supercunundrum theorem

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ninjaman

Joined May 18, 2013
341
I got that wrong it should be 1.6 across the 2000 ohm resistor which is 0.8ma
then 0.0008amps x 500 = 0.4v
then with that voltage across the 5000 ohm resistor I got 0.08ma
I don't know if any of this is right
 

JoeJester

Joined Apr 26, 2005
4,390
Use the diagram I posted in #20. Show the full extent of your calculations.

Attached are my calculations for the 4V supply and the associated diagram. Also attached is how I see the current flow.

You've rounded too much ... keep all the figures consistent.

Kirchoff's voltage law tells us the drop around the circuit = 0. If you know R2 has 1.628V there is no way the voltage across the parallel combination of R1 and R3 is 0.4V.

Do you know scientific notation? X.xxx e-3 Amperes or X.xxx e-1 Volts.
 

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WBahn

Joined Mar 31, 2012
30,082
As JoeJester has said several times already -- SHOW YOUR WORK!

You are basically giving random teases and glimpses at how you are coming up with any of your results and it is constantly shifting and changing. It's impossible for anyone to follow your reasoning and it is your reasoning that we need to get straightened out.

The first step is to provide a diagram that clearly indicates ALL of the various quantities, including polarities, that you are going to refer to. For instance:



Using this as a starting point, you can talk about any of the voltages and any of the currents in the circuit is a clear and consistent way.

Since you are doing superposition, you should take this diagram and make two copies. Then modify each to turn off one of the two voltage sources. Do that and post those. If it's helpful, then redraw one or both of those to make the relationships clearer. Solve each for the voltages and currents, using the same voltages and currents as in the full circuit whenever possible. Then add the results together.

So start off by posting the modified diagrams for each of the two cases.
 

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