Electrical Energy, Power

Deriving the power dissipated

• Take a conductor with ends A and B carrying a current $I$ from A to B, so $V(\text{A}) > V(\text{B})$ and $V = V(\text{A}) - V(\text{B}) > 0$.
• In a time $\Delta t$ a charge $\Delta Q = I\Delta t$ moves from A to B. Its potential energy changes by

$$\Delta U_{\text{pot}} = \Delta Q\left[V(\text{B}) - V(\text{A})\right] = -\Delta Q\,V = -IV\Delta t < 0$$

• If the charges moved without collisions, the loss of potential energy would appear as kinetic energy, $\Delta K = IV\Delta t > 0$, and the carriers would speed up indefinitely.
• They do not: collisions with the ions and atoms hand the energy over to the lattice, the atoms vibrate more vigorously and the conductor heats up. So in an actual conductor the energy dissipated as heat in time $\Delta t$ is

$$\Delta W = IV\Delta t$$

• The energy dissipated per unit time is the power:

$$P = \dfrac{\Delta W}{\Delta t} = VI$$

• Using $V = IR$,

$$P = I^{2}R = \dfrac{V^{2}}{R}$$

• This is the ohmic loss — the power that heats the filament of a bulb to incandescence. The source of this power is the cell: it is the chemical energy of the electrolyte that supplies it, for as long as it can.
• Heat produced in time $t$: $H = I^2Rt$ joule. On a bill, energy is measured in kilowatt-hours, $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$.

Which form of the formula to use

• Nichrome and copper wires of the same length and radius in series: $H = I^2Rt$ with $I$ and $t$ common, so $H \propto R \propto \rho$. Since $\rho_{\text{nichrome}} > \rho_{\text{copper}}$, the nichrome wire gets hotter.
• Two resistors of $4\ \Omega$ and $6\ \Omega$ in parallel: $P_1 : P_2 = R_2 : R_1 = 6 : 4 = 3 : 2$.
• Three resistors in series, $I = 2.0\ \text{A}$ throughout; if $R_3$ dissipates 6.0 W then $6.0 = (2.0)^2R_3$, so $R_3 = 1.5\ \Omega$ and $V_3 = 3\ \text{V}$.

Bulbs and ratings

• A bulb marked $(P, V)$ has resistance $R = V^2/P$, and draws its rated power only at its rated voltage.
• A 220 V, 100 W bulb has $R = 220^2/100 = 484\ \Omega$. Run at 110 V it consumes $P' = 110^2/484 = 25\ \text{W}$ — a quarter of the rating, because $P \propto V^2$ at fixed $R$.
• Two bulbs rated $(P_1, V)$ and $(P_2, V)$ across the same supply $V$: with $R_1 = V^2/P_1$ and $R_2 = V^2/P_2$,

$$\text{series:}\quad \dfrac{V^{2}}{P_s} = \dfrac{V^{2}}{P_1} + \dfrac{V^{2}}{P_2} \ \Rightarrow\ \dfrac{1}{P_s} = \dfrac{1}{P_1} + \dfrac{1}{P_2}$$

$$\text{parallel:}\quad \dfrac{1}{R_p} = \dfrac{P_1}{V^{2}} + \dfrac{P_2}{V^{2}} \ \Rightarrow\ P_p = P_1 + P_2$$

• Heat in a wire at constant applied voltage: $H = V^2t/R$ with $R = \rho l/\pi r^2$. Doubling both the length and the radius gives $R' = \rho(2l)/\pi(2r)^2 = R/2$, so $H' = 2H$.
• If the heat produced per second in a given resistor increases nine-fold, then since $H/t = V^2/R \propto V^2$, the applied voltage has increased by a factor of $\sqrt{9} = 3$.

Power transmission — why the lines are at high voltage

• A device is to receive power $P$ at a voltage $V$, so it draws $I = P/V$. The connecting cables have a finite resistance $R_c$, and the power wasted in them is

$$P_c = I^{2}R_c = \dfrac{P^{2}R_c}{V^{2}}$$

• For a given $P$ and $R_c$ the waste falls as $1/V^2$. Transmission cables run for hundreds of kilometres and $R_c$ is considerable, so the current is carried at enormous voltages — hence the high-voltage warning signs on transmission lines.
• Such voltages are unsafe to use, so a transformer at the far end steps the voltage down to a usable value.

Standard results

• $P = VI = I^2R = V^2/R$; $H = I^2Rt$; $1\ \text{kWh} = 3.6\times10^6\ \text{J}$.
• Series: $P \propto R$. Parallel: $P \propto 1/R$.
• Bulb: $R = V^2/P$ (rated); $P' = V'^2/R$ at another voltage.
• Two appliances on the same supply: $1/P_s = 1/P_1 + 1/P_2$ in series; $P_p = P_1 + P_2$ in parallel.
• Transmission loss $P_c = P^2R_c/V^2$.