allocate23
- Feb 11, 2013
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how to calculate voltage drop at R1 , R2 ,R3??
- Thread starter allocate23
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Use the superposition principle:
1) replace B1 by a short circuit and calculate volategs and currents. This should be easy now: only one source left.
2) re-insert B1 and replace B2 by a short. repeat the calculations. Easy again.
3) superimpose the two solutions to find the total solution.
A thorough explanation is here.
1) replace B1 by a short circuit and calculate volategs and currents. This should be easy now: only one source left.
2) re-insert B1 and replace B2 by a short. repeat the calculations. Easy again.
3) superimpose the two solutions to find the total solution.
A thorough explanation is here.
The straightforward method is to write the node equation(s) for the circuit and solve that. Here there is only one node (at the junction of the 3 resistors) so there is only one equation for the node voltage. Superposition is a gimmick that can reduce the amount of work involved but only if you're doing a numerical solution. For a closed form solution, algebraic superposition can be more complex and error prone than solving the node equation directly, especially so if using a computer algebra system (CAS) to manipulate the node equation. There is really no excuse for not using CAS given the availability of free software such as Maxima. Attached is a solution for Vn using MathCAD.
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hey Laplace, that's a bit hard.
I think it is a useful tool. Homework of this type is meant to teach the principles. Otherwise it would be much more complex. Superposition is just one principle among others.
Of course, solving the node equations directly is also a valid method. In the end, both methods will give the same result.
And of course I didn't mean a numerical solution. I generally prefer closed algebraic solutions.
About CAS: to be able to apply a CAS correctly, you need to know the principle. And it doesn't harm to go the way a few times manually before switching to a CAS. I have the same stance with simulations: If you have no feeling for the expected result and trust the SW too much, you can get lost.
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One of the hard facts of life is that I was taught many such principles for the purpose of reducing the work necessary to solve circuit problems back when the only calculation aid was a slide rule - and I have forgotten them. But node equations I can remember. And I can attest that I have gotten lost more often with shortcuts and manual calculations than I have with node equations and CAS. Simulations are another matter; I still have not learned to trust a simulation engine.
try source transformation on the 2 voltage sources
the resistor will now be parallel to the current source. current source 1= b1/r1
(for the right voltage source)
do the same for the left Vsource. [Csource2 = b2/r3]
the circuit will now be 2 current source and 3 resistors in parallel.
then use KCL for upper node. =)
you can now get the voltage for every resistor
the resistor will now be parallel to the current source. current source 1= b1/r1
(for the right voltage source)
do the same for the left Vsource. [Csource2 = b2/r3]
the circuit will now be 2 current source and 3 resistors in parallel.
then use KCL for upper node. =)
you can now get the voltage for every resistor
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Laplace, don't worry, we're all not getting younger.
Look at it as just another tool in the box. Not every tool fits every situation and personal preferences are just as important. Those starting the game need to be shown the possibilities and need to make up their minds as to which method suits them best.
Look at it as just another tool in the box. Not every tool fits every situation and personal preferences are just as important. Those starting the game need to be shown the possibilities and need to make up their minds as to which method suits them best.
