# Physics question on BMAT Past paper

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Wasn't sure how to work the answer out for this one... I do a level physics so this is slightly alarming! But any help would be much appreciated and hopefully I'll understand any explanations!

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Yes I do, I just don't get why that is the correct answer ( it is B)

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#4

Its because there would be more current in bulb X as the switch is closed and less current in Y so Y would be dimmer and X would be brighter

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Thanks, cud u please explain why with reference to series and parallel?

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#6

(Original post by

Thanks, cud u please explain why with reference to series and parallel?

**Uni12345678**)Thanks, cud u please explain why with reference to series and parallel?

More current flows through a path with less resistence.

Initiallly, most of the current will pass Y, most of it will go down towards switch Q so less current is passing through bulb X. Most of the current will bypass the two bulbs at the bottom and pass through switch Q because that path offers less resistance (it's only a wire).

So because the current bypasses bulb X, it won't be so bright.

Now, when you close switch P and open switch Q, the path with the least resistence after Y is through switch P. Most current after switch P will now pass through bulb X after switch Q. As more current enters X, the bulb is now brighter. The voltage is now shared between X and Y meaning Y will be dimmer.

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#7

With Q closed, bulb X is off - there is no current to it, as there is a zero resistance path around it (Q). Bulb Y receives all the battery's voltage.

With Q open, bulb X receives a current. Bulb Y is dimmer, as there is now a resistance before it, taking some of the battery's voltage.

You don't need to get into resistances to compare bulb brightness for this question.

With Q open, bulb X receives a current. Bulb Y is dimmer, as there is now a resistance before it, taking some of the battery's voltage.

You don't need to get into resistances to compare bulb brightness for this question.

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(Original post by

Think of it in a more qualitative sense (forget equations for now).

More current flows through a path with less resistence.

Initiallly, most of the current will pass Y, most of it will go down towards switch Q so less current is passing through bulb X. Most of the current will bypass the two bulbs at the bottom and pass through switch Q because that path offers less resistance (it's only a wire).

So because the current bypasses bulb X, it won't be so bright.

Now, when you close switch P and open switch Q, the path with the least resistence after Y is through switch P. Most current after switch P will now pass through bulb X after switch Q. As more current enters X, the bulb is now brighter. The voltage is now shared between X and Y meaning Y will be dimmer.

**champ_mc99**)Think of it in a more qualitative sense (forget equations for now).

More current flows through a path with less resistence.

Initiallly, most of the current will pass Y, most of it will go down towards switch Q so less current is passing through bulb X. Most of the current will bypass the two bulbs at the bottom and pass through switch Q because that path offers less resistance (it's only a wire).

So because the current bypasses bulb X, it won't be so bright.

Now, when you close switch P and open switch Q, the path with the least resistence after Y is through switch P. Most current after switch P will now pass through bulb X after switch Q. As more current enters X, the bulb is now brighter. The voltage is now shared between X and Y meaning Y will be dimmer.

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#9

(Original post by

But why does bulb Y get dimmer??

**Uni12345678**)But why does bulb Y get dimmer??

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#10

(Original post by

But why does bulb Y get dimmer??

**Uni12345678**)But why does bulb Y get dimmer??

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(Original post by

It has less voltage across it, as the other bulbs 'take' some of the battery's voltage.

**RogerOxon**)It has less voltage across it, as the other bulbs 'take' some of the battery's voltage.

I did some calculations assuming each bulb had a resistance of 12 ohms. I found that originally, all the bulbs apart from Y had negligible voltage across them and the overall current in the circuit is 0.5 A, so Y has 0.5 A through it. but after the change in switches, X has 4.8v across it and the current in the circuit is now 0.6 A so Y has 0.6 A through it. Therefore I thought the answer was A... what have I done wrong here? its frustrating me so much ahha

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#12

(Original post by

i dont get why it has less voltage across it...

I did some calculations assuming each bulb had a resistance of 12 ohms. I found that originally, all the bulbs apart from Y had negligible voltage across them and the overall current in the circuit is 0.5 A, so Y has 0.5 A through it. but after the change in switches, X has 4.8v across it and the current in the circuit is now 0.6 A so Y has 0.6 A through it. Therefore I thought the answer was A... what have I done wrong here? its frustrating me so much ahha

**Uni12345678**)i dont get why it has less voltage across it...

I did some calculations assuming each bulb had a resistance of 12 ohms. I found that originally, all the bulbs apart from Y had negligible voltage across them and the overall current in the circuit is 0.5 A, so Y has 0.5 A through it. but after the change in switches, X has 4.8v across it and the current in the circuit is now 0.6 A so Y has 0.6 A through it. Therefore I thought the answer was A... what have I done wrong here? its frustrating me so much ahha

With Q open and P closed, we have X in parallel with two series bulbs. That would give a resistance of 1/[1/12 + 1/(12 + 12)] = 8 Ohm (12 in parallel with 24). The total resistance would therefore be that (8) plus Y (12), i.e. 20 Ohm. A current of 0.6 Amp would flow (V/R = 12 / 20 = 0.6).

The voltage drop across X (and the two series bulbs in parallel with it) would be IR = 0.6 * 8 = 4.8V. The voltage drop across Y would be IR = 0.6 * 12 = 7.2 (which equals 12 - 4.8).

With Q closed, Y had a power of P = VI = 12 * 1 = 12W. With Q open, it has P = VI = 7.2* 0.6 = 4.32W, so is less bright.

X previously had zero current and zero voltage, so zero power. It will now have some, so is brighter. (The 0.6A gets split in inverse proportion to the resistance of X and the parallel branch of two bulbs in series - X will get 0.4A and the two series bulbs 0.2A, so that both have the same voltage drop of 4.8V)

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(Original post by

With Q closed, Y has 12V across it. All other bulbs have 0V (as they're in parallel to Q which has zero resistance). A current of 1A would therefore flow (with your 12 Ohm assumption).

With Q open and P closed, we have X in parallel with two series bulbs. That would give a resistance of 1/[1/12 + 1/(12 + 12)] = 8 Ohm (12 in parallel with 24). The total resistance would therefore be that (8) plus Y (12), i.e. 20 Ohm. A current of 0.6 Amp would flow (V/R = 12 / 20 = 0.6).

The voltage drop across X (and the two series bulbs in parallel with it) would be IR = 0.6 * 8 = 4.8V. The voltage drop across Y would be IR = 0.6 * 12 = 7.2 (which equals 12 - 4.8).

With Q closed, Y had a power of P = IR = 1 * 12 = 12W. With Q open, it has P = IR = 0.6 * 12 = 7.2W, so is less bright.

X previously had zero current and zero voltage, so zero power. It will now have some, so is brighter. (The 0.6A gets split in inverse proportion to the resistance of X and the parallel branch of two bulbs in series - X will get 0.4A and the two series bulbs 0.2A, so that both have the same voltage drop of 4.8V)

**RogerOxon**)With Q closed, Y has 12V across it. All other bulbs have 0V (as they're in parallel to Q which has zero resistance). A current of 1A would therefore flow (with your 12 Ohm assumption).

With Q open and P closed, we have X in parallel with two series bulbs. That would give a resistance of 1/[1/12 + 1/(12 + 12)] = 8 Ohm (12 in parallel with 24). The total resistance would therefore be that (8) plus Y (12), i.e. 20 Ohm. A current of 0.6 Amp would flow (V/R = 12 / 20 = 0.6).

The voltage drop across X (and the two series bulbs in parallel with it) would be IR = 0.6 * 8 = 4.8V. The voltage drop across Y would be IR = 0.6 * 12 = 7.2 (which equals 12 - 4.8).

With Q closed, Y had a power of P = IR = 1 * 12 = 12W. With Q open, it has P = IR = 0.6 * 12 = 7.2W, so is less bright.

X previously had zero current and zero voltage, so zero power. It will now have some, so is brighter. (The 0.6A gets split in inverse proportion to the resistance of X and the parallel branch of two bulbs in series - X will get 0.4A and the two series bulbs 0.2A, so that both have the same voltage drop of 4.8V)

Thank you very much for your help- even at the late hour hahha!

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#14

(Original post by

although I must point out I think it should be I^2 *R not IR- but I get the idea!!)

**Uni12345678**)although I must point out I think it should be I^2 *R not IR- but I get the idea!!)

(Original post by

Thank you very much for your help- even at the late hour hahha!

**Uni12345678**)Thank you very much for your help- even at the late hour hahha!

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X

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