# How can I reduce this circuit to a source in series with a single impedance?

How can I Thevenize the following circuit looking into the out terminal:

I want to deduce this circuit to a source and a single series impedance.

Here what I tried:

$$\\frac{R1\cdot R2}{R1+R2}\$$ is in series with C1

This becomes $$\\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 }\$$

And now $$\\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 }\$$ is in parallel with R5

I cannot proceed more. How could this be achieved if mine is wrong as an alternative way?

is the following correct?: $$\\frac{R1\cdot R2}{R1+R2}\$$ is in series with C1

$$\Z_{thevenin} = \frac {\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 } \cdot R5} {\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 } + R5}\$$

And the Thevenin voltage needed to be found as well.

Vth = V1 * ((|R5+(1/jwc)| // R2)/ (R1 + (|R5+(1/jwc)| // R2))) * R5/|R5+(1/jwc)|

• Why can't you proceed more? Do you know how to use phasors? Do you know the frequency of operation? Commented Mar 26, 2019 at 12:02
• @ElliotAlderson Is the stage where I am at correct? Frequency is ω so it is variable. Commented Mar 26, 2019 at 12:03
• You have lost the phase information. Does that matter to you? I would use impedances and phasors. Do you know how to use phasors? Commented Mar 26, 2019 at 12:06
• Yes I know how to use phasors, phase does not matter just the amplitude Commented Mar 26, 2019 at 12:06
• Looks to me like you're overcomplicating things. What if C1 was a resistor, could you then solve this? If yes, replace C1 with an impedance $Z_{C1}$. Then treat $Z_{C1}$ as if it is a resistor. Apply Thevenin. Then fill in $Z_{C1} = 1/j\omega C_1$ and you have your answer. Commented Mar 26, 2019 at 12:31

If I do not make any mistake in the math the correct equation for $$\Z_{th}\$$ will look like this:

$$Z_{th} = \frac {\sqrt{R5^2 + (\omega\ C_1\ R_T\ R_5)^2}}{\sqrt{1+(\omega\ C_1 (R_T+R_5))^2}}$$

Where:

$$R_T = \frac{R_1 \cdot R_2}{R_1+R_2}$$

And the Thevenin voltage

$$Vth = V_{IN}\cdot\frac{R_2||R_5}{R_1 + R_2||R_5}\cdot \frac{\omega\ (R_T +R_5)C_1}{\sqrt{1 +(\omega\ (R_T + R_5)C_1)^2} }$$

• Can you also add Vth? Then it would be complete and to reduced one source and a single impedance. Commented Mar 26, 2019 at 17:08
• I believe it should be handled down a canonical ratio of polynomials to have a better understanding of what this impedance does over frequency Commented Mar 26, 2019 at 17:51
• I add Vth voltage.
– G36
Commented Mar 26, 2019 at 18:11