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Solving a problem, I came across the following source transformation:

      schematic

My question is:

I know that I can do the following:

                                              

But why can I do the first image transformation, even with an impedance between the two voltage sources?

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You can do that because they are all in series. The same current flows through all the components. The terminal voltage is equal to the sum of all the component voltages. Those equalities apply whatever the order of the components.

You would not be able to re-order them if there were any shunt components changing the topology from all-series. As there are always stray capacitances to ground in the real world, it's whether these are significant that controls whether you can do this re-ordering. At mains frequencies, you can, at microwave frequencies, you can't. At intermediate frequencies, you'd need to do the sums to see how much error you're introducing.

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  • \$\begingroup\$ To make sure I understood, I did a more complicated circuit layout. I used current source --> voltage source transformation + association elements + what you said and redesigned the circuit. Is the simplification that I made valid? More complicated circuit simplification \$\endgroup\$ – Vinicius ACP Oct 6 '18 at 10:48
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    \$\begingroup\$ @ViniciusACP what you've done is not a simplification, but a destruction. It's trivially easy to see that red dotted box components are not equivalent. In the 'before' circuit, the presence of the shunt inductor means that the red box is short-circuit at DC. In the after circuit, the series capacitor means the red box is open circuit at DC. Perhaps try to master turning a current source in shunt with a resistor into a voltage source in series with a resistor, before trying to transform LC circuits. \$\endgroup\$ – Neil_UK Oct 6 '18 at 13:41
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    \$\begingroup\$ @ViniciusACP Same issue. R3 gives DC continuity across the first red box, C blocks DC across the second one, so obviously you've broken the equivalence. \$\endgroup\$ – Neil_UK Oct 7 '18 at 6:04
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    \$\begingroup\$ @ViniciusACP Here's a major hint for you. A simple low-pass filter can be made with RC and RL. With RC the R goes to source and C goes to ground. With RL the L goes to source and R goes to ground. Both filters can have the same transfer function. \$\frac{1}{1+s\frac{L}{R}}\$ vs \$\frac{1}{1+sRC}\$. This is a hint. Notice how the shunt capacitor turned into a series inductor. not a series capacitor. \$\endgroup\$ – Harry Svensson Oct 8 '18 at 15:11
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    \$\begingroup\$ @ViniciusACP You seem to be forgetting that a voltage source in series with an impedance can be transformed into a current source in parallel with the same impedance. The operative phrase here is same impedance. In your first and second examples, you had parallel LC and RC initially. In the transformed version, you needed to keep those components parallel, while you combined the sources and moved them from prallel to series with the impedance. \$\endgroup\$ – Neil_UK Oct 8 '18 at 17:01
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If you had a black box with those three components inside (two voltage sources and an inductor) and, although you couldn't see them, you were allowed to measure what output the black box produced using whatever testing method you wanted, you could only conclude that there was a voltage source in series with an inductor.

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