Potential difference and power
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.
Potential difference
- p.d. is the energy transferred per unit charge: $V = \dfrac{W}{Q}$.
- Unit: the volt, $1\ \text{V} = 1\ \dfrac{\text{J}}{\text{C}}$.
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.
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
- Combining $V = \dfrac{W}{Q}$ and $I = \dfrac{Q}{t}$ gives $P = VI$.
- Energy transferred in a time is $E = Pt$.
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 ($V = IR$): $P = VI = I^{2}R = \dfrac{V^{2}}{R}$.
- Choose the form that uses the quantities you already know.
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 $E = Pt$