- #1

scorpa

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I am doing some physics questions and am having trouble with voltage drops. If anyone could help me understand it a bit better I would appreciate it.

The first question is this:

1) Three identical lamps are connected in series to a 6.0 V battery. What is the voltage drop across each lamp?

- All they give you is the voltage of the battery, so do you have to assume the other values are 1 or something like that. I know this one should be easy but I can't seem to get it. The lamps are all identical, so can you just assume that the voltage drop is 6V across each lamp? I really don't know what I am talking about here, as I am sure you can tell.

2) A string of eighteen identical Christmas tree lights are connected in series to a 120V source, the string dissipates 64.0 W.

a) What is the equivalent resistance of the light string?

P = IV

64W/120V

I = 0.533 A

R = V/I

120V/0.533A

R = 225 ohms

Equivalent resistance is 225 ohms?

b) What is the resistance of a single light?

As it is connected in series 225 ohms/18 lights = 12.5 ohms

c) What is the power dissipated by each lamp?

V = IR 120 V/12.5A = 9.6 A

P = IV 9.6A*120V = 1152 W which to me seems way to high

If one of the bulbs in the problem burn out and the lamp has a wire that shorts out the lamp filament when it burns out the resistance of the lamp is dropped to zero.

d) What is the resistance of the light string now?

e) Find the power dissipated by the string.

f) Did the power go up or down when the bulb burned out?

At this point I am pretty lost, and really need help. I know it should be something really easy but I can't think of what to do.

3) Amy needs 5.0V for some integrated circuit experiments. She uses a 6.0V battery, and two resistors to make a voltage divider. One resistor is 330 ohms. SHe decides to make the second resistor smaller. What value should it have?

Once again I have no clue about this one. For some reason voltage drop questions have been causing me a lot of grief today.

Thanks for any help you can give me, I really appreciate the help.