# GCSE physic help

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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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#2

(Original post by

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

**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

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#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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#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'

'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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(Original post by

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.

**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.

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(Original post by

Do you understand why the resistance decreases?

**Mathematics Man**)Do you understand why the resistance decreases?

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#7

**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

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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(Original post by

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,

The

Hopefully that made sense!

**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!

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#9

(Original post by

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

**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

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(Original post by

No current flows through resistor when switch is closed, the whole route is bypassed due to the switch.

**Mehru1214**)No current flows through resistor when switch is closed, the whole route is bypassed due to the switch.

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#11

(Original post by

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?

**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?

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(Original post by

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.

**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.

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**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.

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#14

(Original post by

Why is it in a gcse paper then

**Defence11**)Why is it in a gcse paper then

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#15

**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

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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Wh

Why does the current prefer to take the path with no resistance? In my cgp book it doesn’t say that anywhere

(Original post by

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.

**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.

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(Original post by

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.

**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.

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#18

(Original post by

Wh

Why does the current prefer to take the path with no resistance? In my cgp book it doesn’t say that anywhere

**Defence11**)Wh

Why does the current prefer to take the path with no resistance? In my cgp book it doesn’t say that anywhere

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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(Original post by

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!

**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!

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#20

(Original post by

Can’t it take both As in this example it flows through the voltmeter and lamp

**Defence11**)Can’t it take both As in this example it flows through the voltmeter and lamp

**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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