Hi
Physics Unit 2 is almost over for the year, and I've started revising for the Unit 2 exams. So far, so good; but I'm having doubts on how to do some of the checkpoints questions. Therefore I'll try to outline my method when asking a question so that people can point out where I went wrong. The topics in this thread will be predominantly Motion, Electricity, Light and Waves. Answer only as many questions as you want to, or have time for.
Thanks for your help.
The following information refers to Question 1-3.
A car battery is used in a remote house to power 2 lights and the radio. The circuit is connected in parallel. The lights both have a power rating of 30W, when running off a 12V battery. The battery can be assumed to have zero internal resistance.
1) What is the total power drain from the battery when the radio and both lights are switched on?
2) What is the current drain from the battery when the radio and both lights are switched on?
3) What the resistance of the whole circuit?
The following information refers to Question 4.
The two lights and the radio are connected in parallel. It would be possible to connect them in series.
4) What are at least two good reasons
not to do this?
The following information refers to Question 5-7.
A battery is rated at 12V. When a voltmeter with a very high resistance is used to check it's voltage, however, the voltmeter reads 13.2V. Then the battery is used as a "back up" to run some emergency lights and an air compressor in the storage. Now the battery is supplying 20A, and the voltmeter reads 12V.
5) What is the EMF of the battery.
6) What is the internal resistance of the battery?
7) Explain why the readings of the voltmeter are different.
My answers:1)

2)







3)

=


4) When connected in series the current would be less across each component, limiting them from functioning to their maximum output.
5)





6)

calculated above.
7) The voltmeter shows the reading across the battery terminals. The voltmeter has a very high resistance as mentioned, in particular a resistance higher than the battery's. Therefore it returns a higher reading of voltage.