Physics help on resistors in series and parallel!?

Question

A battery with an emf (a source of electromotive force) of 12.0V shows a terminal voltage of 11.8V when operating in a circuit with two lightbulbs rated at 3.0W(at 12.0V) which are connected in parallel. What is the battery's internal resistance?


Thanks!

Answer 1

The answer above is close, but the current through each bulb won't be 1/4 amp since each bulb isn't getting 12 V (only 11.8). To do this correctly takes a bit more work.

First, the bulbs' resistances can be figured out based upon 3W at 12 V:

P = V^2/R
3 W = (12V)^2/R
R = 48 ohms

Since the two bulbs are identical and in parallel, their effective resistance is 24 ohms.

Now you can find the current through the two bulbs in parallel using I = V/R.

I = 11.8 V/24 ohms = .49167 amps

This is the same current that goes through the internal resistance of the battery, so

I = V / R
.49167 amps = .2 volts / R
R = .2 volts / .49167 amps
R = .4068 ohms or 24/59 ohms

Answer 2

If you measure the battery when not connected you you should read 12.0v after you connect the 2 3w lamps in parallel. therefore when you measure the battery terminals again you are really measuring the voltage drop accross the circuit. so the two formulas are E=I*R and P=E*R

Answer 3

you can find the amount of current for each bulb since they give you the power and voltage
p = vi => i{bulb} = p / v = 3W / 12V
now u can find the resistance of the bulb. since they are in parallel, combine them and then your circuit is just the batter, internal resistance and bulbs. u know the current would be twice the current found in one bulb, by KCL. you have all the info so all you need to do is find the internal resistance which is the unknown

Answer 4

Watts/voltage = current so each bulb is carrying 3/12 = 0.25 amps. Thus total current is 0.5amps
12 volts -11.8 volts = 0.2 volts.
So Resistance = 0.2/.5 = 0.4 ohms