Potential difference and power
| English | Chinese | Pinyin |
|---|---|---|
| potential difference | 电势差 | diàn shì chà |
| volt | 伏特 | fú tè |
| electrical power | 电功率 | diàn gōng lǜ |
| Ohm's law | 欧姆定律 | ōu mǔ dìng lǜ |
What "voltage" really means
- A 240 V supply is dangerous; a 1.5 V cell is harmless.
- The difference is the energy each coulomb of charge carries.
- That energy-per-charge is the potential difference 电势差.
Potential difference
- p.d. is the energy transferred per unit charge: $V = \dfrac{W}{Q}$.
- Unit: the volt 伏特, $1\ \text{V} = 1\ \dfrac{\text{J}}{\text{C}}$.
Charge carriers drifting inside a conductor
Potential difference is the energy transferred per unit:
$V = \dfrac{W}{Q}$ — energy per unit charge, in joules per coulomb (volts).
One volt equals one joule per ____.
$1\ \text{V} = 1\ \dfrac{\text{J}}{\text{C}}$ — one joule of energy moved per coulomb of charge.
e.m.f. vs p.d.
- e.m.f. = energy a source gives to each coulomb.
- p.d. = energy a coulomb gives up to a component. Same unit, opposite roles.
Voltage is energy per charge. A 240 V supply gives each coulomb 160 times more energy than a 1.5 V cell — that's why it's dangerous.
Both e.m.f. and potential difference are measured in volts.
Yes — both are energy per unit charge (J/C). The difference is whether energy is given to, or taken from, the charge.
Electrical power 电功率
- Combining $V = \dfrac{W}{Q}$ and $I = \dfrac{Q}{t}$ gives $P = VI$.
- Energy transferred in a time is $E = Pt$.
I-V characteristic of a filament lamp
Electrical power
P = VI
At a fixed voltage, power is proportional to the current it drives.
A device runs at $12\ \text{V}$ and draws $2.0\ \text{A}$. What is its power?
$P = VI = 12 \times 2.0 = 24\ \text{W}$.
Three forms of power
- With Ohm's law 欧姆定律 ($V = IR$): $P = VI = I^{2}R = \dfrac{V^{2}}{R}$.
- Choose the form that uses the quantities you already know.
Resistance of an LDR decreases as light intensity increases
Which form of electrical power uses only the voltage and the resistance?
$P = \dfrac{V^{2}}{R}$ needs only $V$ and $R$ — handy when you do not know the current.
A $100\ \text{W}$ lamp is on for $60\ \text{s}$. How much energy does it transfer?
$E = Pt = 100 \times 60 = 6000\ \text{J}$.
You've got it
- p.d. is energy per charge: $V = \dfrac{W}{Q}$ (volt = joule per coulomb)
- e.m.f. is energy given to charge by the source; p.d. is energy given up by charge
- power $P = VI = I^{2}R = \dfrac{V^{2}}{R}$; energy $E = Pt$