# RMS current of a circuit driven by a white noise voltage source

Note: I am an applied mathematician, not an electrical engineer, so apologies if this comes across as silly. I am analyzing the following circuit as part of some modeling related to energy storage systems:

simulate this circuit – Schematic created using CircuitLab

Where the voltage source is gaussian white noise (hence my use of "mixed Hz" in the schematic software this site uses). This means that the fourier transform of the voltage waveform is flat across all frequencies.

I was wondering if, given the rms Voltage, I could calculate an "equivalent capacitative reactance" $X_{c,eq}$ such that:

$$I_{rms} = \frac{V_{rms}}{X_{c,eq}}$$

I've heard of Johnson-Nyquist noise and kTC noise for capacitors, but I don't see how temperature applies here (since they are modelling a physical phenomena).

• And your actual question is? – PlasmaHH Jun 30 '16 at 12:56
• @PlasmaHH oops..yes, added my question. – user115412 Jun 30 '16 at 12:58
• Actually kTC is not the noise generated by a capacitor as ideal capacitors are noiseless. From the voltage and the current you can indeed detemine Xc and from that the capacitance. – Bimpelrekkie Jun 30 '16 at 13:38
• @FakeMoustache thanks for clarifying. How would one calculate $X_c$ for my specific example of gaussian noise? – user115412 Jun 30 '16 at 13:39
• Xc is like a resistor so apply Ohm's law: R = V/I where instead of R you use Xc. As long as you define the magnitudes of Current and voltage in the same way (like both RMS or both peak value) then this should work. EEs normally only apply this using sinusoidal signals but it should apply to noise as well (and as noise can considered the sum of many sinewaves...) – Bimpelrekkie Jun 30 '16 at 13:45

## 1 Answer

It doesn't really make sense to talk about a single or "effective" value of $X_C$ in this context, since as you say, you don't have a single frequency that you're dealing with.

Instead, you need to take a step back and apply a more fundamental relationship:

$$I_C(t) = C\frac{d V_C(t)}{dt}$$

Since you're the mathematician here, I'm not going to tell you how to differentiate a random signal, other than to make a few qualitative comments.

A differentiator functions as a high-pass filter that deemphasizes low frequencies and emphasizes high frequencies. Therefore, if your voltage source is "white" (equal amplitudes at all frequencies), the current value will be very "blue", with arbitrarily high values at arbitrarily high frequencies.

The RMS value of this current (relative to the RMS value of the voltage) will depend on exactly what bandwidth you're considering.