Circuit question

Stuck with question 6ci, how does the resistance increase won't it be the same, can anyone explain, here's in the link to the paper:

http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JAN12.PDF


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With lamp X still in circuit, the total parallel resistance of (Y+R2) and X is less than the series resistance of Y+R2 alone. (i.e. after X has gone open circuit).

Since the resistance between the battery (-ve) and R1 has therefore increased (after X has blown), the voltage dropped across Y+R2 must now be greater than the original parallel combination of Y+R2 and X.

The voltage across R1 will therefore decrease because all the voltages dropped around the circuit must sum to the supply voltage.

I thought the lamps didn't have any resistance so the total resistance would be the same. But now i understand that all filament lamps have resistance, which we could calculate by R= V^2/P. Am I right.

You seem like a physics God, you have answered all my questions, thanks soo much


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Also R = P/I² as well as V²/P

You got it! (resistance bit not the God part lol!)

One last question if you had two resistors which had different values in parallel would the current in each resistor be different?


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