Kirchhoff's laws for voltage in parallel circuits

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cassielle
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#1
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#1
I think I understand his current law fine, but how do you apply his voltage law in parallel circuits? For example, in this circuit

Image

If you know the battery is 9V, then you could look at the "top loop" and figure out that the bulb also has 9V across it
But if you look at the "bottom loop", since voltage is the same everywhere, and there are two bulbs in that loop, wouldn't it add up to 18V? Since they're both taking up voltage, isn't the sum of the potential differences in that loop not equal to 0?
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Stonebridge
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(Original post by cassielle)
I think I understand his current law fine, but how do you apply his voltage law in parallel circuits? For example, in this circuit

Image

If you know the battery is 9V, then you could look at the "top loop" and figure out that the bulb also has 9V across it
But if you look at the "bottom loop", since voltage is the same everywhere, and there are two bulbs in that loop, wouldn't it add up to 18V? Since they're both taking up voltage, isn't the sum of the potential differences in that loop not equal to 0?
Well firstly, you don't need Kirchhoff's Law's to actually answer a question like this. There is only one source of emf, the cell, and you just have 2 resistors in parallel in the circuit.
The voltage law says that the (vector) sum of the pd's in a closed loop is zero. But the important point is you have to take the direction of the pd into account in the loop.
Is the pd pointing clockwise or anticlockwise within the closed loop?
Assuming no internal resistance.
In the lower part of this circuit, the pd across the lower bulb is in the clockwise direction in the lower loop. (The Plus on the left)
The pd across the upper bulb is pointing in an anticlockwise direction in the lower loop.
The 'sum' of these two is zero as they are in opposite directions in the lower loop.
This is correct as there is no source of emf inside the lower loop.
Last edited by Stonebridge; 4 weeks ago
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Ethanity007
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Bottom loop. If u take an anticlockwise loop, the top bulb is + V while the bottom bulb is - V. Add up to zero. From the bottom loop can only deduce the pd across each bulb is numerically equal to each other.

But since from the top loop u deduced the top bulb is 9 V, That will suggest the pd across both light bulbs are 9 V.
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cassielle
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(Original post by Ethanity007)
Bottom loop. If u take an anticlockwise loop, the top bulb is + V while the bottom bulb is - V. Add up to zero. From the bottom loop can only deduce the pd across each bulb is numerically equal to each other.

But since from the top loop u deduced the top bulb is 9 V, That will suggest the pd across both light bulbs are 9 V.
in an anticlockwise loop, why is the top bulb positive? aren't they all using up voltage?
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Ethanity007
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The right side of the bulb is negative and left side positive (same as the batt). Going anti clockwise, top bulb you are “moving” from negative to positive and thus its a rise in potential . Bottom u r “moving” from positive to negative and thus a drop in potential
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cassielle
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(Original post by Ethanity007)
The right side of the bulb is negative and left side positive (same as the batt). Going anti clockwise, top bulb you are “moving” from negative to positive and thus its a rise in potential . Bottom u r “moving” from positive to negative and thus a drop in potential
oh my god I finally get it
thanks so much!
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cassielle
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(Original post by Stonebridge)
Well firstly, you don't need Kirchhoff's Law's to actually answer a question like this. There is only one source of emf, the cell, and you just have 2 resistors in parallel in the circuit.
The voltage law says that the (vector) sum of the pd's in a closed loop is zero. But the important point is you have to take the direction of the pd into account in the loop.
Is the pd pointing clockwise or anticlockwise within the closed loop?
Assuming no internal resistance.
In the lower part of this circuit, the pd across the lower bulb is in the clockwise direction in the lower loop. (The Plus on the left)
The pd across the upper bulb is pointing in an anticlockwise direction in the lower loop.
The 'sum' of these two is zero as they are in opposite directions in the lower loop.
This is correct as there is no source of emf inside the lower loop.
(I know I didn't need to use Kirchhoff's law for that question, I was just using a random circuit to illustrate ^^)
I get it now! I didn't think about taking into account the vector sum of the pds lol
Thanks!
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Ethanity007
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Glad i could be of some help! Anw, it may help to know that potential is a scalar quantity and thus algebraic sum may be a better choice when u are stating KVL
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Stonebridge
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(Original post by Ethanity007)
Glad i could be of some help! Anw, it may help to know that potential is a scalar quantity and thus algebraic sum may be a better choice when u are stating KVL
It's not 'potential' it's potential difference here.
And as such, in a circuit loop, has an implied direction.
The use of the word 'vector' here is simply saying that you need to take this direction into account.
Thanks for helping to answer this question.
Last edited by Stonebridge; 4 weeks ago
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Ethanity007
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(Original post by Stonebridge)
It's not 'potential' it's potential difference here.
And as such, in a circuit loop, has an implied direction.
The use of the word 'vector' here is simply saying that you need to take this direction into account.
Thanks for helping to answer this question.
Thanks for clarification. May i just check, so potential difference is a vector quantity?
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Stonebridge
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(Original post by Ethanity007)
Thanks for clarification. May i just check, so potential difference is a vector quantity?
Not in the same sense as normally implied. No.
The point with pd in a loop, as we have here, is that it can be clockwise or anticlockwise, and as such has 'direction' for the purpose of addition in Kirchhoff's law.
It's exactly the same consideration as 2 cells with emf 1.5 V placed in series. What is the total emf here? 3V or 0V?
You have to take into account the 'direction' of the cells (and therefore of their emf) in order to 'add' them together.
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Ethanity007
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(Original post by Stonebridge)
Not in the same sense as normally implied. No.
The point with pd in a loop, as we have here, is that it can be clockwise or anticlockwise, and as such has 'direction' for the purpose of addition in Kirchhoff's law.
It's exactly the same consideration as 2 cells with emf 1.5 V placed in series. What is the total emf here? 3V or 0V?
You have to take into account the 'direction' of the cells (and therefore of their emf) in order to 'add' them together.
Thanks! One last qn, this definition can be found in which exam board learning objectives/syllabus?
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Stonebridge
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(Original post by Ethanity007)
Thanks! One last qn, this definition can be found in which exam board learning objectives/syllabus?
It's not a definition. It is just the way in which you solve questions using Kirchhoff's laws.
In my experience there are slightly different ways of setting up the equations to solve a circuit using them.
Any text book will give an example of how to do this.
I don't think exam boards care which way you set them up so long as what you do is correct/consistent and you get the correct answer.
Of course, you need to know Kirchhoff's Laws if they are in your syllabus.. The exam board syllabus will state, then, what you need to know.
Btw
It's a long time since I did my A-Levels and Kirchhoff wasn't in the syllabus then, though a lot of other stuff was.
It was first year uni when I did them.
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