Removing a Light Bulb: How Does it Affect Resistance and Current in a Circuit?

[AdkinsJr]

Homework Statement

Homework Equations

Ohm's Law, Equations for parallel and series combinations of resistors[/B]

The Attempt at a Solution

Working this problem I obtain a solution that doesn't make sense, which is that the bulbs are dimmer.

If I assign 10 ohms to each bulb, I believe the total resistance would be:

$$R_D+R_C+\left(\frac{1}{R_A}+\frac{1}{R_B}\right)^{-1}=25\Omega$$

After removing A, it would just be 30 ohms.

Before: 25 ohms
After: 30 ohms

So now there's more resistance and less current? [/B]

 

[phinds]

So now there's more resistance and less current?

That's an unhelpfully vague statement. WHERE would there be more or less of something and what would the effect be?

 

[TomHart]

Less overall current - true. So figure out each bulb individually to see how much current - before and after.

 

[AdkinsJr]

Less overall current - true. So figure out each bulb individually to see how much current - before and after.

Right, and I still find that they are dimmer. If there's less current after removing the bulb A, then the current to the bulbs next to the battery are going to be less (since they should have the same current as the overall current going over them) and therefore they should be dimmer? I'm told they should be brighter.

That's an unhelpfully vague statement. WHERE would there be more or less of something and what would the effect be?

I was referring to overall current and resistance.

 

[TomHart]

Right, and I still find that they are dimmer. If there's less current after removing the bulb A, then the current to the bulbs next to the battery are going to be less (since they should have the same current as the overall current going over them) and therefore they should be dimmer? I'm told they should be brighter.

I think they should be dimmer if they have less current. What about the other bulb not next to the battery?

 

[magoo]

So you said that before removing a bulb,

Rbefore = 10 + 10 + 5 = 25 ohms

After removing one,

Rafter = 10 + 10 + 10 = 30 ohms

What if the battery was 100 V. What would the current be through each bulb?
I'm particularly interested in the Before condition.

 

[phinds]

I was referring to overall current and resistance.

Yes, I assumed that your were, and I was making the point that that is not really all that helpful. What you need to do is be completely specific, and I see that you STILL are not doing that, just talking in generalities when you could in fact be solving for exact numerical values for everything in the circuit by assigning arbitrary values to the elements the way you started out doing and which you should have followed through with.