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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 Uni12345678)
Thanks, cud u please explain why with reference to series and parallel?
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 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.
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 Uni12345678)
But why does bulb Y get dimmer??
But why does bulb Y get dimmer??
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#10
(Original post by Uni12345678)
But why does bulb Y get dimmer??
But why does bulb Y get dimmer??
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(Original post by RogerOxon)
It has less voltage across it, as the other bulbs 'take' some of the battery's voltage.
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 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
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 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)
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 Uni12345678)
although I must point out I think it should be I^2 *R not IR- but I get the idea!!)
although I must point out I think it should be I^2 *R not IR- but I get the idea!!)
(Original post by Uni12345678)
Thank you very much for your help- even at the late hour hahha!
Thank you very much for your help- even at the late hour hahha!
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