Question about parallel circuit- OCR AS physics
*I can't provide images unfortunately, there's something wrong with my tsr, but it's ok because the textbook answers only give the answers in word format rather than a diagrams
A parallel circuit contains three identical lamps connected to a battery with an e.m.f. of 12V. Two lamps are on the first branch and one is on the second.
a) Sketch a circuit for this arrangement.
Already done; answers not shown in the textbook, but I have 6V for each lamp on the first branch (to make 12V in total for the first branch) and 12V on the second branch, I'm pretty sure that's correct
b) If the current of 6A is drawn from the cell, determine the current through each lamp in the circuit.
TEXTBOOK ANSWER:
1 mark: First branch has twice the resistance as the second branch. (Side note, because there's two lamps in the first branch hence the internal resistances add up to twice the second branch with only one lamp thus only one internal resistance, right?)
1 mark: Therefore, the lamps on this branch have have half the current on the second branch through them.
1 mark: First branch has 2A through them. (side note- how? see next line)
1 mark: Second branch has 4A. (Side note- 4 + 2 + 2 = 8A, and 8A isn't equal to the total 6A the textbook says in the question)
MY ANSWER:
So, I understood why first branch (not lamps) has twice the resistance and thus half the current
First branch: 6A x 1/3 = 2A for each branch,
and 2A/2 = 1A for each lamp of the first branch
Second branch: 6A x 2/3 = 4A for the single lamp on the second branch
This makes more sense to me as 1 + 1 + 4 = 6A
Where did I go wrong in my thinking? And is the textbook wrong- 2 +2 +4 isn't 6A? Is both me and the textbook wrong and the answer is something different?
I looked into part c) and the textbook does indeed treat each lamp in the first loop as having 2A (2 + 2 + 4 = 8A for the whole circuit which isn't 6A) so it's not because I misunderstood the language
A parallel circuit contains three identical lamps connected to a battery with an e.m.f. of 12V. Two lamps are on the first branch and one is on the second.
a) Sketch a circuit for this arrangement.
Already done; answers not shown in the textbook, but I have 6V for each lamp on the first branch (to make 12V in total for the first branch) and 12V on the second branch, I'm pretty sure that's correct
b) If the current of 6A is drawn from the cell, determine the current through each lamp in the circuit.
TEXTBOOK ANSWER:
1 mark: First branch has twice the resistance as the second branch. (Side note, because there's two lamps in the first branch hence the internal resistances add up to twice the second branch with only one lamp thus only one internal resistance, right?)
1 mark: Therefore, the lamps on this branch have have half the current on the second branch through them.
1 mark: First branch has 2A through them. (side note- how? see next line)
1 mark: Second branch has 4A. (Side note- 4 + 2 + 2 = 8A, and 8A isn't equal to the total 6A the textbook says in the question)
MY ANSWER:
So, I understood why first branch (not lamps) has twice the resistance and thus half the current
First branch: 6A x 1/3 = 2A for each branch,
and 2A/2 = 1A for each lamp of the first branch
Second branch: 6A x 2/3 = 4A for the single lamp on the second branch
This makes more sense to me as 1 + 1 + 4 = 6A
Where did I go wrong in my thinking? And is the textbook wrong- 2 +2 +4 isn't 6A? Is both me and the textbook wrong and the answer is something different?
I looked into part c) and the textbook does indeed treat each lamp in the first loop as having 2A (2 + 2 + 4 = 8A for the whole circuit which isn't 6A) so it's not because I misunderstood the language
(edited 1 month ago)
"hence the internal resistances add up to twice the second branch with only one lamp thus only one internal resistance, right?" - Yes thats correct
"Second branch has 4A. (Side note- 4 + 2 + 2 = 8A, and 8A isn't equal to the total 6A the textbook says in the question)" - I think you have a misunderstanding about current - in the first branch the current in is 2A, so the current out is 2A - you do not add the current for each lamp.
"2A/2 = 1A for each lamp of the first branch" - no each lamp has 2A since current is same in series.
"4A for the single lamp on the second branch" - correct but not because it is a single lamp, even if there were more lamps in that branch they would all have the same current as each other.
You should watch a video about Kirchhoff's laws to clarify your misunderstanding.
I attached working out to help you as well.
"Second branch has 4A. (Side note- 4 + 2 + 2 = 8A, and 8A isn't equal to the total 6A the textbook says in the question)" - I think you have a misunderstanding about current - in the first branch the current in is 2A, so the current out is 2A - you do not add the current for each lamp.
"2A/2 = 1A for each lamp of the first branch" - no each lamp has 2A since current is same in series.
"4A for the single lamp on the second branch" - correct but not because it is a single lamp, even if there were more lamps in that branch they would all have the same current as each other.
You should watch a video about Kirchhoff's laws to clarify your misunderstanding.
I attached working out to help you as well.
Original post by mosaurlodon
"hence the internal resistances add up to twice the second branch with only one lamp thus only one internal resistance, right?" - Yes thats correct
"Second branch has 4A. (Side note- 4 + 2 + 2 = 8A, and 8A isn't equal to the total 6A the textbook says in the question)" - I think you have a misunderstanding about current - in the first branch the current in is 2A, so the current out is 2A - you do not add the current for each lamp.
"2A/2 = 1A for each lamp of the first branch" - no each lamp has 2A since current is same in series.
"4A for the single lamp on the second branch" - correct but not because it is a single lamp, even if there were more lamps in that branch they would all have the same current as each other.
You should watch a video about Kirchhoff's laws to clarify your misunderstanding.
I attached working out to help you as well.
"Second branch has 4A. (Side note- 4 + 2 + 2 = 8A, and 8A isn't equal to the total 6A the textbook says in the question)" - I think you have a misunderstanding about current - in the first branch the current in is 2A, so the current out is 2A - you do not add the current for each lamp.
"2A/2 = 1A for each lamp of the first branch" - no each lamp has 2A since current is same in series.
"4A for the single lamp on the second branch" - correct but not because it is a single lamp, even if there were more lamps in that branch they would all have the same current as each other.
You should watch a video about Kirchhoff's laws to clarify your misunderstanding.
I attached working out to help you as well.
I basically said the total current is equal to 6A and labelled current through first branch as I_1 and second branch as I_2 - I then used equivalent resistance to find out total resistance of circuit (I recommend watching a video on equivalent resistance as well) and then used V=IR for whole circuit and V=IR for each lamp.
I would say a good understanding of Kirchhoff's laws and equivalent resistance will serve you well as a good foundation for electricity - dont worry about the difficulty at the start it will start to make sense. Hope this helps!
I would say a good understanding of Kirchhoff's laws and equivalent resistance will serve you well as a good foundation for electricity - dont worry about the difficulty at the start it will start to make sense. Hope this helps!
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