# Help with a circuit question!

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I looked at the answer but have no idea what the reasoning behind it is..someone please help!

Thanks.

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#2

Ok, I believe I know how to do this.

First off applying Kirchoff's laws you know that current is shared in a parallel circuit, so that eliminates A and D as the brightness must INCREASE through L as current isn't being shared anymore, therefore the value for V=IR is bigger meaning the brightness level be bigger as well.

That leaves B and C, I believe the answer is B as the current re-joins at the end of a parallel component, this means before and after regardless of the arrangement the current is the same so the brightness must therefore stay the same.

I might be wrong but I'm fairly confident that is correct.

Hope your revision is going well

First off applying Kirchoff's laws you know that current is shared in a parallel circuit, so that eliminates A and D as the brightness must INCREASE through L as current isn't being shared anymore, therefore the value for V=IR is bigger meaning the brightness level be bigger as well.

That leaves B and C, I believe the answer is B as the current re-joins at the end of a parallel component, this means before and after regardless of the arrangement the current is the same so the brightness must therefore stay the same.

I might be wrong but I'm fairly confident that is correct.

Hope your revision is going well

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#3

(Original post by

I looked at the answer but have no idea what the reasoning behind it is..someone please help!

Thanks.

**Hersh_K**)I looked at the answer but have no idea what the reasoning behind it is..someone please help!

Thanks.

No values are given, however the question does state all resistances are equal.

1) With all lamps working, what can you say about the ratio between the currents through each lamp M and lamps L & N?

2) Given the deduction of 1), what can you now say about the power developed (hence relative brightness) by each of the lamps M and lamps L & N?

3) If lamp N is now open circuit (breaks), what happens to the voltage developed across lamp M and lamp L?

4) Again, what can you say has happened to the power developed across the remaining lamps?

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#4

(Original post by

Ok, I believe I know how to do this.

First off applying Kirchoff's laws you know that current is shared in a parallel circuit, so that eliminates A and D as the brightness must INCREASE through L as current isn't being shared anymore, therefore the value for V=IR is bigger meaning the brightness level be bigger as well.

That leaves B and C, I believe the answer is B as the current re-joins at the end of a parallel component, this means before and after regardless of the arrangement the current is the same so the brightness must therefore stay the same.

I might be wrong but I'm fairly confident that is correct.

Hope your revision is going well

**GammaNick**)Ok, I believe I know how to do this.

First off applying Kirchoff's laws you know that current is shared in a parallel circuit, so that eliminates A and D as the brightness must INCREASE through L as current isn't being shared anymore, therefore the value for V=IR is bigger meaning the brightness level be bigger as well.

That leaves B and C, I believe the answer is B as the current re-joins at the end of a parallel component, this means before and after regardless of the arrangement the current is the same so the brightness must therefore stay the same.

I might be wrong but I'm fairly confident that is correct.

Hope your revision is going well

The first part is OK.

The reasoning in the second part is incorrect. When lamp N blows, the total circuit resistance has changed. That means the total current in the circuit has changed and the p.d.'s across the remaining lamps has also altered.

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**GammaNick**)

Ok, I believe I know how to do this.

First off applying Kirchoff's laws you know that current is shared in a parallel circuit, so that eliminates A and D as the brightness must INCREASE through L as current isn't being shared anymore, therefore the value for V=IR is bigger meaning the brightness level be bigger as well.

That leaves B and C, I believe the answer is B as the current re-joins at the end of a parallel component, this means before and after regardless of the arrangement the current is the same so the brightness must therefore stay the same.

I might be wrong but I'm fairly confident that is correct.

Hope your revision is going well

Revision is going eh, there's just so much to remember and the exam requires a lot of application..

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#6

No problem, electricity is the hardest for me on Unit 1, what exam board are you doing?

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(Original post by

Incorrect.

The first part is OK.

The reasoning in the second part is incorrect. When lamp N blows, the total circuit resistance has changed. That means the total current in the circuit has changed and the p.d.'s across the remaining lamps has also altered.

**uberteknik**)Incorrect.

The first part is OK.

The reasoning in the second part is incorrect. When lamp N blows, the total circuit resistance has changed. That means the total current in the circuit has changed and the p.d.'s across the remaining lamps has also altered.

How does exactly does the total circuit resistance change though - does it increase or decrease, and why?

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(Original post by

No problem, electricity is the hardest for me on Unit 1, what exam board are you doing?

**GammaNick**)No problem, electricity is the hardest for me on Unit 1, what exam board are you doing?

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#9

i would put in some numbers... 12 volts battery, 2 ohms each lamp.

the two in parallel L & N have a combined resistance of 1 ohm... so the whole circuit has 3 ohms... thus 12/3 = 4 amps... this splits in two for the parallel lamps so they get 2 amps each, the other one gets all 4 amps.

with the broken link the total resistance is 2 + 2 = 4 ohms.... current is 12/4 = 3 amps throughout the circuit. thus the right hand lamp M experiences a drop from 4 amsps to 3 amps... it gets dimmer. the other lamp L experiences an increase from 2 amps to 3 amps so gets brighter.

so C

the two in parallel L & N have a combined resistance of 1 ohm... so the whole circuit has 3 ohms... thus 12/3 = 4 amps... this splits in two for the parallel lamps so they get 2 amps each, the other one gets all 4 amps.

with the broken link the total resistance is 2 + 2 = 4 ohms.... current is 12/4 = 3 amps throughout the circuit. thus the right hand lamp M experiences a drop from 4 amsps to 3 amps... it gets dimmer. the other lamp L experiences an increase from 2 amps to 3 amps so gets brighter.

so C

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#10

(Original post by

Ahh, I think I understand it now..

How does exactly does the total circuit resistance change though - does it increase or decrease, and why?

**Hersh_K**)Ahh, I think I understand it now..

How does exactly does the total circuit resistance change though - does it increase or decrease, and why?

The total circuit resistance with all the lamps working must be:

When lamp N blows, the total circuit resistance changes to:

The total circuit resistance has increased.

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#11

I'm doing AQA A, waves is on the other exam and that is ridiculous as well. Is a lot of OCR Phyiscs multiple choice then?

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(Original post by

I'm doing AQA A, waves is on the other exam and that is ridiculous as well. Is a lot of OCR Phyiscs multiple choice then?

**GammaNick**)I'm doing AQA A, waves is on the other exam and that is ridiculous as well. Is a lot of OCR Phyiscs multiple choice then?

I've heard grade boundaries are generally low for this paper, so that's a relief!

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#14

Yeah mine is 70% raw marks for an A so that's good for us

I think it's just people not bothering to revise, most of it isn't particularly taxing (except electricity).

Anyway wish you the best of luck for your exams!

I think it's just people not bothering to revise, most of it isn't particularly taxing (except electricity).

Anyway wish you the best of luck for your exams!

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