GCSE physic help

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Thread starter 1 year ago
#1
Hello guys.. It would be great if u helped me. I don’t understand this physics exam question. http://filestore.aqa.org.uk/resource...84631H-SQP.PDF That is the link for the paper. The question is 7.3. I do not understand what the mark scheme says what why it is the answer. It would be much appreciated if u answered. Thanks http://filestore.aqa.org.uk/resource...84631H-SMS.PDF That is link to mark scheme. Thanks
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1 year ago
#2
(Original post by Defence11)
Hello guys.. It would be great if u helped me. I don’t understand this physics exam question. http://filestore.aqa.org.uk/resource...84631H-SQP.PDF That is the link for the paper. The question is 7.3. I do not understand what the mark scheme says what why it is the answer. It would be much appreciated if u answered. Thanks http://filestore.aqa.org.uk/resource...84631H-SMS.PDF That is link to mark scheme. Thanks
Do you understand why the resistance decreases?
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1 year ago
#3
When the switch is closed, the current has a new path of zero resistance that's through the switch and it will travel on that path instead of the 60ohm, so the total resistance goes down as the 60ohm resistor is no longer in play and because V=IR, I=V/R, so current increases when resistances decreases.
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1 year ago
#4
because you closed the switch the current increase as it can flow through there now and therefore the total resistance decreases as it is a parallel circuit.
'As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit decreases and the total current of the circuit increases. Adding more resistors in parallel is equivalent to providing more branches through which charge can flow'
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Thread starter 1 year ago
#5
(Original post by Radioactivedecay)
When the switch is closed, the current has a new path of zero resistance that's through the switch and it will travel on that path instead of the 60ohm, so the total resistance goes down as the 60ohm resistor is no longer in play and because V=IR, I=V/R, so current increases when resistances decreases.
But can’t the current still flow through the 6ohm resistor, just not as much as before as there is a new zero resistance loop. Is that right?
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Thread starter 1 year ago
#6
(Original post by Mathematics Man)
Do you understand why the resistance decreases?
No I don’t get that bit
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1 year ago
#7
(Original post by Defence11)
Hello guys.. It would be great if u helped me. I don’t understand this physics exam question. http://filestore.aqa.org.uk/resource...84631H-SQP.PDF That is the link for the paper. The question is 7.3. I do not understand what the mark scheme says what why it is the answer. It would be much appreciated if u answered. Thanks http://filestore.aqa.org.uk/resource...84631H-SMS.PDF That is link to mark scheme. Thanks
When the switch is closed, the 60 ohm resistor is short-circuited, so current doesn't flow through it, only through the switch's wire. So the total resistance decreases due to one resistor being out of action.

By Ohm's Law, V = IR.

The voltage remains constant, and dividing it by a smaller value of R results in the current, I, increasing.

Hopefully that made sense!
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Thread starter 1 year ago
#8
(Original post by Mehru1214)
When the switch is closed, the 60 ohm resistor is short-circuited, so current doesn't flow through it, only through the switch's wire. So the total resistance decreases due to one resistor being out of action.

By Ohm's Law, V = IR.

The voltage remains constant, and dividing it by a smaller value of R results in the current, I, increasing.

Hopefully that made sense!
Last bit does but the first bit doesn’t. I still would have thought current would flow through the 60ohm resistor but just not as much as before. Am I right. Or is there no current flowing through the 60ohm resistor when the switch is closed
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1 year ago
#9
(Original post by Defence11)
Last bit does but the first bit doesn’t. I still would have thought current would flow through the 60ohm resistor but just not as much as before. Am I right. Or is there no current flowing through the 60ohm resistor when the switch is closed
No current flows through resistor when switch is closed, the whole route is bypassed due to the switch.
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Thread starter 1 year ago
#10
(Original post by Mehru1214)
No current flows through resistor when switch is closed, the whole route is bypassed due to the switch.
But that Doesnt make sense cause normally in a circuit let’s say a voltmeter is connected in parallel to a lamp. Current flows through the lamp and voltmeter right?
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1 year ago
#11
(Original post by Defence11)
But that Doesnt make sense cause normally in a circuit let’s say a voltmeter is connected in parallel to a lamp. Current flows through the lamp and voltmeter right?
This is a different case. When the current has of choice of going through one of two paths, one with zero resistance and another with non zero resistance, all of it will always go through the path with zero resistance, that's just a property of current. It can be shown mathematically, but it's not need for GCSE level.
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Thread starter 1 year ago
#12
(Original post by Radioactivedecay)
This is a different case. When the current has of choice of going through one of two paths, one with zero resistance and another with non zero resistance, all of it will always go through the path with zero resistance, that's just a property of current. It can be shown mathematically, but it's not need for GCSE level.
As this was in the GCSE paper how was I supposed to know that current will always pick the zero resistance path. Now I know thanks to you but seriously don’t u think that was hard because not many students would pick up on that theory
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Thread starter 1 year ago
#13
(Original post by Radioactivedecay)
This is a different case. When the current has of choice of going through one of two paths, one with zero resistance and another with non zero resistance, all of it will always go through the path with zero resistance, that's just a property of current. It can be shown mathematically, but it's not need for GCSE level.
Why is it in a gcse paper then
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1 year ago
#14
(Original post by Defence11)
Why is it in a gcse paper then
It's GCSE level knowledge, its just that there is a mathematical way to show this also which you don't need (for this level). For GCSE you just need to be able to explain it with words, that's what it means.
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1 year ago
#15
(Original post by Defence11)
Hello guys.. It would be great if u helped me. I don’t understand this physics exam question. http://filestore.aqa.org.uk/resource...84631H-SQP.PDF That is the link for the paper. The question is 7.3. I do not understand what the mark scheme says what why it is the answer. It would be much appreciated if u answered. Thanks http://filestore.aqa.org.uk/resource...84631H-SMS.PDF That is link to mark scheme. Thanks
As soon as the switch is closed, the current won't flow through the 60 ohm resistor anymore because there is a new path of no resistance which the current prefers to take.

In a series circuit (which this is), the total resistance of the circuit is the sum of the individual resistances of the resistors. So this means as soon as the switch is closed, the 60 ohm resistor can be excluded and so your new total resistance is smaller by 60 ohms.

Using Ohm's law V=IR and hence I=V/R . We know V is fixed as the supplied voltage is 12 V but because the resistance is now smaller this means the current will now increase.
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Thread starter 1 year ago
#16
Wh
(Original post by Anonymouspsych)
As soon as the switch is closed, the current won't flow through the 60 ohm resistor anymore because there is a new path of no resistance which the current prefers to take.

In a series circuit (which this is), the total resistance of the circuit is the sum of the individual resistances of the resistors. So this means as soon as the switch is closed, the 60 ohm resistor can be excluded and so your new total resistance is smaller by 60 ohms.

Using Ohm's law V=IR and hence I=V/R . We know V is fixed as the supplied voltage is 12 V but because the resistance is now smaller this means the current will now increase.
Why does the current prefer to take the path with no resistance? In my cgp book it doesn’t say that anywhere
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Thread starter 1 year ago
#17
(Original post by SlashaRussia)
It's GCSE level knowledge, its just that there is a mathematical way to show this also which you don't need (for this level). For GCSE you just need to be able to explain it with words, that's what it means.
Why does the current prefer to take the path with no resistance? In my cgp book it doesn’t say that anywhere
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1 year ago
#18
(Original post by Defence11)
Wh

Why does the current prefer to take the path with no resistance? In my cgp book it doesn’t say that anywhere
The current cannot take both paths as there will be a potential difference across the 60 ohm resistor but no potential difference across the 0 resistance path. This means the when the electrons meet at the 2nd junction they wont have the same amount of energy and this isn't allowed.

Hence it must take either take the 60ohm resistance path or the no resistance path. It just so happens to be the electrons always take the path of 0 resistance. There are mathematical explanations for this but you don't need to worry about that!
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Thread starter 1 year ago
#19
(Original post by Anonymouspsych)
The current cannot take both paths as there will be a potential difference across the 60 ohm resistor but no potential difference across the 0 resistance path. This means the when the electrons meet at the 2nd junction they wont have the same amount of energy and this isn't allowed.

Hence it must take either take the 60ohm resistance path or the no resistance path. It just so happens to be the electrons always take the path of 0 resistance. There are mathematical explanations for this but you don't need to worry about that!
Can’t it take both As in this example it flows through the voltmeter and lamp
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1 year ago
#20
(Original post by Defence11)
Can’t it take both As in this example it flows through the voltmeter and lamp
This is a different scenario because there is resistance to both the lamp and voltmeter. So the current will flow through both directions and the potential difference across the lamp and voltmeter will be same.

However an ideal voltmeter had infinite resistance when we deal with problems we assume no current flows through the voltmeter (even thought a tiny amount will).
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