘덴서가 12 V로 충전되었습니다. 저장된 에너지는 mJ 단위로 얼마입니까?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
W = C V^2 / 2 = 1.2e-4 x 144 / 2 = 8.6e-3 J입니다. 10 V일 때는 에너지가 전압의 제곱에 비례하므로 오직 6.0 mJ만 저장됩니다.
콘덴서 양단 전압을 두 배로 하면 저장되는 에너지도 두 배가 됩니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
W는 V의 제곱에 비례하므로 전압을 두 배로 하면 에너지는 네 배가 됩니다. 이것이 플래시 및 심박재진동기 회로가 고전압을 사용하는 이유입니다.
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.
충전된 콘덴서를 비충전 콘덴서에 연결했습니다. 추론의 순서를 바르게 나열하십시오.
난이도 3/5 · 200 코인 + 오늘 첫 완료 시 소액 보너스
전하는 보존되지만 에너지는 그렇지 않습니다. 손실된 에너지의的去處(장소)를 지칭하는 것이 최종 점수를 위한 것입니다.
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.
콘덴서가 정확히 한 시간 상수 동안 방전되었습니다. 저장된 에너지의 몇 퍼센트가 남아있습니까?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
전압은 37%까지 떨어지지만, 에너지는 V의 제곱에 비례하므로 e^-2 = 13.5%가 남습니다. 에너지를 37%라고 인용하는 것은 이 문제가 설계한 함정입니다.
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