BMAT Section 2 Help! Watch

champion1
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Hi! Please could someone explain the answer to Question 15 of 2010 Section 2 BMAT attached. Answer is B but I just can't figure out how!

Thanks!
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champion1
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(Original post by champion1)
Hi! Please could someone explain the answer to Question 15 of 2010 Section 2 BMAT attached. Answer is B but I just can't figure out how!

Thanks!
I found an explanation to this question online from:

http://www.bmatcrashcourse.com/electricity-notes.pdf

"BEFORE: All of the charge will bypass the 4 parallel bulbs (through switch Q) and only flow through bulb Y. Bulb Y will therefore have all 12V running through it, and will be at maximum brightness.

AFTER: Switch Q is now closed, so we no longer have 4 bulbs being short circuited. Bulb X has some charge flowing through it now, so regardless of how much actually is, the bulb will be brighter than it was before. Bulb Y now has less than 12V flowing through it, as some (doesn't matter how much) voltage is being taken up by the parallel combination, so it must be dimmer than it was before.

The answer therefore, has to be B. Notice we didn't have to do any calculations or anything, we just had to recognize that X must be somewhat brighter than before, and Y must be dimmer."

But I don't understand how before all charge bypasses the 4 parallel bulbs.
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Stonebridge
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Report 4 years ago
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(Original post by champion1)
I found an explanation to this question online from:

http://www.bmatcrashcourse.com/electricity-notes.pdf

"BEFORE: All of the charge will bypass the 4 parallel bulbs (through switch Q) and only flow through bulb Y. Bulb Y will therefore have all 12V running through it, and will be at maximum brightness.

AFTER: Switch Q is now closed, so we no longer have 4 bulbs being short circuited. Bulb X has some charge flowing through it now, so regardless of how much actually is, the bulb will be brighter than it was before. Bulb Y now has less than 12V flowing through it, as some (doesn't matter how much) voltage is being taken up by the parallel combination, so it must be dimmer than it was before.

The answer therefore, has to be B. Notice we didn't have to do any calculations or anything, we just had to recognize that X must be somewhat brighter than before, and Y must be dimmer."

But I don't understand how before all charge bypasses the 4 parallel bulbs.

Because the switch forms a "short circuit" which is explained on page 10 of that file.
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champion1
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(Original post by Stonebridge)
Because the switch forms a "short circuit" which is explained on page 10 of that file.
Thanks for your help! I understand now.
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