Talk:RC circuit

shouldn't 'calculations' be used in the title

or, at least rc calculation formulas.. 12.146.12.2 (talk) 09:10, 23 November 2024 (UTC)

RC filter not −3 dB

Hello :)

Contrary to what almost everyone claims, a first-order RC filter does not attenuate amplitude, voltage, or power by −3 dB at the cutoff frequency!

BECAUSE: The electric current is the same at every point in a series circuit, which is the case for the RC filter—this resistive and capacitive series divider.

The definition of the bel is clear: The bel is the logarithmic measure, base ten, of the ratio between two physical quantities of the same nature expressing power.

This means that in the gain calculation, we must use only physical quantities that express power. Examples:

Gain = log (P_output ÷ P_input) Gain = log (|S_output| ÷ |S_input|)

Or physical quantities expressing fields that, when combined, result in physical quantities expressing power—this within the logarithmic function. Example:

Gain = log ((U_rms_output × I_rms_output) ÷ (U_rms_input × I_rms_input))

THEREFORE: Since the effective electric current is the same at every point in a resistive and capacitive series divider, the effective output voltage, i.e., across C at the cutoff frequency, being U_rms_RC ÷ √2, here is an example calculation:

U_rms_input = 1 V U_rms_output = 1 V ÷ √2 ≈ 0.707 V Gain = log ((0.707 V × 1 A) ÷ (1 V × 1 A)) ≈ −0.15 bel

The attenuation of an RC filter is therefore about 0.15 bel at the cutoff frequency, whether in effective voltage or apparent power. Example:

|S_RC| = 1 V × 1 A = 1 VA |S_C| = 0.707 V × 1 A = 0.707 VA Gain = log (0.707 VA ÷ 1 VA) ≈ −0.15 bel

In this example, the apparent power, i.e., the modulus or magnitude of the complex power, allows us to retain all the information regardless of the system—not just a part of the power, meaning active and/or reactive power.

If I express it in watts, meaning active power, the active power of C would be 0 W, so in active power, the attenuation would be −∞ bel, which is not very useful for the example.

As shown by an experiment I conducted with an RC filter using R = 1000 Ω and C = 100 nF at the cutoff frequency:

P_RC ≈ 499 μW P_R ≈ 499 μW P_C = 0 W Q_RC ≈ −499 μvar Q_R = 0 var Q_C ≈ −499 μvar |S_RC| ≈ 706 μVA |S_R| ≈ 499 μVA |S_C| ≈ 499 μVA

SO TO SUMMARIZE: Let's avoid confusion. My advice is to include, along with the effective voltage in the logarithmic function, either the effective current or the apparent electrical impedance. Example: 50 Ω if the input and output of a device have this impedance.

OTHER EXAMPLES: If we use only electrical voltages without additional information in this logarithm, it no longer means anything because another physical quantity is missing to define power. Examples:

U_rms_input = 1 V I_rms_input = 1 A U_rms_output = 2 V I_rms_output = 1 A Gain = log ((U_rms_output × I_rms_output) ÷ (U_rms_input × I_rms_input)) ≈ 0.301 bel

U_rms_input = 1 V I_rms_input = 1 A U_rms_output = 2 V I_rms_output = 2 A Gain = log ((U_rms_output × I_rms_output) ÷ (U_rms_input × I_rms_input)) ≈ 0.602 bel

The examples show that if we use only a single physical quantity expressing a field instead of power, like voltage alone, the information about current is missing—which, by the way, can be identical from input to output in a system. This gives:

1 V × 1 A to 2 V × 1 A ≈ 0.301 bel 1 W to 2 W ≈ 0.301 bel

Whereas:

1 V × 1 A to 2 V × 2 A ≈ 0.602 bel 1 W to 4 W ≈ 0.602 bel

Below, I have gathered equations that help avoid any errors.

Calculation of gain in bel based on the output apparent power relative to the input apparent power:

Gain = log ((E_output ÷ t) ÷ (E_input ÷ t)) Gain = log ((U_rms_output × I_rms_output) ÷ (U_rms_input × I_rms_input)) Gain = log ((U_rms_output² ÷ |Z_output|) ÷ (U_rms_input² ÷ |Z_input|)) Gain = log ((I_rms_output × |Z_output|) ÷ (I_rms_input × |Z_input|)) Gain = log (|S_output| ÷ |S_input|)

Calculation of input apparent power based on output apparent power and gain in bel:

|S_input| = (E_output ÷ t) ÷ 10 ^ Gain |S_input| = (U_rms_output × I_rms_output) ÷ 10 ^ Gain |S_input| = (U_rms_output² ÷ |Z_output|) ÷ 10 ^ Gain |S_input| = (I_rms_output × |Z_output|) ÷ 10 ^ Gain |S_input| = |S_output| ÷ 10 ^ Gain

Calculation of output apparent power based on input apparent power and gain in bel:

|S_output| = (E_input ÷ t) × 10 ^ Gain |S_output| = (U_rms_input × I_rms_input) × 10 ^ Gain |S_output| = (U_rms_input² ÷ |Z_input|) × 10 ^ Gain |S_output| = (I_rms_input × |Z_input|) × 10 ^ Gain |S_output| = |S_input| × 10 ^ Gain

If we step outside the bel scale, meaning the logarithmic measure, base ten, of the ratio between two physical quantities of the same nature expressing power, and we use only the logarithm function, then we can use this function for anything we want, just like any other mathematical function. Sylvainmahe (talk) 13:31, 20 March 2025 (UTC)

This article uses the voltage ratio of input to output.

dBs = 20 log_10 (voltage ratio) + 20 log_10(0.707) = 3.01. dBs = 20 log_10 (voltage ratio) = 20 log_10(0.707) = 3.01. Constant314 (talk) 15:14, 20 March 2025 (UTC)

This is false and mistake, it makes no sense, reread my explanation. Sylvainmahe (talk) 12:07, 21 March 2025 (UTC)

It is clear that 20 log_10(voltage ratio) makes no sense to you. So the sensible thing for you to do is to find out where it is used and why. It is widely and reliably used in electrical engineering, and it is completely compatible with the 10 log_10(power ratio) usage. You are just missing the context. Gwideman (talk) 09:33, 25 December 2025 (UTC)

Hard to see an audience for this article as it stands

"is well-suited to be analyzed in the Laplace domain, which the rest of this article requires a basic understanding of."

Do the authors of this article imagine that there are readers who have mastered Laplace transform and s-domain analysis, yet somehow still need an explanation of the R-C circuit, the simplest of all circuits that exhibit time- or -frequency-dependent behavior?

Or to put it another way, sure, you can analyze an R-C circuit economically using Laplace domain. But it is an utterly useless way to explain it to a person who did not previously understand how an R-C circuit behaves and would like to gain such understanding.

In my view, an accessible starting point would be to portray an RC circuit as a simple 2-component voltage divider, in which the "resistance" of one of the components is frequency dependent. This gives a clear picture of the amplitude response versus frequency, corresponding to the Bode plot of amplitude. It also makes clear why there's a knee at fc (capacitor resistance gets low enough to dominate). It can be noted that the phase is also affected, and that's a more advanced topic, and maybe give a general intuition corresponding to the Bode phase plot, and point the way to the more complex analysis. Gwideman (talk) 10:51, 25 December 2025 (UTC)

The target audience should be incoming college first year students. You are welcome to improve the article. The ratio of articles on English Wikipedia to experienced, active editors is about 1000 to 1. See WP:BEBOLD. Constant314 (talk) 13:36, 25 December 2025 (UTC)