# Solving three impedances in series

I have a circuit of two capacitors and three resitors, where two pairs of the components are in parallel combinations:

I'm trying to calculate the total impedance.

Since $C_1$ and $R_1$ are in parallel and likewise $C_2$ and $R_2$ i would think the solution would be:

$Z_E = R_1 + \frac{1}{j\omega C_1} + R_2 + \frac{1}{j\omega C_2} + R_3$

However my textbook tells me that the answer is what's given on the picture.

Can anyone clarify this for me? Thank you.

• This kind of conceptual error happens when a professor gets to talking too fast. Capacitance in parallel is added to form a larger overall capacitance, but when analyzing parallel impedance, bear in mind that you are effectively looking at a resistor network. – Sean Boddy Apr 8 '14 at 12:23

Let's translate your words into an equation for the equivalent impedance.

Since C1 and R1 are in parallel

$$Z_1 = R_1||\frac{1}{sC_1}$$

likewise C2 and R2

$$Z_2 = R_2||\frac{1}{sC_2}$$

Solving three impedances in series

$$Z_E = Z_1 + Z_2 + R_3$$

• Thank you Alfred. It seems obvious now, but was kinda confused. – Attaque Apr 8 '14 at 12:43

Since C1 and R1 are in parallel

$$Z_1 = R_1||\frac{1}{sC_1}$$

$$Z_1 = \frac{R_1 \cdot \frac{1}{s \ C_1}}{R_1 + \frac{1}{s \ C_1}} = \frac{R_1}{1+s \ C_1 \ R_1}$$

likewise C2 and R2

$$Z_2 = \frac{R_2}{1+s \ C_2 \ R_2}$$

Solving three impedances in series

$$Z = \frac{R_1}{1+s \ C_1 \ R_1} + \frac{R_2}{1+s \ C_2 \ R_2} + R_3$$

• Thank you Warren. I don't know which answer to accept now. Your answer is a direct answer to my question, but Alfreds clarification solved my problem.. – Attaque Apr 8 '14 at 12:47
• Accept either one Id probably go with Alfred's as he was first though and I haven't added much except to give a final formula. – Warren Hill Apr 8 '14 at 12:50