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Parallel Lightbulbs- resistance total

  • Thread starter GDGirl
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  • #1
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Homework Statement


Eight lights are connected in parallel to a 110-V source by two long leads of total resistance 1.97 Ω.
a) If 405 mA flows through each bulb, what is the resistance of each?


Homework Equations


V=IR
P=IV=I2R=V2/R
Rtotal(for a series)= (R1-1+R2-1)-1
Rtotal(for parallel)= R1+R2


The Attempt at a Solution


So I tried just setting what I assumed was the total resistance (1.97) equal to my derived equation (Rtot=(8R-1)-1) and came up with 15.76 Ohms. This was wrong.
So then I tried using V=IR, and got 271.605. This was also wrong.
Upon re-looking at the problem, I see that (I think) it means the total resistance 1.97 Ohms is actually just of the two long leads, in which case I'm not sure how to work this into the problem... :/

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
I'm not 100% sure about this, but i will work as such:

first, change the circuit into series.
so it consists of battery-wire-bulbs-wire

since the wire has a resistance, it must take up some voltage of the 110V.
V=IR. Current (I) will be 8*0.405.

The solve the total resistance of the each bulb
R=V/I. V is 110V minus whatever is got from the previous step. Current is 0.405

This is the almost the same as what you second trial is (when you got 271 ohms), the only difference is the voltage used.
 
  • #3
50
0
I'm not 100% sure about this, but i will work as such:

first, change the circuit into series.
so it consists of battery-wire-bulbs-wire

since the wire has a resistance, it must take up some voltage of the 110V.
V=IR. Current (I) will be 8*0.405.

The solve the total resistance of the each bulb
R=V/I. V is 110V minus whatever is got from the previous step. Current is 0.405

This is the almost the same as what you second trial is (when you got 271 ohms), the only difference is the voltage used.
I just tried this, and it did not work.
However, I want to know why you made the current in the equation for the wire 8*.405. I understand that 8 probably came from the lightbulbs, but I don't know how that would apply to the wire. :/
 
  • #4
Because the the each light bulb has 0.405A flowing through them,
so the current of the circuit in series will be 8*0.405

Itotal(in parallel) = I1+I2+I3...

because in series, the current throughout the circuit is the same, hence 8*0.405 would apply to the wires (and also to the 8 bulbs as a whole, with each bulb having 0.405A because the bulbs are identical in resistance).

But after all, the answer is wrong having it done this way. I'm stuck too then..
This is the only way i could think to solve the problem.
 
  • #5
50
0
I figured it out! You were completely right with the multiplying .405 times 8. :) ThenI applied it to my equation for the total resistance (110/(8*.405)) and used that total for this:

33.951=R/8+1.97 (R/8 is my equivalent for what the total resistance of just the lightbulbs are) And I got it, so you did help, thanks! :)
 

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