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Concept Development Practice Page 34-1 and Power and Electricity worksheet answers:

Wkst 34-2:

1) 240W    2) .50A    3) .83A    4) 2V    5) energy=Powerxtime

6) kilowatt=unit of power; kilowatt-hour=unit of energy

7) .060kW x 720hr x $.08/kW-hr = $3.46

Power and Electricity:

1. Resistance of the bulb (the filament) decreases as you go from 40W to 100W bulbs, and this causes more current to flow in the circuit.  P=IV, and since V=110V for both, when the current increases for the 100W bulb, its power will be greater, and thus its brightness.

2. a) Brownout.  They expect 220V and get 110V, which means they are getting half the energy per charge that they need to operate.

b) Burnt out/fried.  They expect 110V and get twice as much.  

3. The adapter must be basically a resistor in series which will take half the voltage available, so the appliance will get the 110V it's used to.  This means it is taking half the power that the appliance would have gotten.

4. When a lamp is first turned on, its filament is cold and allows current to move through it easily.  That is, its resistance is low, so its current is high.  Sometimes this current spike (which drops when the appliance warms up) is enough to burn out the bulb.

5. The 100W bulb has less resistance and will allow more current through it.  However, the lamp is only expecting a 60W bulb, which would allow less current.  The larger amount of current may be enough to heat plastics or cardboards within the lamp to the point where a fire starts.  If not, it is still not a good idea, since you are putting more current through an appliance than it was designed for.  It may burn out the lamp.