Dec 3, 2007 #1 R rocko75 Newbie level 1 Joined Dec 3, 2007 Messages 1 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,290 how do find the charge time of a capacitor in series with a resistor? Question for school
Dec 3, 2007 #2 P pabloec Member level 2 Joined Sep 20, 2007 Messages 47 Helped 9 Reputation 18 Reaction score 3 Trophy points 1,288 Activity points 1,551 capacitor charging Very good page http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html
Dec 3, 2007 #3 B barrybear Full Member level 4 Joined Nov 22, 2001 Messages 239 Helped 14 Reputation 28 Reaction score 4 Trophy points 1,298 Activity points 2,035 Re: capacitor charging CxR=T C= capacitance in Farads R= resistance in ohms T= time in seconds In t time the voltage across the capacitor wil be .6v
Re: capacitor charging CxR=T C= capacitance in Farads R= resistance in ohms T= time in seconds In t time the voltage across the capacitor wil be .6v
Dec 4, 2007 #4 E electronics_kumar Advanced Member level 2 Joined Nov 29, 2004 Messages 657 Helped 34 Reputation 68 Reaction score 9 Trophy points 1,298 Location Tamilnadu Activity points 5,552 capacitor charging it charging duration depends on its time constant RC ..Usually five times RC sec...
capacitor charging it charging duration depends on its time constant RC ..Usually five times RC sec...
Dec 5, 2007 #5 V vivek_bahire Member level 1 Joined Aug 30, 2005 Messages 40 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,470 Re: capacitor charging Vc = Vf-(vf-Vi)*e^(-t/(R*C)) Vf= vfinal Vi=Vinitial
Dec 6, 2007 #6 K Kral Advanced Member level 4 Joined Mar 28, 2005 Messages 1,319 Helped 280 Reputation 558 Reaction score 84 Trophy points 1,328 Location USA Activity points 13,401 Re: capacitor charging rocko75, Assuming that the capacitor is initially dischharged, the following equation works: Let Vin = the input voltage to the resistor Vout = the capacitor voltage at the time of interest Then t = -ln[(Vin - Vout)/Vin] Regards, Kral
Re: capacitor charging rocko75, Assuming that the capacitor is initially dischharged, the following equation works: Let Vin = the input voltage to the resistor Vout = the capacitor voltage at the time of interest Then t = -ln[(Vin - Vout)/Vin] Regards, Kral
Dec 6, 2007 #7 M miky_boy Newbie level 4 Joined Feb 23, 2006 Messages 6 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,309 Re: capacitor charging Fine thank's