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1. HIII
HOW DO I SOLVE THIS??

I thought that the power dissipated in the parallel circuits is 12 - 4 = 8W

But I did not get the correct answer! Help!!
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2. (Original post by nwmyname)
HIII
HOW DO I SOLVE THIS??

I thought that the power dissipated in the parallel circuits is 12 - 4 = 8W

But I did not get the correct answer! Help!!
8W is dissipated in resistor P.

12W - 8W = 4W is dissipated in the parallel combination of resistors Q and R.

Resistors Q and R are equal values and therefore dissipate 2W each.
3. (Original post by uberteknik)
8W is dissipated in resistor P.

12W - 8W = 4W is dissipated in the parallel combination of resistors Q and R.

Resistors Q and R are equal values and therefore dissipate 2W each.
Why 8W? Please respond! My apologies!
4. (Original post by nwmyname)
Why 8W?
If all resistors are the same value, then the circuit forms a potential divider of (R + (R || R) = R + R/2.

The p.d. developed across resistor P is 2V/3 whilst the p.d. across the parallel combination is V/3

The current (I) flowing through P is also split between Q and R. (I splits into I/2 + I/2)

Power = VI = 12W

For resistor P, power = 2V/3 x I = 2/3 x 12W = 8W

For resistors Q and R, power = V/3 x I/2 = 1/6 x 12W = 2W each.
5. (Original post by uberteknik)
If all resistors are the same value, then the circuit forms a potential divider of (R + (R || R) = R + R/2.

The p.d. developed across resistor P is 2V/3 whilst the p.d. across the parallel combination is V/3

The current flowing through P is also split between Q and R. (I splits into I/2 + I/2)

Power = VI

For resistor P, power = 2V/3 x I = 2/3 x 12W = 8W

For resistors Q and R, power = V/3 x I/2 = 1/6 x 12W = 2W

LOVELY! Can i ask you for a favour and help me out on this question as well? I thought that it would halve, so be B but apparently I am wrong...
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6. (Original post by nwmyname)
LOVELY! Can i ask you for a favour and help me out on this question as well? I thought that it would halve, so be B but apparently I am wrong...
Hi, I did this question for practice and got the answer B, my physics teacher said that the mark scheme is wrong and the answer should actually be B. Other people in my class seemed to get B too. If you put a value in for D, say 100, into wavelength = ax/D, then it should point to the answer being B.
7. (Original post by Abeh)
Hi, I did this question for practice and got the answer B, my physics teacher said that the mark scheme is wrong and the answer should actually be B. Other people in my class seemed to get B too. If you put a value in for D, say 100, into wavelength = ax/D, then it should point to the answer being B.
D is actually correct.
You know that d sin x = nlambda

so d = nlambda / sin x

but d stays the same, so if the wavelength is increased, so must the angle to keep the same value of d.

increasing the angle increases the spacing.

I understand where you are coming from, but that equation is not for this question... tut tut tut ocr these days.

does that help?

Good luck to you tmrw too!
8. (Original post by nwmyname)
D is actually correct.
You know that d sin x = nlambda

so d = nlambda / sin x

but d stays the same, so if the wavelength is increased, so must the angle to keep the same value of d.

increasing the angle increases the spacing.

I understand where you are coming from, but that equation is not for this question... tut tut tut ocr these days.

does that help?

Good luck to you tmrw too!
Haha now I'm confused, I had a look back at it and I put D in the first place, then the mark scheme said D and then my teacher said B! What equation is for this question then? and why are you talking about changing the angle? the angle doesnt change?

The value of D is the distance from sources to the screen, so this stays the same no matter what the angle?
9. (Original post by Abeh)
Haha now I'm confused, I had a look back at it and I put D in the first place, then the mark scheme said D and then my teacher said B! What equation is for this question then? and why are you talking about changing the angle? the angle doesnt change?

The value of D is the distance from sources to the screen, so this stays the same no matter what the angle?
wait lmao i read it in reverse

you are right.

it only works in the opposite direction.

damn this exam board ergh.
10. (Original post by nwmyname)
wait lmao i read it in reverse

you are right.

it only works in the opposite direction.

damn this exam board ergh.
what do you mean? what only works in the opposite direction? good luck tomorrow also!
11. (Original post by Abeh)
what do you mean? what only works in the opposite direction? good luck tomorrow also!
increasing wavelength increases the angle so the fringe spacing increases.

haha I know this is confusing.

watch.

d sinx = n * lambda

so d = n * lambda / sin x

but d STAYS THE SAME. YOU CANNOT CHANGE THE FRINGE SPACING

So, if i increase lambda, to keep d the same, I must also increase the sin x.

increasing sin x increases the angle (look at the sine graph)

so fringe spacing increases
12. (Original post by nwmyname)
HIII
HOW DO I SOLVE THIS??

I thought that the power dissipated in the parallel circuits is 12 - 4 = 8W

But I did not get the correct answer! Help!!
what exam board is this?
13. (Original post by voltz)
what exam board is this?
ocr
14. (Original post by nwmyname)
ocr
Im also doing OCR but dont understand this...can you explain this by any chance in simple terms?
15. (Original post by nwmyname)
increasing wavelength increases the angle so the fringe spacing increases.

haha I know this is confusing.

watch.

d sinx = n * lambda

so d = n * lambda / sin x

but d STAYS THE SAME. YOU CANNOT CHANGE THE FRINGE SPACING

So, if i increase lambda, to keep d the same, I must also increase the sin x.

increasing sin x increases the angle (look at the sine graph)

so fringe spacing increases
Isn't that equation to do with diffraction gratings? we aren't tested on that until A2
16. (Original post by uberteknik)
If all resistors are the same value, then the circuit forms a potential divider of (R + (R || R) = R + R/2.

The p.d. developed across resistor P is 2V/3 whilst the p.d. across the parallel combination is V/3

The current (I) flowing through P is also split between Q and R. (I splits into I/2 + I/2)

Power = VI = 12W

For resistor P, power = 2V/3 x I = 2/3 x 12W = 8W

For resistors Q and R, power = V/3 x I/2 = 1/6 x 12W = 2W each.
This is a really stupid question but I dont get how you worked out that P is 8W, the way I see it, it has I current whereas the other two have I/2? Is this not correct?
17. (Original post by voltz)
This is a really stupid question but I dont get how you worked out that P is 8W, the way I see it, it has I current whereas the other two have I/2? Is this not correct?
the ratio of resistance is 2 : 1 so one will have 8w and the other 4w
18. uberteknik Abeh here, in this question, why is x positive but not negative?
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19. (Original post by voltz)
This is a really stupid question but I dont get how you worked out that P is 8W, the way I see it, it has I current whereas the other two have I/2? Is this not correct?
Yes that is part of it.

The other part is the p.d. developed across resistor P and the parallel resistors Q and R.

The p.d. across P is 2V/3 whilst the p.d. across Q and R is V/3.

Power = V x I = 12W

i.e. Power = 2VI/3 = 12 x 2/3 for resistor P

And

Power = V/3 x I/2 = VI/6 = 12/6 for each of the resistors Q and R.
20. (Original post by uberteknik)
Yes that is part of it.

The other part is the p.d. developed across resistor P and the parallel resistors Q and R.

The p.d. across P is 2V/3 whilst the p.d. across Q and R is V/3.

Power = V x I = 12W

i.e. Power = 2VI/3 = 12 x 2/3 for resistor P

And

Power = V/3 x I/2 = VI/6 = 12/6 for each of the resistors Q and R.
So essentially we are just finding the ratio of voltage in the resistors and using that? I managed t get the ration of 2/3 by assigning the same resistance to each and finding the total resistance. Then, worked out ratio of P and multiplied by 12 to get 8

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