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
