I'm looking through some old IB exams, and I came across this (supposedly fairly easy)question. I'm having some problems, and I feel there's something really basic I'm missing.
http://img52.imageshack.us/img52/5139/physicsproblem.jpg [Broken] http://g.imageshack.us/img52/physicsproblem.jpg/1/ [Broken]
V = IR
P = VI
R of parallel = (1/R1) + (1/R2)
The Attempt at a Solution
My first instinct was that the correct answer is B is the correct answer; the brightness of Lamp L increases, the brightness of lamp M stays the same. My reasoning there was that since there is no longer a parallel, all the current will be forced to go through lamp L. Since the lamps have equal resistance, the current going through lamp L should double.
Initially, I didn't think M would change. But since the resistance of what used to be the parallel (lamp L) now is R rather than (2/R), the total resistance of the circuit has increased, and thus the current has decreased. Which would make lamp M less bright.
I actually thought of the last part while writing this post, so now I'm wondering if there's more to it than that, and if I'm correct. Thanks a lot.
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