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It is the nature of the capacitor. There can be current through the capacitor only if the voltage across it is changing. The defining equation is: $$i_C=C\frac{dv_C}{dt}$$ …

The symbols do indicate that these capacitors are polarity sensitive. Elecrolytic, for example, appear as \$0\Omega\$ when reverse biased. Therefore the reverse voltage should be zero, definately not …

the term \$V_{0}e^\frac{-t}{\tau}\$ represents a discharging of the capacitor. … Reusing \$V\$ in the second equation in the question as the voltage the capacitor is sourcing is incorrect. The voltage across the capacitor is always \$v_{C}(t)\$ (equ 1). …

The leakage between the plates is essentially zero. You can calculate using the resistivity of air and the volume between the plates. By reducing the air pressure, the resistivity will increase. The p …

I got this far: $$Xc=\frac{v}{i}=\frac{sin(\omega t)}{C\omega cos(\omega t)}$$ Almost right. The reactance is the ratio of magnitudes, not the complete functions, as shown in the link provided in …

A capacitor can switch its behaviour in an instant. … During the interval when either D1 or D3 is conducting, the capacitor acts as a passive element. So the current through D1/D3 supplies both the capacitor and the load. …

This voltage is coupled by capacitor \$K_{QR}\$ to Q's cathode driving it high enough to turn Q off. Every pulse then causes each thyratron to turn on in sequence. …

At higher frequencies the capacitor looks like an inductor. So a non-inductive ceramic capacitor is placed in parallel. …

But since the diode is a current source, the current must be diverted around the capacitor while maintaining the capacitor voltage. … During the hold interval (or at least at the end of it), the capacitor is discharged. Buffering a grounded capacitor with an op-amp is a good choice. …

There is a small capacitance (sometimes called a distributed capacitance) between the coils of an inductor. By using a capacitance probe and an ac signal the effective capacitance of the inductor can …

Placing a constant voltage source from one node to another isolates (decouples) all branches connected to the nodes from each other. Because the voltage across each branch is known, each can be analyz …

A 1V or 2 V \$\Delta V\$ may be better but will need a bigger capacitor. It is best to measure the current, then multiply in a safety margin. …

For the circuit in the diagram, and for the relay in the OP, a 2200μF capacitor should work. The capacitor will charge to 12V less one diode drop. … When the capacitor discharges, the initial coil voltage will be 12V less two diode drops. The relay datasheet will provide values for \$V_{\text{dropout}}\$ and coil resistance. …

A 500uF electrolytic capacitor has an ESR < 0.5 ohms. The FET must supply a minimum (probably higher) 200 A pulse. The PTC isn’t fast enough nor does it limit the current sufficiently. … Currents this high are probably not healthy for the capacitor either. Update: There is another reason that the FET may fail . The max VDS of the 9120 is 100V. …

The circuit provided is the definitive 1st order low pass filter. The unwanted frequencies are dropped across the resistor. If R is zero then there would be no filtering at all. The time constant that …