Kirchhoffs second law
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Original post by Zestful
0.033A !!
Are you sure?
Is this the answer given?
What is the current (in the 4V cell) in the upper loop?
What is the current (in the 10V cell) in the lower loop?
The current at the point X asked for in the question is the vector sum of these two. I.E. it is the one plus or minus the other depending on the relative directions.
(edited 11 years ago)
Original post by Stonebridge
Are you sure?
Is this the answer given?
What is the current (in the 4V cell) in the upper loop?
What is the current (in the 10V cell) in the lower loop?
The current at the point X asked for in the question is the vector sum of these two. I.E. it is the one plus or minus the other depending on the relative directions.
Is this the answer given?
What is the current (in the 4V cell) in the upper loop?
What is the current (in the 10V cell) in the lower loop?
The current at the point X asked for in the question is the vector sum of these two. I.E. it is the one plus or minus the other depending on the relative directions.
I'm not positive, hope it's 0.033A or else I'm confused
I've not done this kind of question for a few years so wanted to check against the actual answer before confusing anyone further...I made an assumption about the direction of the current, then evaluated it to see whether it was positive (assumption correct), or negative (assumption incorrect - hence opposite direction is true).
I calculated the value of the current through the 10V cell to be 0.1333V.
What did you get?
I never gave my direction in my answer - I'm saying left to right (clockwise).
-edit- I'm full of typos today. I meant 10V cell, I didn't do the current in the 4V cell as it wasn't necessary how I solved it but it would be exactly 0.1A
(edited 11 years ago)
Original post by Zestful
I'm not positive, hope it's 0.033A or else I'm confused I've not done this kind of question for a few years so wanted to check against the actual answer before confusing anyone further...
I made an assumption about the direction of the current, then evaluated it to see whether it was positive (assumption correct), or negative (assumption incorrect - hence opposite direction is true).
I calculated the value of the current through the 10V cell to be 0.1333V.
What did you get?
I never gave my direction in my answer - I'm saying left to right (clockwise).
-edit- I'm full of typos today. I meant 10V cell, I didn't do the current in the 4V cell as it wasn't necessary how I solved it but it would be exactly 0.1A
I've not done this kind of question for a few years so wanted to check against the actual answer before confusing anyone further...I made an assumption about the direction of the current, then evaluated it to see whether it was positive (assumption correct), or negative (assumption incorrect - hence opposite direction is true).
I calculated the value of the current through the 10V cell to be 0.1333V.
What did you get?
I never gave my direction in my answer - I'm saying left to right (clockwise).
-edit- I'm full of typos today. I meant 10V cell, I didn't do the current in the 4V cell as it wasn't necessary how I solved it but it would be exactly 0.1A
Just checking.
Well, I've just had a go at this and I get 0.033A
The current in the upper loop is 0.166A clockwise and in the lower loop 0.133A anticlockwise
At X it's the difference between these which is 0.033A left to right. So we agree at least.
(edited 11 years ago)
Original post by Stonebridge
Just checking.
Well, I've just had a go at this and I get 0.033A
The current in the upper loop is 0.166A clockwise and in the lower loop 0.133A anticlockwise
At X it's the difference between these which is 0.033A left to right. So we agree at least.
Well, I've just had a go at this and I get 0.033A
The current in the upper loop is 0.166A clockwise and in the lower loop 0.133A anticlockwise
At X it's the difference between these which is 0.033A left to right. So we agree at least.
I now also agree with your 0.166A for the upper loop. That sounds good to me
Original post by Stonebridge
Just checking.
Well, I've just had a go at this and I get 0.033A
The current in the upper loop is 0.166A clockwise and in the lower loop 0.133A anticlockwise
At X it's the difference between these which is 0.033A left to right. So we agree at least.
Well, I've just had a go at this and I get 0.033A
The current in the upper loop is 0.166A clockwise and in the lower loop 0.133A anticlockwise
At X it's the difference between these which is 0.033A left to right. So we agree at least.
I wonder if you can describe how you got these currents. I know they are correct, but I'm interested to know if there is a method more straightforward than the one I outlined earlier. Thanks!
Original post by Pangol
I wonder if you can describe how you got these currents. I know they are correct, but I'm interested to know if there is a method more straightforward than the one I outlined earlier. Thanks!
I have seen a number of ways of doing these over the years. I was taught different methods by different teachers/lecturers. Whichever way will always involve simultaneous equations. It should be one equation for each closed loop. And one current circulating in each loop. The way I do them may not be the same as others have been taught, but here goes.
Let current in upper loop be i1 and in lower loop i2
Take the directions as shown in the positive direction of the emfs.
Then the current at X will be i1 - i2
This is how I do these. It may differ from how others have been taught.
Original post by Pangol
I think that this is logically equivelant to my method, but is a bit quicker. I'll use that from now on!
There is a link below to an on-line circuit editor and calculator I found, but to be quite honest, in the time it took me to figure out how to get it to work I could have done the calculation by hand several times.
Still, if you are prepared to persevere it's a useful tool for checking your circuit calculations - simple or more complex - AC or DC.
https://www.circuitlab.com/editor/
Here's a screenshot of how the solution looked with the answer ringed in red.
(edited 11 years ago)
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