Confusion about AC current relative to frequency

Question

Having trouble conceptually understanding why AC peak current (capacitor voltage = 0) is different between circuits where all variables are the same, except the frequency.

Looking at the pictures I attached, I understand that in the 10 Hz circuit, the capacitor has more time to charge/discharge meaning less current in the circuit, compared to the 1 kHz circuit.

But intuitively, I cant wrap my head around why the current would be different when capacitor voltage = 0 (peak current) between circuits with the same variables except frequency.

Please don't just answer the math formula Xc=1/2pifC. Of course it makes sense that way.

Thanks

Answer 1

The current into or out of a capacitor is related to the rate of change of the applied voltage since the voltage is what "pushes" charge onto the plates or "pulls" it off. The equation is I = C(dV/dt). In your circuit, when the applied frequency is higher, the rate of voltage change is higher - at all points in the waveform, not just at V = 0 (differentiate the applied waveform), so the current into or out of the capacitor is higher.

Answer 2

Consider a 1 µF capacitor in series with a 1 kΩ resistor with resultant time constant of 1 ms.

• If a steeply-rising square wave was input, with a rise-time of 1 µs, the capacitor is effectively a dead short, and maximum current is limited only by the series resistor, i.e., 1 kΩ.
• with a slowly-rising waveform, e.g., a 1 Hz sine wave, the capacitor has ample time to charge and discharge, offering an effective impedance of ~160 kΩ. The maximum current is limited by that "160 kΩ" capacitor.

Answer 3

It sounds like you're trying to understand an AC circuit using DC behavior. If I read your question right, you're expecting a capacitor with zero voltage across it to act exactly like a short circuit in that instant:

^{simulate this circuit – Schematic created using CircuitLab}

The assumptions here are that 1) the capacitor voltage is always 90 degrees out of phase with the source voltage, and 2) the current really does always vary with frequency. Neither of these assumptions are true. At very low frequencies \$(f \ll \frac 1 {2 \pi R C})\$ the source voltage and capacitor voltage are almost exactly in phase. When \$V_C\ = 0\$, \$V_S \approx 0\$, so the current through the resistor is very small. At high frequencies \$(f \gg \frac 1 {2 \pi R C})\$ the source and capacitor voltage are about 90 degrees out of phase. The peak current maxes out at \$V_S/R\$ and is roughly independent of frequency.

I think part of your confusion comes from the fact that in your circuit:

$$f_{transition} = \frac 1 {2 \pi \cdot 1 \mathrm{k\Omega} \cdot 10 \mathrm{\mu F}} = 15.9 \mathrm {Hz}$$

You're testing at 10 Hz, in the frequency range where the circuit transitions between low-frequency and high-frequency behavior. Try 0.1 Hz instead and you'll see the low-frequency behavior emerge.

If you want to sharpen your understanding, consider the following two circuits. First, to convince yourself that there's nothing special about \$V_C = 0\$, try adding a DC offset to your source. This circuit has the same current as yours:

^{simulate this circuit}

Second, to better see the behavior of the capacitor by itself, try removing the resistor. This circuit's current never maxes out -- it always depends on frequency:

^{simulate this circuit}

Answer 4

For intuition regarding this behaviour, it might be easier to consider capacitor voltage rather than current, and then relate that to current through both resistor and capacitor. The voltage \$V\$ across a capacitor is:

$$ V = \frac{1}{C}\int{I \cdot dt} $$

That is, for any given current through the capacitor, the voltage is going to rise (or fall) at a rate proportional to that current.

Now consider that the time between peaks of higher frequency signals will necessarily be less than for lower frequencies. That means that the capacitor voltage has less time to "integrate", less time during which it can rise (or fall). Therefore, at higher frequencies, you can expect smaller amplitude voltage variations across the capacitor.

This is easy to see when one directly drives an alternating current through a capacitor. Using square wave current sources of equal amplitude (±1A), but at different frequencies, we can compare how capacitor voltage differs:

^{simulate this circuit – Schematic created using CircuitLab}

You can see how the "slopes" (rate of change of capacitor voltage) are identical, but one of the capacitors simply has more time between instants where current changes direction, in order to accumulate charge.

It's fair to say, then, that the amplitude of voltage signal across a capacitor is inversely proportional to frequency of current through it.

Returning to the RC circuit, you must realise that whatever voltage exists across the capacitor is voltage that isn't across the resistor. That's KVL at work; the instantaneous sum of resistor voltage and capacitor voltage is equal to the source voltage.

At lower frequencies, the capacitor's voltage amplitude is greater, and therefore resistor's voltage amplitude is reduced. At higher frequencies, it's the other way around, the resistor gets the lion's share of the source's voltage.

Therefore, since it is the resistance that determines current, as frequency increases, and resistor voltage amplitude increases, peak current will increase.

This is a very contrived explanation, and I'm not super happy with it, but that might be good enough. However, you must also consider that in your own example, the source is sinusoidal, smoothly changing between positive and negative values. This adds another consideration: by the time capacitor voltage has peaked, the source voltage is already on its way back down again, which will necessarily influence the peak voltage attainable by the capacitor. That's very hard to describe purely intuitively, and it only becomes clearer (in my opinion) when examined mathematically.

I can't think of a way to describe this response to a sinusoid other than to say that the total source voltage available to be shared between both resistor and capacitor is changing continuously, which not as trivial to envisage or explain, compared to if the source were square. Suffice to say that you must consider this too; you can't just assume that the maximum voltage across the resistor coincides with the minimum voltage across the capacitor, or with peak source voltage, as you could with a square wave.

More pictures to illustrate:

^{simulate this circuit}

Here is a plot of source voltage \$V_A\$ and voltage \$V_R=V_A-V_B\$ across the resistor:

It's important to notice that the peak voltage across R1 (corresponding to peak current) does not coincide with peak source voltage. For this reason it is inappropriate to assume that maximum current should be the same regardless of frequency, since that varying frequency is also causing a phase shift which separates those peaks in time.

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One thing I find myself repeating, is that when you apply KCL and KVL and Ohm's law to these circuits, what you end up with is a set of simultaneous equations, and the word simultaneous is not there for nothing. It implies that at any instant in time, there is only one solution for all parameters, impedance (resistance), current and voltage.

You wouldn't be wrong, therefore, to say that those three things aren't actually different things, since their values are intimately related. It might be better to think of them all as different measurements, different "aspects" of the same underlying thing, electricity, three sides of the same three-sided coin, if that makes any sense.

The issue with trying to get intuition about even simple circuits involving capacitors and inductors, is that not only does everything depend on everything else, all at once, there is now an added variable, time. The capacitor here is developing a voltage which is not only dependent on the current through it, but also the time for which that current has been passing. The introduction of time into the equations means that there are now three variables: voltage, current, and time. I omit impedance, since that's just the ratio of voltage to current.

Capacitors and inductors consequently, introduce behaviour that is not just a function of current and voltage, but also of time, and this produces delays and other time-related phenomena, such as the phase shift between current through and voltage across them.

The fact that the capacitor's current is not in phase with its voltage (and vice versa) means that the voltage across the capacitor is not (necessarily) in phase with the voltage source. Consequently, resistor voltage which is dependent upon both, will not be in phase with the voltage source or the capacitor! You find, then that every single element's voltage is out of phase with every other, and that's enough to have me throw my hands up in resignation, and revert to just trusting in the maths.

Answer 5

Since that, in your circuit, the current \$i(t)\$ is always in phase with the voltage across the resistor \$v_r(t)\$, the instantaneous values of these two quantities reach the maximum (and minimum) at the same time. Also, at these moments, the voltage on the capacitor \$v_c(t)\$ is zero (but its derivative \$dv_c(t)/dt\$ is maximum). But all this does not mean that the input voltage \$v_i(t)\$ will be at its maximum (\$V_{pk}\$) at these moments. As the frequency increases, the instantaneous value of \$v_i(t)\$ is at an increasingly higher value when \$v_c(t) = 0\$ (note the graphs below when \$v_c(t)\$ crosses the time axis). And since \$v_r(t) = v_i(t)\$ at these moments, the instantaneous value of the current \$i(t)\$ ends up having a greater value. Also, with frequency increasing, the entire shape of \$v_r(t)\$ approaches to that from \$v_i(t)\$.

Table for the relevant quantities in three frequencies (assuming \$V_{pk} = 5V\$):

When vc = 0 V @5 Hz:
vr:  1.498584 V, vi:  1.498584 V, i:  0.001499 A
vr: -1.498584 V, vi: -1.498584 V, i: -0.001499 A
    
When vc = 0 V @10 Hz:
vr:  2.660090 V, vi:  2.660090 V, i:  0.002660 A
vr: -2.660090 V, vi: -2.660090 V, i: -0.002660 A
    
When vc = 0 V @1000 Hz:
vr:  4.999367 V, vi:  4.999367 V, i:  0.004999 A
vr: -4.999367 V, vi: -4.999367 V, i: -0.004999 A

$)

The equations for the steady-state condition are:

$$ \begin {cases} v_i(t)&=V_{pk}\sin(\omega t)\\ \\ v_c(t)&=\frac{V_{pk}}{\sqrt{1+(\omega RC)^2}}\sin(\omega t - \arctan\left(\omega RC \right )) \\ \\ v_r(t)&=\frac{V_{pk}}{\sqrt{1+\frac{1}{(\omega RC)^2}}}\sin(\omega t + \arctan\left(\frac{1}{\omega RC} \right ))\\ \\ i(t)&= \frac{v_r(t)}{R} \end{cases} $$

Answer 6

Intuitively? Okay, think about an audio amplifier in a car. When a low bass note hits it dims the headlights. That happened because some part of the system couldn't provide the current that the bass note demanded— and he demands a lot being so long an all. He might eat up everything the capacitors have and still need more, and in the case of the car he said "You better pull the rest from somewhere because I'm coming through." So the system has to accommodate for that, making the voltage dip which will of course dim the lights. Kind of selfish, those low frequencies. You already know that more capacitance would prevent this if it's enough capacitance to accommodate, and in this case, that's not so much determined by reactance but it helps you to see how low frequencies are a lot harder to accommodate for in terms of current. Think about how many times you've seen headlights dim for mid frequencies (i.e. 1kHz). It doesn't happen because they're a lot easier to accommodate for. They take what they need and GTFO real fast. So if you think about that way, then in your case that's all going to be determined by reactance. Your lowest expected frequency is 10Hz but you haven't really accommodated for him. With 10uF he's going to be impeded quite a bit but he'll squeeze through. Meanwhile, as he's pulling the last of his 10 cycles through, 1kHz comes flying up and see's your cap as almost a short circuit so it's like "thank you I brought 99 of my friends too kbye." So you see you're passing a lot more 1k current than you are 10Hz current.