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I am new to electronics and I have a little trouble applying the basics I learnt so far. My problem is basically that I don't understand why we consider the second circuit below as the Thevenin equivalent, even though the behaviour of the transfer function is different.

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Here you can see the parallel RC circuit.

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The supposed Thevenin equivalent.

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Blue is the parallel RC-Circuit and green the Thevenin equivalent. I think the behaviour of the two circuits is apparent. Also, if we derive the transfer function for the circuits, we get two different solutions.

Transfer function for parallel RC-Circuit:

formula

formula

Transfer function for thevenin equivalent:

formula

formula

formula

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    \$\begingroup\$ Yes it should, did you forget to halve the voltage to get the same curve? \$\endgroup\$
    – Justme
    Commented Jun 16, 2021 at 20:48
  • \$\begingroup\$ That's right. I should have seen that. \$\endgroup\$
    – jedect_13
    Commented Jun 17, 2021 at 11:35

1 Answer 1

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The voltage of the Thevanized circuit is only half that of the original.

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  • \$\begingroup\$ Specifically, V2=V1/2 for proper representation. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Jun 16, 2021 at 20:27

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