Lamp Brightness in Parallel Circuit

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SUMMARY

The discussion centers on the behavior of lamps connected in a parallel and series circuit configuration. When lamp N breaks, the brightness of lamp L increases while the brightness of lamp M decreases. This is due to the change in current distribution; lamp L no longer shares its current with lamp N, leading to an increase in brightness. The correct answer to the question posed is C: lamp L increases in brightness, and lamp M decreases.

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  • Knowledge of series and parallel resistor configurations
  • Familiarity with voltage drop calculations in circuits
  • Ability to analyze equivalent resistance in circuits
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  • Study the concept of equivalent resistance in parallel circuits
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I'm looking at a problem where a battery of negligible internal resistance is connected to two lamps L and N in parallel with each other, then through in series to a 3rd lamp M. The question asks if the filament of one of the lamps in parallel N breaks then what happens to the brightness of the remaining lamps L and M.

The choice was

Lamp L , Lamp M
A) stays the same , decreases
B) increases , stays the same
C) increases , decreases
D) decreases , increases

The markscheme says its C. I understand that L would increase as it no longer shares its current but I thought the answer would be B as I didn't think M would be affected? Where am I going wrong?
 
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Imagine them all as equal resistors 'R'.
If you have two resistors in parallel what is the total resistance ?
If the pair and M are in series what is the voltage drop across the pair and across M?

Now if one breaks, you have the same resistance L and M in series.
What happens to the voltage drop across M?
 
Think equivalent resistance before the break and current passing through the resistances, and then again after the break.

Before (R*R/R+R)+ R


After R+ R

substitute a value of your choice for all resistances in the circuit, and see how the current changes in both L and M
 

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