capacitors in series

A capacitor store electrical charge when capacitors are in series, the total capacitance drops and the voltage drops as pass to each capacitors the total charge is measure as Columbus= Qt= (Ct )(Vt) to find the total capacitance in series you have to use the same formula that we use in resistors in parallel that is (divide the value of each capacitors by one adding together and then divide the total by one) (1/5f+1/8f+1/12f)-1= Ct=2.44f Qt = the total charge of the capacitance is measure as Q=coulombs Qt= (Ct )(Vt) Qt=(2.44f)(5v) Qt= total charge of 12.2 coulombs find the voltage across each capacitor? Qt/C1 12.2/5f= 2.44v Qt/C2 12.2/8f= 1.52v Qt/C3 12.2/12=1.08v adding the voltage from each capacitors 2.44v +1.52v +1.08v= Vt=the total voltage of a power supply of 5volt all of this circuit is create by Jenko022 on via youtube

published 9 years ago