Electrical energy and power

Where does the energy go?

Narrated video. Transcript:

Switch on an electric heater and its coil glows red hot. The battery or the mains is handing energy to the coil. Let us find out how much, every second.

Take a conductor between two points A and B. A current I flows from A to B, and the potential at A is higher than at B by V.

In a short time Δt, a charge ΔQ equal to I Δt moves from A to B. Its potential energy falls by ΔQ times V, which is I V Δt.

If the charges moved freely, this energy would become kinetic energy. But inside the conductor the electrons keep colliding with the ions, so their average speed stays at the steady drift speed.

So in each collision the energy passes to the ions, which vibrate more strongly. The conductor heats up. This is Joule heating.

The energy dissipated per unit time is the power: P equals ΔW over Δt, which is V I. It is measured in watts, joules per second.

Using Ohm’s law, V equals I R, we can write the same power two more ways: P equals I² R, and P equals V² over R.

Worked example. A heater is rated 1000 watt at 220 volt. Its resistance is V² over P: 220² over 1000, which is 48.4 ohm. It draws 1000 over 220, about 4.55 ampere.

Now run it on 110 volt. The resistance stays about 48.4 ohm, so P equals 110² over 48.4, which is 250 watt. Half the voltage gives only a quarter of the power.

Units of energy, and sending power far away

Narrated video. Transcript:

The electricity bill does not count joules. It counts units. One unit is one kilowatt hour: the energy used by a 1 kilowatt appliance running for one hour.

1 kilowatt hour is 1000 watt times 3600 second, which is 3.6 × 10⁶ joule.

So our 1000 watt heater, left on for 2 hours, uses 2 units: 7.2 × 10⁶ joule. At 8 rupees a unit, that costs 16 rupees.

Now a bigger problem. A power station must send electrical power P to a town far away, through long cables of total resistance R c.

The current in the cables is I equals P over V, where V is the voltage at which the power is sent. The cables waste power as heat: P c equals I² R c.

Put in I equals P over V, and P c equals P² R c over V². For a given power and given cables, the loss falls as the square of the transmission voltage.

Example: 800 kilowatt sent through cables of 4 ohm at 11 kilovolt. The current is 800,000 over 11,000, about 72.7 ampere, and the loss is 72.7² times 4, about 21 kilowatt: roughly 2.6 percent.

Raise the voltage three times, to 33 kilovolt. The current drops to about 24.2 ampere, and the loss drops nine times, to about 2.35 kilowatt.

Try sending the same 800 kilowatt at 220 volt: the current would be over 3600 ampere, and the cables would waste far more than the whole 800 kilowatt. It simply cannot be done.

That is why power is sent at very high voltages, stepped up by transformers at the station and stepped down again before it reaches your home.

Try it: run an appliance

Interactive simulation. Things to try:

• Check the readouts: VI, I²R and V²/R always give the same power.
• Lower the supply to 110 V. The heater’s power falls to a quarter, 250 W.
• Change the hours and watch the energy in joules and in kWh (units), and the cost.
• Compare the 100 W bulb and the 1000 W heater at 220 V. Which has the larger resistance?

Try it: send power to a town

Interactive simulation. Things to try:

• At 11 kV about 2.6% of the power is lost in the cables. Raise the voltage to 33 kV: the loss falls nine times.
• Bring the voltage down towards 0.22 kV (220 V) and watch the cables glow: the loss becomes enormous.
• Double the cable resistance: the loss doubles. Double the voltage: the loss falls to a quarter.

Same bulb, lower voltage

A bulb is rated 100 W at 220 V. Assuming its resistance does not change, what power does it use on a 110 V supply?

1. 100 W
2. 50 W
3. 25 W
4. 200 W

Doubling the transmission voltage

A power station sends a fixed power to a town through the same cables. If the transmission voltage is doubled, the power wasted in the cables becomes:

1. twice as much
2. the same
3. half as much
4. a quarter as much

Summary

• Power: P = VI = I²R = V²/R
• Energy: W = Pt; 1 kWh = 3.6 × 10⁶ J
• Power lost in cables: P_c = I²R_c = P²R_c / V²

• When a charge ΔQ falls through a potential difference V it loses energy ΔQ V. In a conductor this goes to the ions through collisions, as Joule heat.
• Power is measured in watts; the energy we pay for is measured in kilowatt hours (units).
• For a fixed resistance, power goes as V²: half the voltage gives a quarter of the power.
• Power is transmitted at high voltage so the current, and the I²R_c loss in the cables, is small. Transformers step the voltage up and down.