When replacing a capacitor in a circuit, it is generally safe to increase the capacitance value while keeping the voltage rating the same or higher. In this case, you can replace a 1000uF 50V capacitor with a 1200uF 50V capacitor without any issues. However, using a 2000uF capacitor may not be ideal as it significantly deviates from the original capacitance value, potentially affecting the circuit's performance. It is recommended to stick as close as possible to the original specifications for optimal results.
NO dont do it if you devalue a capacitors voltage parameter what will it do when it is face to face with 20V ? The voltage will destroy the thing. Remember it is OK for you to use a higher voltage parameter for a capacitor but never lower
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.)
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.
Well it depends on several things, does it need to be AC or DC? how much amperage must it be able to supply? For a very basic power supply you could us: A Step-down transformer A bridge rectifier (4 diodes of sufficient voltage and amp rating) Two capacitors; one at 1000uF and one at 0.01uF A 9 Volt voltage regulator (LM7809 is good) Depending on transformer and rectifier, this'll deliver 9V at up to 1A
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.
1000 microfarads is its rated capacitance, while 35 volts is its rated voltage.
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.
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
NO dont do it if you devalue a capacitors voltage parameter what will it do when it is face to face with 20V ? The voltage will destroy the thing. Remember it is OK for you to use a higher voltage parameter for a capacitor but never lower
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.
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.)
Filtering: When you see the output of a bridge rectifier fed with ac, you will find that the output is actually a dc + ac. The ac can be filtered out reasonably using a filter. The filters used often are capacitive input filters, and in its simplest form has just one capacitor. the capacitor charges to the peak value of the ac, and discharges when the diodes do not conduct, due to a fixed load. The capacitor is usually selected as 1000uF per ampere of load if a ripple of about 6V pp can be tolerated. For a low voltage supply (such as 9V), this is far too high, and one may use 3000uF/ amp of load current and get 2Vpp ripple, at mains of 60 Hz. the ripple frequency is twice, at 120 Hz. There is no regulating diode. One can have a regulator, and if it has large ripple rejection, one can use a lower value capacitor such as 1000uF for a 9V supply, and regulate the output to get a dc of 5V with low ripple of the order of 100mVpp or even lesser.
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.
first you need a analog multimeter. put in your meter in the ohm section x10. then remove the cap in the ckt and discharge that. then connect the multimeter -ve probe to the cap +ve terminal and connect the other terminals.the needle comes to low value then com to infinity, remove the probs then connect multimeters -ve prob to the cap -ve terminal that time look carefully the meter's needle comes to the one value and go back to the infinity that the low value is capacitors equivalent resistance value. the cap is 1000uf the equivalent resistance value is around 15 at x10 .the cap is 100uf the equivalent resistance value is around 15 at x100
Well it depends on several things, does it need to be AC or DC? how much amperage must it be able to supply? For a very basic power supply you could us: A Step-down transformer A bridge rectifier (4 diodes of sufficient voltage and amp rating) Two capacitors; one at 1000uF and one at 0.01uF A 9 Volt voltage regulator (LM7809 is good) Depending on transformer and rectifier, this'll deliver 9V at up to 1A