**Watts**

The power, called wattage, delivered
to an electric device (a power saw, for example) can

be increased by (more than one answer may be true)

a) increasing the current flow (amperage) it draws but not the potential difference (voltage)

b) increasing the voltage applied to it but not the amperage

c) increasing both amperage and voltage

d) increasing the time the circuit is left on

e) by use of a transformer

**Electric Power**

Recall that the rate energy is converted from one form to another is

power.

power = energy converted / time = voltage x charge/time = voltage x currentThe unit of power is the

watt ..(1000 watts = 1kilowatt). So in units form

Electric power (watts) = current (amperes) x voltage (volts),

where 1watt= 1amperex 1volt.1. What is the power when a voltage of 120 V drives a 2-A current through a device?

2. What is the current when a 60-W lamp is connected to 120 V?A3. How much current does a 100-W lamp draw when connected to 120 V?

4. If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

5. The equationpower = energy converted / timerearranged gives

energy converted = power x time

**Power Production**

Does it take a lot of water to light a llight bulb? That depends on its wattage and

how long it glows. In this practice page, you are to calculate the mass and

volume of water that must fall over a 10-m high dam in order to keep a 100-W

light bulb glowing for 1 year.

1. First, calculate how many joules are required to keep the bulb lit for 1 year.2. What mass of water elevated 10 m has this much PE? Recall that

gravitational PE = mgh:3. But this assumes 100% efficiency. A hydroelectric plant is typically 20% efficient. This means only 1 part in 5 of the PE of the falling water ends up as electricity. So the mass above must be multiplied by 5 to get the actual amount of water that must fall to keep the 100-W bulb lit.

4. This is an impressive number of kilograms! To visualize this amount of wa ter, convert it to cubic meters. (Recall 1 kg of water occupies 1 liter, and there are 1000 liters in 1 cubic meter.)

5. For comparison, an Olympic-size swimming pool holds about 4000 m3 of water. How many such poolfuls of water are required to keep a 100-W bulb lit for one year?

Does it take a lot of water to light a light bulb? To light a city full of light bulbs? Now you have a better idea!

Teachers NotesDoes it take a lot of water to light a llight bulb? That depends on its wattage and how long it glows. In this practice page, you are to calculate the mass and

volume of water that must fall over a 10-m high dam in order to keep a 100-W light bulb glowing for 1 year.1. First, calculate how many joules are required to keep the bulb lit for 1 year.

Energy = power x time = 100 W * 1 yr = 365d/1yr * 24h/1d * 3600s/1h

= 3 150 000 000 J2. What mass of water elevated 10 m has this much PE? Recall that gravitational PE = mgh: m = PE/gh = 3 150 000 000/(9.8m/s2)(10m) J = 32 100 000 kg

3. But this assumes 100% efficiency. A hydroelectric plant is typically 20% efficient. This means only 1 part in 5 of the PE of the falling water ends up as electricity. So the mass above must be multiplied by 5 to get the actual amount of water that must fall to keep the 100-W bulb lit. 5 x 32 100 000 kg = 160 000 000 kg

4. This is an impressive number of kilograms! To visualize this amount of wa ter, convert it to cubic meters. (Recall 1 kg of water occupies 1 liter, and there are 1000 liters in 1 cubic meter.) Volume = 160 000 000 kg ¥ 1 L/kg * 1m3 /1000 L = 160 000 m3

5. For comparison, an Olympic-size swimming pool holds about 4000 m3 of water. How many such poolfuls of water are required to keep a 100-W bulb lit for one year?

Number of poolfuls = 160 000 m3/4000 m3/poolful = 40 poolfuls