# Physics question on BMAT Past paper

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#1
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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4 years ago
#2
Do you have the correct answer?
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#3
Yes I do, I just don't get why that is the correct answer ( it is B)
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4 years ago
#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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#5
Thanks, cud u please explain why with reference to series and parallel?
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4 years ago
#6
(Original post by Uni12345678)
Thanks, cud u please explain why with reference to series and parallel?
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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4 years ago
#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.
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#8
(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.
But why does bulb Y get dimmer??
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4 years ago
#9
(Original post by Uni12345678)
But why does bulb Y get dimmer??
It has less voltage across it, as the other bulbs 'take' some of the battery's voltage.
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4 years ago
#10
(Original post by Uni12345678)
But why does bulb Y get dimmer??
Since the current is now flowing through both bulb X and bulb Y, the voltage between them is shared. Before, bulb Y would get almost all the voltage.
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#11
(Original post by RogerOxon)
It has less voltage across it, as the other bulbs 'take' some of the battery's voltage.
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
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4 years ago
#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
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 = 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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#13
(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)
Got it! I had got the initial current incorrect, thank you very much I understand it much better, the power thing especially helped me understand what's actually happening (although I must point out I think it should be I^2 *R not IR- but I get the idea!!)

Thank you very much for your help- even at the late hour hahha!
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4 years ago
#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!!)
Ooops. You are right.
(Original post by Uni12345678)
Thank you very much for your help- even at the late hour hahha!
Not in California
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#15
(Original post by RogerOxon)
Ooops. You are right.

Not in California
OMG you're in California? LITERALLY my fav place in the world ahaha want to live there some day! Anyway lol I'm weird forgive me
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