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Is it possible, once I've converted the circuit diagram into its Laplace equivalent, to replace a current source which is in parallel with, in this case, a resistor and inductor by the usual V=IR method considering I treat the R in the equation as a combination of the impedances from both the resistor and the inductor?

Here is an example circuit for what I mean: enter image description here

Here is what I've converted it to. Is this correct / allowed? enter image description here

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    \$\begingroup\$ Including a schematic (no matter how simple) might make your question more clear? Are you asking about doing a Norton to Thevenin equivalent circuit transformation? \$\endgroup\$ – The Photon Nov 16 '17 at 3:07
  • \$\begingroup\$ Components in series with a current source don't affect the rest of the circuit. The indicated parts can't be replaced with a Thevenin equivalent. If you also included the vertically aligned 3-ohm resistor, we could produce a thevenin equivalent. \$\endgroup\$ – The Photon Nov 16 '17 at 23:55
  • \$\begingroup\$ @ThePhoton Okay, thanks! I think I have tried what you have suggested. Does this appear to look okay? \$\endgroup\$ – Clement Nov 17 '17 at 1:00
  • \$\begingroup\$ Your conversion doesn't look right to me, but I haven't done the math for myself. \$\endgroup\$ – The Photon Nov 17 '17 at 2:31
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Yes, you can convert the circuit diagram by replacing the impedance in parallel to the current source even after converting to the Laplace domain( This is because Laplace transform is simply domain transformation for simplification of calculation and has nothing to do with the circuit itself).

But, the way you've transformed the circuit isn't correct(on the L.H.S voltage source, R.H.S. is fine), since the voltages across the L.H.S and R.H.S circuit aren't the same which you can verify yourself.

Alternatively, if you really want to convert to a voltage source, first convert the 3 ohm connected to the ground as two 6 ohm resistors connected in parallel to the ground and then you can use one 6 ohm for converting current to voltage source like you've done above.

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