#### Conservation of Energy and Electrical Efficiency

Electrical energy is useful to us mostly because it is easy to transport and can be easily converted to or from other forms of energy. Of course, conversion involves waste, typically as heat.

Electrical energy consumed can be determined by multiplying power by time \begin{align*}(E = P \Delta t)\end{align*}. Recall the equations for mechanical and thermal energy/work \begin{align*}(PE=mgh, KE=1/2mv^2, Q=mc \Delta T)\end{align*}. An important idea is the *efficiency* of an electrical device: the fraction of electrical energy consumed that goes into doing useful work \begin{align*}(E_{out}/E_{in})\end{align*}, expressed as a percentage.

\begin{align*} P = V \cdot I \end{align*} ; Power in electricity is the voltage multiplied by the current

\begin{align*}E = P \cdot \Delta t \; \end{align*} ; the electrical energy used is equal to the power dissipated multiplied by the time the circuit is running

\begin{align*} Eff = \frac{P_{out}}{P_{in}} \end{align*} ; Efficiency is the Power out divided by the Power input

Assuming same time periods: \begin{align*} Eff = \frac{E_{out}}{E_{in}} = \frac{Work}{E_{in}} \end{align*}

#### Example

You use a 100 W electric motor to lift a 10 kg mass 5 m, and it takes 20 s.

The electrical energy consumed is \begin{align*}E_{elec} = P \cdot t = (100 \ W) (20 \ s) = 2000 \ J\end{align*}

The work done is against gravity, so we use \begin{align*}PE = mgh = (10 \ kg) (10 \ m/s^2) (5 \ m) = 500 \ J\end{align*}

The efficiency is \begin{align*}E_{out}/E_{in} = (500 \ J) / (2000 \ J) = 0.25 = 25\% \ \text{efficient}\end{align*}

### Review

- The useful work a light bulb does is emitting light (duh). The rest is “wasted” as heat.
- If a standard incandescent 60 W bulb is on for 1 minute and generates 76 J of light energy, what is its efficiency? How much heat energy is produced?
- If the efficiency of a CFL (compact fluorescent) bulb is 20%, how many joules of light energy will a 60 W bulb produce?

- Most microwave ovens are 1000 W devices. You heat 1 cup (8 fl. oz., or 30 ml) of room temperature water \begin{align*}(20^\circ C)\end{align*} for 30 seconds in a microwave.
- What current does the oven draw?
- If the water heats up to \begin{align*}90^\circ C\end{align*}, what is the heating efficiency of the oven?

- The 2010 Toyota Prius \begin{align*}(m=1250 \ kg)\end{align*} battery is rated at 201.6 V with a capacity of 6.5 Ah.
- What is the total energy stored in this battery on a single charge?
- If you used the battery alone to accelerate to 65 mph one time (assuming no friction), what percentage of the battery capacity would you use?

### Review (Answers)

- a. 2.1%, 3524 J b. 720 J
- a. 8.3A b. 29%
- a. \begin{align*}4.7 \times 10^6 \ J\end{align*} b. 11.2%