1000 microfarads is its rated capacitance, while 35 volts is its rated voltage.
Take two electrolytic capacitors of the same voltage and capacity, connect the positive leads together and connect the negative leads to the circuit. Just keep in mind that this will reduce the cap. value by half (2, 1000uF caps = 500uF) Also the voltage of the circuit should not exceed the voltage of one of the caps.
While it is possible that the company producing the capacitor is using the NF as an identification tool, usually we would expect nF to specify the capacitance or value of the capacitor. For example, a capacitor with the value of 10nF ought to have a capacitance of about 10 nano Farads. This is equivalent to 10*10-9 Farads.
You can not by-pass the capacitor in an electric motor. Most are capacitor-start motors which require the capacitor to be operational in order to start. If the capacitor is not working then it will need to be replaced.
what is flying capacitor
The C represents the capacitance (in farads) of the capacitor. It is a measure of how much charge a capacitor can hold. This is needed to know how much energy the capacitor is holding.
You can use a larger capacitor (more uF) in the decoupling or smoothing parts of the circuit, but not in anything that is used to control the frequency response, but that is unlikely.
You can use a larger capacitor (more uF) in the decoupling or smoothing parts of the circuit, but not in anything that is used to control the frequency response, but that is unlikely.
Capacitance and voltage may be varied independently of one another, but the outcome depends exactly what the circuit is and what role this capacitor plays. If it's a timing capacitor, the time constant will increase. Certainly the impedance will change in a circuit where it's required to decouple a rail with ripple present; as a filter or as a coupling capacitor.
Put a 690 to 1000uf capacitor across the switch. It should start in the off mode instead of the on mode
You might blow something up! I wouldn't do it. Even if it doesn't smoke everything immediately, in time it will ruin the rest of whatever you're putting it in.
the answer is 7v * 5
in a psu a general yes but the peek load with go up on the diodes before it you could go for a 47uF and the peek load with go down but you might get some hum.. if it is the start up cct from a switch mode and the thing starts with a 100uF or a 47uF then fine.. you can get 68uF 50V caps from some suppliers. and in any case the tolerances on these caps is + or - 20% ish anyway.. hope this helps.. Simple answer,,,Yes, you can.
5*7*v
Take two electrolytic capacitors of the same voltage and capacity, connect the positive leads together and connect the negative leads to the circuit. Just keep in mind that this will reduce the cap. value by half (2, 1000uF caps = 500uF) Also the voltage of the circuit should not exceed the voltage of one of the caps.
The answer to this question depends on the application. Generally no - manufacturers of devices design the components precisely, and if they've used a 35uf cap, they've probably done it for a reason. In an audio application the change of a capacitor value will give you a different tonal quality compared to the 35uF cap, though this isn't always a bad thing, though I wouldn't recommend changing it unless you're sure of the effect on the signal. In a power supply application, the capacitors are used to limit something known as ripple current. In this application the higher the capacitor value the lower the ripple current, which is something that is usually a good thing. If you want a more detailed answer i'd need to know more about the circuit you are working on.
While it is possible that the company producing the capacitor is using the NF as an identification tool, usually we would expect nF to specify the capacitance or value of the capacitor. For example, a capacitor with the value of 10nF ought to have a capacitance of about 10 nano Farads. This is equivalent to 10*10-9 Farads.
It will take slightly less than one second (0.92 seconds) to charge a 1000uF capacitor to 12 volts through a 1000 ohm resistor if your power source is 20 volts.The time constant of a 1000uF capacitor in series with a 1000 Ohm resistor is 1 second. (1x10-3 Farads times 1x10+3 Ohms = 1 second) It takes 1 time constant to reach 63% of a step change, 2 time constants to reach 86% of step change, and so forth using the equation VT = V0 (1 - e (-T/RC)). See notes below12 is 60% of 20, so it will take about 0.92 seconds for the capacitor to reach 12 volts. In two seconds the capacitor will reach about 17 volts. In five seconds, five time constants, the capacitor will be considered to be fully charged, 99.3%, to 20 volts.Notes:VT is voltage after a given number of seconds. V0 is the total initial voltage, in this case, 20 volts. -T/RC is the negative number of time constants for the exponential equation e-TC, which a charging capacitor exhibits. (More specifically, e-TC is the proportion of voltage across the resistor, while 1 - e-TC is the proportion of voltage across the capacitor.)This equation is based on the fundamental equation of a capacitor...dv/dt = i/c... which states that the slope of the voltage is proportional to current and inversely proportional to capacitance. Plug this into an initial state differential equation, for the case of charging an initially discharged capacitor through a resistor, and you getVT = V0 (1 - e (-T/RC))(Derivation requires calculus, and that seems a little bit out of scope for this question.)