Is a microphone a current source? An ideal current source would produce the same current no matter what the resistance connected across it. I don't know a lot about microphones, but I think they're closer to ideal voltage sources. An example of a current source would be the collector (or emitter) current of a transistor operating in its active region. If the base current is kept constant, the collector current will be almost constant over a wide range of load resistances. You can make an even better current source using an opamp instead of a single transistor. Perhaps current sources are difficult to envision because they're somewhat difficult to create, unlike voltage sources. It's fairly easy to make a battery. But imagine if you could just open your parts bin and grab a current source. It would be a dangerous device. You'd have to be very careful to store it with its leads connected together. Otherwise, an arc of current would flow through the air from one lead to the other. And I suppose all that constant current flow would rapidly deplete all its energy. Ultimately, of course, any linear circuit has both a Thevenin equivalent and a Norton equivalent, the first using an ideal voltage source and the second an ideal current source. So a simple battery can be viewed as either a voltage source with a small series resistance, or a current source with a very large parallel resistance.

I am sorry but your formula makes no sense. By definition C=Q/V for a capacitor. Then it follows up that V=1/C * Q. Assuming the charge on the capacitor changes over time it is the result of accumulation of charges brought by current traveling through the circuit and voltage on the capacitor is the sum of all charges brought by i(t) over dt time intervals.

@@dan-florinchereches4892 How does that formula makes no sense? I (capacitor current) = C (capacitance in F) * dV/dt (rate of change of voltage across the capacitor). If dV/dt is close to zero (voltage is not changing) then current (I) will be close to zero. If dV/dt is high, then current through capacitor (I) will be high. Current can also be thought of as a rate of change of charge I = dQ/dt.

@@oriole8789 I see the differential equations are pretty much equivalent. My problem would be more along the lines of chicken or egg problem. I see it as the voltage at the capacitor plates is changing as a result of current depositing charge. The formula you are using is from the perspective of the charge existent on capacitor producing a current.

​@@dan-florinchereches4892 Your perspective of the voltage between the capacitor plates changing as a result of current depositing charge is perfectly correct, and consistent with all the formulas provided (V = Q/C, I = C*dV/dt, I = dQ/dt etc). In that sense, V = Q/C directly leads to dV/dt = 1/C * dQ/dt. The perspective depends on what you're measuring and controlling. A voltage (potential) difference causes current to flow. Current cannot flow if there is no potential difference. Current depositing charge is a process that happens only as long as a potential difference exists. Once the system is in equilibrium (a capacitor is fully charged), there is no current flow, because there is no potential difference (with respect to either side of the capacitor, and the circuit nodes those sides are connected to, respectively). Between the legs of the capacitor, a potential difference is naturally high since the capacitor is charged. If a load is then applied to the capacitor, a new potential difference now exists with respect to that load, and if we assume that the load is 0ohm short circuit let's say, charge will flow out of the capacitor in the form of current (limited by the ESR of that capacitor in the real world), until the potential difference between the capacitor and the load reaches 0. Current flowing in or out of the capacitor doesn't change the formulas as long as sign conventions are followed. Hope this helps!

A capacitor is similar to a battery in that regard. It can be charged up so that there's a voltage difference between its two plates. But, like a battery, even with a large voltage, there may or may not be any current flowing into or out of it. It all depends on what else it's connected to. Conductors and resistors follow Ohm's Law. They're sometimes called "Ohmic devices". But a lot of devices aren't Ohmic. In fact, most of the interesting ones aren't, including capacitors, inductors, diodes, transistors, etc. For a capacitor, the current flowing through it isn't proportional to the voltage across it, but rather to the *rate of change* of the voltage. So when the voltage reaches its maximum, for a brief moment the voltage isn't changing, and thus the current is zero. Also, when the voltage is zero, it's changing at its maximum rate, and so the current is flowing at a maximum. So if the voltage takes the form of a sine wave, oscillating between positive and negative, the current will also be an oscillating sine wave, but the oscillations will occur with a 90° phase shift relative to the voltage. Hope that helps.

@@zenbum2654 I think I understand. Thank you.

because the circuit looks like it is open.

I think you misinterpreted Ohm's law. This law defines a relationship between the current value given a certain voltage applied to a closed circuit with a certain *impedance*. You don't need a current (whose unit is given in amps) to have a voltage. A voltage is created, for instance, when you bring together opposing electrical charges separated by a certain distance (aka insulator). The result of doing that is that you have created a disturbance in the electric potential. In our universe, everything tends to seek a more stable state. In the same way that a body will tend to fall to the ground if it is lifted, the opposed charges will be delighted to meet each other, and if you give them that chance by connecting a wire, they will take it for sure. The value of the current will depend on the *resistance* , in the case of DC power source, or on the *impedance* of the wire in the case of a AC power source. Hope this helps you to understand the matter better. The electric and electronic world is a very tricky one.