- #1
queenspublic
- 59
- 0
Finding Unknowns in Circuits: Using Loop Rules and Substitution
- Thread starter queenspublic
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Homework Help Overview
The discussion revolves around solving a circuit problem using Kirchhoff's loop rules and substitution methods to find unknown currents. Participants are exploring the application of circuit analysis techniques in the context of electrical circuits.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- Participants inquire about the methods used to approach the problem, specifically whether Kirchhoff's rules were applied. There is a suggestion to first apply loop rules and then use substitution for current relationships.
Discussion Status
The discussion is ongoing, with participants providing hints and questioning assumptions about current relationships in the circuit. There is no clear consensus yet, but some guidance on the approach has been offered.
Contextual Notes
There is a mention of potential issues with the assumption that two currents are equal, indicating that participants are critically evaluating the setup and relationships within the circuit.
- #2
rl.bhat
Homework Helper
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Show your calculations.
- #3
ordirules
- 25
- 0
I'm not sure, what have you used to solve the problem?
Did you use Kirchoff's loop rules and the junction rule?
Did you use Kirchoff's loop rules and the junction rule?
- #4
queenspublic
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rl.bhat said:
Hello?
Last edited:
- #5
ordirules
- 25
- 0
I don't know what you did but the assumption that i2 = i3 may not be correct.
I would do the loop rules first (the Vb -Va part you wrote) then at the last moment do a substitution i3 = i1 - i2 (assuming the same directions you chose for your current)
you should then end up with two equations and two unknowns. (the potential diff by going through left loop and the right loop.
From the looks of this circuit, this should have some annoying algebra, but I don't believe there's a better way.
So as another hint in case you still can't get it:
Good luck!
I would do the loop rules first (the Vb -Va part you wrote) then at the last moment do a substitution i3 = i1 - i2 (assuming the same directions you chose for your current)
you should then end up with two equations and two unknowns. (the potential diff by going through left loop and the right loop.
From the looks of this circuit, this should have some annoying algebra, but I don't believe there's a better way.
So as another hint in case you still can't get it:
The loop rule is that the potential difference by going on in a loop is zero.
Going around the loop to the left, assuming the current i1 is going clockwise on the left part of loop and i3 is going up for the middle part of the circuit:
E1 - i1R1 +i3R2 -E2 -i1R1 = 0
You can get the equation for the second loop the same way.
Then you can use i1 -i2 = i3 (junction rule) to get three equations, three unknowns (i1, i2, i3)
It is important you understand how the equation resulting from the loop rule is obtained. When you move along the current across a resistor, you lose I * R in potential difference, and when you move opposite to the current, you gain I * R in current. For batteries (Emf), it's the opposite, you gain potential difference when you're moving with the battery's arrow, and negative when you're moving against (current does not matter here).
Please note this is just a trick and what is really happening in the circuit is not exact.
Going around the loop to the left, assuming the current i1 is going clockwise on the left part of loop and i3 is going up for the middle part of the circuit:
E1 - i1R1 +i3R2 -E2 -i1R1 = 0
You can get the equation for the second loop the same way.
Then you can use i1 -i2 = i3 (junction rule) to get three equations, three unknowns (i1, i2, i3)
It is important you understand how the equation resulting from the loop rule is obtained. When you move along the current across a resistor, you lose I * R in potential difference, and when you move opposite to the current, you gain I * R in current. For batteries (Emf), it's the opposite, you gain potential difference when you're moving with the battery's arrow, and negative when you're moving against (current does not matter here).
Please note this is just a trick and what is really happening in the circuit is not exact.
Good luck!
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