Question is in the title.
Capacitor uF rating: 1000uF, voltage rating: 25V.
Circuit is here: http://gyazo.com/4d1de54869f2fcf16d4a65a74ea85504
Thanks!
Yes. It will several time constants to charge up.
one time constant is 5k x 1 mF = 5 seconds, so it will take about 30 seconds to charge.
But you can reduce the 4.7k as low as you want to decrease the time. Even to zero.
edit: if you reduce R to zero, the charge time will be limited by the internal resistance of the cap, probably a few ohms. It will take less than 1 second. It will still charge only to 18 volts.
Yes. Only the voltage of the batteries is important here. The value of the resistor and capacitor don't matter except in determining the time it takes to charge the cap. At present, the cap will take about 20 seconds to fully charge.
Time constant is calculated as the resistance times the capacitance. In your case 1mx4700 or 4.7 seconds. And then it takes 4-5 time constants depending on what one considers "fully charged" So choose values and recalculate accordingly. If you reduce the value of the resistor to zero the cap won't charge immediately, as the batteries and wires will have some internal resistance and stray inductance. The resistor smooths the charging, reducing large instantaneous currents and LC ringing.
Those saying the voltage across the capacitor will 'never' reach the battery voltage don't live in the real world.
After 10 time constants (50s), it will be 99.995% of the supply voltage. After 2 minutes, 99.999999996% of the supply voltage. Other physical factors like the battery draining and the fact that voltage cannot be measured to this level of accuracy come into effect.
After just 4 minutes it will be 99.99999999999999999985%
Strictly speaking, the voltage across the capacitor rises asymptotic to 18V, so it never fully reaches 18V. In practice we would take it to be charged after 5CR seconds, which is approximately the time to reach 99% of the asymptotic limit.
a capacitor is within 1% of being fully charged in 5 time constants, that's 5 x 1000^-6 x 4.7^3 = 23.5 sec
with a 0.1 ohm resistor its 0.0005 seconds.
the capacitor achieves a rise time between 10 and 90% in 2.2 (cr) time constants which is called tau (greek letter)
People who say that the capacitor will charge to 18 v are incorrect. Given enough time the capacitor will charge as close, even infinitesimally close, as you like to the battery voltage; however, the capacitor will never reach 100% of the supply voltage. This is because the voltage across the capacitor approaches the supply voltage in an exponential fashion, meaning the capacitor voltage never quite reaches the supply voltage..
Yes, and the resistor value only changes how long it takes to charge.
Yes, but it does not need to use any resistor in series.
Hi,
Question is in the title.
Capacitor uF rating: 1000uF, voltage rating: 25V.
Circuit is here: http://gyazo.com/4d1de54869f2fcf16d4a65a74ea85504
Thanks!
