The discussion revolves around analyzing a circuit with a battery of emf 12.7 V and three resistors with values of 2000 Ω, 3000 Ω, and 4000 Ω. The main question posed is about determining the current through one of the resistors, R2.
Several participants have provided equations and expressed their reasoning regarding the currents in the circuit. There is an ongoing exploration of symmetry and its effects on current distribution, with some participants questioning the assumptions made about current behavior in the circuit.
Some participants note the potential for simpler methods to solve the problem, while others emphasize the importance of labeling currents effectively to reduce complexity in the algebra involved. There is also mention of a failed image upload that may have contained relevant circuit information.
There may be simpler ways to solve this, but just go ahead and write the KCL equations and try it that way. Please show your work...Sho Kano said:
berkeman said:There may be simpler ways to solve this, but just go ahead and write the KCL equations and try it that way. Please show your work...
Did you read through the thread that SammyS linked to in post #4?Sho Kano said:
It is usually preferable to not involve more current variables than are needed, to minimise the amount of algebra it generates. Once you label 3 currents, the others are already defined in terms of those 3 so they can be marked on the schematic accordingly.Sho Kano said:
So in the end, I only need 5 equations?NascentOxygen said:
Yes, the currents through R2 are the same and so are the ones for R1 because reversing the polarities would give an identical result.berkeman said:Did you read through the thread that SammyS linked to in post #4?
Symmetry says so. If you can flip or rotate a circuit about one or more axes of symmetry and the result is identical to the original circuit, then the resulting circuit must behave identically to the original and the symmetric pairs of circuit elements must display the same behavior (current, potential difference).Sho Kano said:
Are you trying to say that the current into point A for the non-rotated circuit is the same as the current flowing into point D for the rotated circuit?gneill said:
The magnitude is certainly the same. Change the polarity of the voltage source and the signs of the currents will agree too.Sho Kano said:
Right, charge is conserved. If all the resistors were different, would the current still be the same?gneill said:
Which current? Certainly the current entering (or leaving) node A via the battery is the same as the current leaving (or entering) node D via the battery.Sho Kano said:
I see, so only currents through same resistors will have the same magnitude?gneill said:
Rephrase that as "The currents through the symmetrically paired resistors will be the same" and you would be right. Don't overspecify by saying "only". Some circuits may have resistors of different values that happen to have the same magnitude of current.Sho Kano said:
The currents through the symmetrically paired resistors will be the same.gneill said:
Will there still be the same symmetry?Sho Kano said:
In the un-rotated figure: top left resistor is R1 top right is R2. Bottom left is R3, bottom right is R4. The bridge resistor is R5.SammyS said:
Then the currents will be affected. No more assumptions about one current being the same as another.Sho Kano said: