How to solve R-parallel RC circuit?

Question

How to calculate the Vout when the input voltage Vin is = 20V DC + 10V AC. Here is the circuit:

Answer 1

Due to some controversy, I will work the AC component of the answer two different ways.

(1) The parallel combination of the R and C is:

\$Z_{EQ} = 100 || Z_C = 100||\dfrac{1}{j2\pi(2000Hz)(8\mu F)} = 100||(-j9.95) = (0.980 - j9.85) \Omega \$

By voltage division, the AC signal is reduced by:

\$|\dfrac{Z_{EQ}}{Z_{EQ} + 100}| = 0.0976 = 9.76\% \$

Since the AC input is 10Vpp, the AC output is 0.976Vpp.

(2) Using the schematic in Olin's answer, the AC output is:

\$5V_{pp} |\dfrac{Z_C}{Z_C + 50}| = 0.976V_{pp} \$

Answer 2

This is very simple if you first reduce the voltage source and the two resistors to a single Thevenin source first. That source then has the capacitor on its output. In other words, what you have is equivalent to,

which is now trivial to solve.

Answer 3

You can treat resistors, capacitors and inductors as resistors with a complex resistance (impedance):

Z = R+ j*X

• resistor: R = Ohms resistance, X = 0
• capacitor: R = 0, X = -1/(omega*C)
• inductor: R = 0, X = omega*L

Then use common formulas for parallel and serial circuits to calculate the resulting impedances like you are used to do for ordinary (=real, non-complex) resistances.

Answer 4

Original answer deleted

I can appreciate the frustration. I too failed. We know

\$ \frac{1}{Z_{\text{eq}}} = \frac{1}{Z_1} + \frac{1}{Z_2} = \frac{Z_1 + Z_2}{Z_1 Z_2} \$

Checking Alfred's formula, we agree on Thevenin reduction \$ Z_C = \dfrac{1}{j2\pi(2000Hz)(8\mu F)} =(-j9.95)\Omega \$

and agree that ... \$ V_{out} = 5 V_{pp} * \dfrac{Z_{C}}{Z_{C} + 50} \$ \$ = 5V_{pp} |\dfrac{-j9.95}{50 - j9.95} | \$

My mistake was taking to absolute values of each part instead of conjugating and squaring each term, then taking absolute. You can only approximate scalar ratio terms but the full conjugation gives the correct answer.

With complex alegbra the result in general terms is; The equivalent impedance \$ Z_{\text{eq}} \$ can be calculated in terms of the equivalent series resistance \$ R_{\text{eq}} \$ and reactance \$ X_{\text{eq}}. \$ \begin{align} Z_{\text{eq}} &= R_{\text{eq}} + j X_{\text{eq}} \\ R_{\text{eq}} &= \frac{(X_1 R_2 + X_2 R_1) (X_1 + X_2) + (R_1 R_2 - X_1 X_2) (R_1 + R_2)}{(R_1 + R_2)^2 + (X_1 + X_2)^2} \\ X_{\text{eq}} &= \frac{(X_1 R_2 + X_2 R_1) (R_1 + R_2) - (R_1 R_2 - X_1 X_2) (X_1 + X_2)}{(R_1 + R_2)^2 + (X_1 + X_2)^2} \end{align}

Of course Z1 = R1 and Z2 = X2 ( the cap impedance) and R2=X1=0

thus the result is ;

\$ \begin{align} Z_{\text{eq}} &= R_{\text{eq}} + j X_{\text{eq}} \\ R_{\text{eq}} &= \frac{R_1 X_2^2}{R_1^2 + X_2^2} \\ X_{\text{eq}} &= \frac{R_1^2 X_2}{R_1^2 + X_2^2} \end{align} \$

Now I know why I use nomographs... http://www.testecvw.com/carl/images/ImpedanceNomograph.pdf