physics, please help?

Question

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


Please show work so I will understand how to work similar problems.

Answer 1

First, find the load resistance (the two 3W bulbs)

The formula to use is: P = E^2 / R
3.0 = 12^2 / R
3 = 144 / R
R = 48 Ohms

But there are 2 of them in parallel, so the resistance is halved.
R = 24 Ohms

There is a more general formula for resistors in parallel when they are not of equal value:
Rparallel = (R1 * R2) / (R1 + R2)
the result is the same for equal value R1 and R2 ---> R / 2


Next step: find the current flowing through the load.
The formula is Ohms law
E = I * R
11.8 = I * 24
I = 0.49166 amps

Final step: find the internal resistance
E = I * R
E is the small voltage drop: 12 - 11.8 = 0.2 Volts
I is 0.49166.. Amps
0.2 = 0.49166 * R
R = 0.40678 Ohms

.

Answer 2

Rs will be battery internal resistance

each bulb is rated 3 watts at 12v. But remeber, its not 12v going across them, only 11.8. so you cant calculate theirr current directly. Instead, you have to calculate resistance first.

Rbulb = rated volts^2/rated watts = 144/3 = 48 ohms

total resistance in the two bulbs(they're in parallel) = Rbulb^2/2Rbulb = 24 ohms

current I = 11.8v/24 ohms = .49167 amps

now you can calculate the Rs using ohms law.

12v - 11.8v = .2 volts across this resistance

Rs = 0.2v/.49167 amps = .4068 ohms