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These are probably decoupling capacitors, and their value generally isn't critical. Risk is that too large a value (e.g. 5x the 'correct' value) could cause large inrush currents and weaken something, …

At that time, in an ideal circuit, the voltage cannot be determined as there is an indeterminate condition -- the capacitor voltages are equal which implies that they changed instantaneously. … inductance or resistance in the switch, the instantaneous infinite current would not happen and the voltages could be analyzed normally -- there would be an exponential approach to the final (equal) capacitor …

Yes, you can use superposition -- once for the V source, and once for the cap decay. The cap decay is (set the other V to 0) Vc=2+(0−2)(1−e(-t/RC)), or 2.e(-t/RC) [you missed a '-' in the exponential …

Choose R2 << R1, but at least 10 ohm (to protect the switch from arcing when you use it to discharge the capacitor. So to charge in 10 seconds (consider that a long time), choose R1 = 100k. …

For 'only' 400 V, you can build a denser capacitor using the capacitance between adjacent fingers of metal (like two combs intertwined). …

Use C = I*T/deltaV Assuming that the relay holds in until 9 V and starting from 12 V, you need C = 20m*8/(12-9) = 53 mF. These are not readily available or practical. It is possible to get capacitors …

When you analyze this circuit with zero resistance, you get some infinite values that make it impossible to get a meaningful answer. However, if you analyze with a resistance and then take the limit a …

In a parallel plate capacitor, \$A\$ and \$d\$ are clearly obvious. Imagine that \$d\$ increases by separating the plates -- the capacitance decreases. …

Remember that the current in a capacitor is proportional to the derivative of the capacitor voltage. … The net result if you plotted voltage and capacitor current vs. time is that the capacitor current would follow the slope of the V source. …

Connect the top of the capacitor to +5 V (not Q) Move the 3.5 k (why do you have that value ?) to the collector of the NPN Insert 10k between Q and the base of the NPN. … It will remain charged that way, even if a load is connected If you then press the button on R, the NPN will turn off; the capacitor will remain charged until a load is connected to it. …

It makes no electrical difference if the caps are chosen to have no influence on the AC signals. If the capacitors cause a small impedance mismatch on the lines, then as long as the rest of the lines …

Take an n-channel MOSFET, connect drain to the power source; source to the motor//capacitor. … Connect a 10 uF capacitor from gate to ground, 1 MΩ from gate to the switch (other end of the switch at 12V), and 100k from the switch to ground. …

Or just put a 3k R from the junction of the switch and the 10k to GND -- you can use the switch you have. For faster discharge add a diode across the 10k (opposite polarity to the NPN's B-E diode).

If you want to charge the capacitor quickly, but not overcharge it, then just use a large power FET and a comparator that turns it off if the capacitor is > 2.7 V. … It will take 13 s to charge the capacitor, and the FET would need a heatsink. Therefore I suggest that you do add a series resistance. …

Individual transistors are not very precise devices -- their parameters (e.g. gain) vary widely, and the operating voltage and current vary dramatically with temperature. Conversely, resistors are r …