# Find resistance of resistor connected in series

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1. Mar 3, 2016

### PhysicsBoyMan

1. The problem statement, all variables and given/known data

Two 1.80 V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of R1 = 0.280Ω, the other an internal resistance of R2 = 0.155Ω. When the switch is closed, a current of 0.600 A passes through the lamp.

2. Relevant equations

R = V/I resistance = electric potential / current

3. The attempt at a solution

Underlined in black is the data I was given, as you can see in the question. I drew two batteries and a lightbulb connected in series. I multiplied 0.6 amps times resistance for each battery, and found the electric potential in volts for each battery. Knowing only the current in the lightbulb, but not the electric potential or the resistance, how am I supposed to solve this problem? I only have current in the lightbulb and resistance in the batteries, and I don't know how to use that to determine resistance in the lightbulb.

Thanks.

2. Mar 3, 2016

### Qwertywerty

I would've also mentioned what I wanted to actually find, in the original problem. This would be for any and all reading your OP.

3. Mar 3, 2016

If you examine a flashlight in detail, there is a connection to the outer edge of the base of the light bulb that usually returns along the barrel of the metal flashlight to the bottom end of the lower battery. This makes it so the filament of the light bulb is a resistor/resistance in a complete series circuit. (Tried responding to your sketch below, but my reply didn't work)...You need to include two 1.8 Volt batteries in series. Then use ohm's law, V=I*R to solve for R. The calculations are kind of fun, but you do need to make an effort.

Last edited: Mar 3, 2016
4. Mar 3, 2016

### PhysicsBoyMan

Here is my updated sketch to match the feedback from you guys.

With electric potential completely unknown, and resistance of the system only partially solved, I have no idea how to find RL AKA the lamp's resistance. Can someone point me in the right direction, maybe what to solve for first or something.

5. Mar 3, 2016

### PhysicsBoyMan

The question is "What is the lamp's resistance" just to be clear.

6. Mar 3, 2016

### PhysicsBoyMan

I forgot to add that the batteries have 1.8 V of electric potential each. Does anyone have any clues? I'm clueless and would really like to learn how to do this kind of problem.

7. Mar 3, 2016

### ehild

You did not include the voltage sources. Without voltage source, no current flows. A real battery is equivalent with an ideal voltage source in series with its internal resistance. The problem says, there are two 1.8 V batteries.

8. Mar 3, 2016

### Staff: Mentor

What's the total voltage supplied by the battery EMF's? What's an expression for the total resistance? (You can use a variable for the unknown resistance of the light bulb). You know the current. You should be able to write Ohm's law using those quantities.

9. Mar 3, 2016

### PhysicsBoyMan

I figured it out. I found the voltage reduction by each battery by going 0.6 amps x resistance value for battery. Then I took 3.6 V combined total from both batteries, subtracted both of the hits the voltage takes from the batteries that I calculated before, and I had the voltage for the lamp. Then I did the new voltage divided by the same current and found the resistance of the lamp.

10. Mar 4, 2016

### Qwertywerty

Yes, that's correct.