No, the value is far too small. If it is the capacitor used for the timing, the time/s will be reduced to one tenth of the deisred value.
The unit of capacitor is farad. 1 farad =10 to the power of 6 microfarad and also = 10 to the power of 12 picofarad Therfore if you are replacing one picofarad capcitor into one microfarad capacitor you are increasing the capcitance to 1000000 times. If it is in an oscillator circuit you are changing the frequency drastically which will be of no use.
The same as the time constant of a 2.7 microfarad capacitor and a 33 ohm resistor connected in series.
tuned circuit consists of resistance and capacitor so this one RC circuit formula to be used f = 1/ 2 pi RC
The capacitive reactance is approximately 4 kΩ .
No! This is a term for capacitance. A capacitor will store a voltage up to it's breakdown limit plus cause a voltage reaction to a following circuit.
2*103*10-5 = 2*10-2 Seconds = 20 milliseconds
Not a good idea, without knowing more about the circuit in which it's installed. Presumably, the 440v capacitor was selected because its max voltage rating (440v) is higher than the instantaneous voltage to be expected at that point in the circuit. By that criterion, the voltage at that point in the circuit may exceed 370 volts, and your proposed replacement component won't hold it.
A small capacitor can be part of an integrated circuit.
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
Any circuit using a capacitor will not work if the cap is short-circuited.
paper capacitor
when we replace the resistor with a capacitor ,the current will flow until the capacitor charge when capacitor will fully charged there is no current through the circuit because now capacitor will act like an open circuit. for more info plz E-mailt me at "zaib.zafar@yahoo.com"