Lamp Brightness in Parallel Circuit

AI Thread Summary
In a parallel circuit with lamps L and N connected to a battery and a third lamp M in series, if lamp N breaks, the brightness of lamp L increases while the brightness of lamp M decreases. This occurs because lamp L no longer shares current with lamp N, leading to increased current through L. Meanwhile, the overall resistance of the circuit changes, affecting the voltage drop across lamp M. The mark scheme confirms that the correct answer is C, as the current distribution alters significantly when one lamp in parallel fails. Understanding the changes in equivalent resistance and current flow is crucial for analyzing the circuit's behavior.
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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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