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Circuit question Watch

1. 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...1-QP-JAN12.PDF

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2. because lamp X has broken, the parallel part of the cart becomes series. Therefore, the resistance due to R2 increases and hence R2 takes up more of the voltage than it did before. This leaves less voltage across R1.
3. (Original post by Jimmy20002012)
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...1-QP-JAN12.PDF

Posted from TSR Mobile
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.
4. (Original post by uberteknik)
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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5. (Original post by Jimmy20002012)
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!)
6. (Original post by uberteknik)
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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7. (Original post by Jimmy20002012)
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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Yes, yes, yes! The current in parallel resistors is only the same if the resistances are the same.

It's only the voltage across both parallel resistors which is the same whether their resistances are the same or not.

I = V/R if the resistors are different, then the current is different.

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Updated: May 13, 2013
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