# How to solve R-parallel RC circuit?

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

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if an "R-parallel RC circuit" is a well known term, I'm not familiar with it... I see exactly two resistors in the picture, and they aren't even in parallel... – vicatcu Dec 6 '12 at 2:48
For the DC solution, replace the capacitor with an open circuit and use voltage division. For the AC solution, treat the parallel resistor and capacitor as an equivalent impedance and use voltage division. Add the two results together for the total solution. – Alfred Centauri Dec 6 '12 at 2:49
@vicatcu an R is in series with a parallel RC. thats why i wrote R-"parallel RC". – user16307 Dec 6 '12 at 2:50
@Alfred I have problem with "AC solution". How can I calculate equivalent impedance of RC? And should I sum the DC and AC solutions arithmetically to calculate Vout? – user16307 Dec 6 '12 at 2:54
It's actually easier for this type of circuit to treat the voltage source and the two resistors as their Thevenin equivalent and proceed from there with the capacitor. – Dave Tweed Dec 6 '12 at 2:59

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:

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

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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.

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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.

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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

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In the AC calculation, the equivalent impedance of the parallel RC is 100||-j10 instead of 100||10. The AC output term is then 976 mVpp – Alfred Centauri Dec 6 '12 at 13:00
True if using Complex Algebra notation and peak voltage. but question was given in scalar terms with peak to peak... 976? – Tony Stewart Dec 6 '12 at 17:15
so if I measure Vout with an oscilloscope will I see Vout = 10Vdc + 0.833 Vpp ?? – user16307 Dec 6 '12 at 18:14
@user16307, no. The 2kHz sinusoidal component at the output is 0.976Vpp, not 0.833Vpp, as can be quickly verified with SPICE (as I have). Put another way, the 2kHz AC signal is reduced to just 9.76% of the value at the input You cannot combine the impedances like Richman has done in his answer. – Alfred Centauri Dec 6 '12 at 18:18
i'm still trying to calculate without using Thevenin memorization thing. my approach is to find total impedance Z of the circuit then the main current and then Vout=Vin-IR by using phasors. I hate Thevenin thing.. – user16307 Dec 6 '12 at 18:25