0
$\begingroup$

Since capacitors allow charges to pack on the negative plate, there is some current flowing for a short time. Also the dielectric does not prevent the EMF otherwise there would be no voltage at any time. So I wonder, given an alternating current having a frequency fast enough and a current low enough to not saturate the capacitors if a series of capacitors can actually power something.

My main interest is to know if conductors can actually be discontinuous, with dielectric gaps between actual metallic conductors, in the case of alternating current.

$\endgroup$

4 Answers 4

1
$\begingroup$

Yes. The complex impedance of a capacitance $C$ is $Z=1/j\omega C$, where $j=\sqrt{-1}$ is the imaginary unit and $\omega$ is the angular frequency of the signal. For non-sinusoidal signals, one way to analyze the circuit is the to use Fourier analysis to split the signal into sinusoidal parts. Another way is to imagine each change in the voltage across the capacitor as a step-like change which causes the capacitor to charge or discharge, as you would in a simple circuit where a capacitor is connected to a DC supply by a switch.

When my undergraduate electronics lab was discussing this subject, the lecturer set up the scenario this way:

Suppose that your experiment is connected to your data acquisition by a long cable. You're expecting to get some kind of a square wave. However, somewhere in the middle of the long cable, a mouse has chewed a tiny gap into one of the conductors. The tiny gap acts like a capacitor in series, and turns the cable into a high-pass filter. What you see at the other end is the derivative of the signal you expected: positive- and negative-going spikes instead of a square wave.

I thought this scenario was absurd for ten years, until the first time I saw that kind of behavior unexpectedly and knew to trace it to a broken cable connector. (No rodents were involved that time.)

$\endgroup$
1
$\begingroup$

Yes. You can think of a capacitor as a filter that allows AC but blocks DC, and an inductor as a filter that allows DC but blocks AC.

$\endgroup$
1
$\begingroup$

In a word, yes. You only need one capacitor connected to a load with large inductance compared to its resistance to power the load with alternating current.

In the ideal case, you have an LC circuit (https://en.wikipedia.org/wiki/LC_circuit), in which you have an ideal capacitor and an ideal inductor (i.e. no resistance across either component). Electrical energy oscillates between being stored in the electric field of the capacitor, and being stored in the magnetic field of the inductor, with a frequency

$$\omega = \frac{1}{\sqrt{LC}}$$

for a capacitance $C$ and an inductance $L$. Since any configuration of multiple capacitors is equivalent to a single capacitor, and any configuration of multiple inductors is equivalent to a single inductor, you can use the LC circuit for nearly any situation where you have some capacitors that power some inductive load. Current will flow back and forth in this ideal circuit "forever" (i.e. until the energy lost to electromagnetic radiation becomes significant).

Realistically, though, nothing has zero resistance. Your capacitor is going to have a nonzero resistance across its plates, the wires have some resistance, and so does your load (which has to have a nonzero resistance to do basically anything useful anyway). In this case, we have an RLC circuit (https://en.wikipedia.org/wiki/RLC_circuit). This behaves similarly to an LC circuit, except that the alternating-current oscillations are now damped by the resistance.

The behavior of this circuit is dependent on how large the resistance is compared to the capacitance and inductance in the circuit. Specifically, it depends on the ratio $\zeta=\frac{R}{2}\sqrt{\frac{L}{C}}$, called the damping factor. If $\zeta\geq1$, the circuit doesn't oscillate. The current flows in one direction, increases, and then decreases exponentially toward zero. The closer $\zeta$ is to $1$, the faster this decay happens. If $\zeta<1$, then the capacitor produces alternating current with a frequency

$$\omega_d=\sqrt{\frac{1-\zeta^2}{LC}}$$

The magnitude of the alternating current decays exponentially, at a rate proportional to $e^{-\alpha t}$, where $\alpha=\frac{R}{2L}$.

$\endgroup$
1
$\begingroup$

I wonder, given an alternating current having a frequency fast enough and a current low enough to not saturate the capacitors if a series of capacitors can actually power something.

The capacitors don't supply any power to the load. But AC current from some generator can be delivered to a load through a capacitor.

Two capacitors in series with values $C_1$ and $C_2$ are equivalent to one capacitor with value $1/(1/C_1 + 1/C_2)$. Therefore current from a generator can be delivered to a load through a combination of capacitors connected in series.

This is used, for example, in those circuit testers that look like a screwdriver with an LED in the handle. These testers only need to contact the wiring being tested at one point because the capacitance between the user and the ground is used to complete the circuit.

enter image description here

My main interest is to know if conductors can actually be discontinuous, with dielectric gaps between actual metallic conductors, in the case of alternating current.

Yes, but if you don't design that gap to have fairly high areas of conductor facing each other across a fairly narrow gap, the capacitance value will be very small and so only fairly high AC frequencies will be able to pass through.

$\endgroup$
1
  • $\begingroup$ I didn't explicitely write that the series of capacitors was plugged to a power supply but I didn't think of capacitors as a power source. $\endgroup$
    – Winston
    Sep 30, 2019 at 20:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.