Energy stored in a capacitor
| English | Chinese | Pinyin |
|---|---|---|
| defibrillator | 除颤器 | chú chàn qì |
| capacitor | 电容器 | diàn róng qì |
| energy stored | 储存的能量 | chǔ cún de néng liàng |
| electrolytic | 电解 | diàn jiě |
The defibrillator 除颤器 jolt
- A defibrillator charges a capacitor 电容器, then releases the energy in one quick jolt.
- A capacitor is an energy store as well as a charge store.
- How much energy? It depends on charge and voltage.
Energy stored 储存的能量
- $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$.
- Pick the form with the quantities you know.
Real capacitors come in many sizes: large electrolytic 电解 cans store the most charge; the small ones below store very little
Energy in a capacitor
E = ½C·V²
Stored energy grows with the square of the voltage.
The energy stored in a capacitor is:
Equivalent forms: $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$.
A $2.0\ \mu\text{F}$ capacitor is charged to $1000\ \text{V}$. How much energy does it store?
$W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 2.0 \times 10^{-6} \times (1000)^{2} = 1.0\ \text{J}$.
Area under the V–Q graph
- A graph of $V$ against $Q$ is a straight line; the energy is the area beneath it.
- That area is the triangle $\tfrac{1}{2}QV$ — the $\tfrac{1}{2}$ is there because the average p.d. while charging is only $\tfrac{V}{2}$.
The energy stored is the area under the potential-charge line (the triangle 1/2 QV)
Why is there a factor of ½ in the stored-energy formula?
The p.d. rises from 0 to V as it charges, so the average is V/2 — hence the triangle area $\tfrac{1}{2}QV$.
The energy stored equals the ____ under the V–Q graph.
The triangle under the line has area $\tfrac{1}{2}QV$ — the stored energy.
Charging is "half efficient"
- Charging a capacitor through a wire delivers charge $Q = CV$ from the battery, giving out $QV$.
- But the capacitor only stores $\tfrac{1}{2}QV$ — the other half is lost as heat in the wire.
Capacitors store electric charge in a circuit
Charging a capacitor through a wire wastes about half the supplied energy as heat.
The battery gives out $QV$ but the capacitor stores only $\tfrac{1}{2}QV$; the rest heats the wire, whatever its resistance.
You've got it
- energy stored $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$
- it is the area under the V–Q line (the $\tfrac{1}{2}$ from the average p.d. $\tfrac{V}{2}$)
- charging through a wire wastes about half the energy as heat