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I was doing some exercises on transfer functions and ran into this RC Circuit.

RC Circuit

I thought that I was allowed to add up the two resistors since they are in series. Which makes that the voltage across all elements would be equal and Vin = Vout.

Although I am not completely sure if this is correct, could someone please verify or point me out my mistake?

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  • \$\begingroup\$ If the Vin is DC, you can add the resistors together for a Rtotal.(provided the capacitor is charged). The Vout is a voltage from a voltage divider. \$\endgroup\$
    – sparky Al
    Jan 11 '15 at 2:34
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Although you've awarded the answer points, I think you've missed a point. This is an excellent example of when to use a Thevenin equivalent, to wit:

schematic

simulate this circuit – Schematic created using CircuitLab

ETA: Bugger. Got the V2 equation wrong. Fixed it.

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  • \$\begingroup\$ @WhatRoughBeast I didn't even consider using a Thevenin Equivalent, thanks for this answer! \$\endgroup\$ Jan 11 '15 at 14:04
  • \$\begingroup\$ Yeah I already figured that out to be honest haha, thanks for the input! \$\endgroup\$ Jan 12 '15 at 17:36
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You could add them in series - if there was nothing in parallel with R2.

Like you said, if you could add them in series, Vin = Vout which is clearly wrong because it disregards the voltage drop across R1. Essentially you've replaced R1 with a short.

In this circuit, R1 is in series with parallel combination of R2 and C1. R1 is not in series with R2 alone, so you can't combine them.


EDIT: @WhatRoughBeast makes an excellent point in that you can combine the resistors by transforming the circuit into a Thevenin Equivalent. You could also use a Norton Equivalent and replace V, R1, and R2 with a current source and a resistor in parallel, however in my short experience studying EE I find that Thevenin is used far more than Norton.

Thevenin's Theorem allows us to simplify any linear network consisting of voltage sources, current sources, and only resistances into a single Thevenin voltage source \$V_{TH}\$ (not to be confused with the threshold voltage of a MOS transistor) and a Thevenin resistance \$R_{TH}\$.

How to calculate a Thevenin Equivalent courtesy of HyperPhysics

For your circuit (you can follow along with the schematic that @WhatRoughBeast posted) you ignore C1 and calculate the Thevenin Equivalent of V1, R1, and R2. There are two steps:

  1. Find \$V_{TH}\$: The open circuit voltage is simply the output voltage at R2, which is the voltage divider equation. Thus \$V_{TH} = \frac{R_2}{R_1 + R_2} V_1 \$.
  2. Find \$R_{TH}\$: Short the voltage source V1 and look at the equivalent resistance through the output terminals. You just have \$R_1 || R_2\$, so \$R_{TH} = \frac{R_1 R_2}{R_1 + R_2}\$.

Training yourself to see opportunities to use Thevenin Equivalents will come in very useful as an EE.

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Same current will be flowing through the components if they are connected in series.
The voltage across the components will be same if they are connected in parallel.

From the diagram, it is clear that the current through R1 and R2 are not the same and hence they are not in series. But same voltage appears across R2 and C and they are in parallel connection. Now the resulting equivalent impedance R2||C is in series with R1.

Solve the parallel connection first to calculate the final output from it.

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