posted by Sara on .
I have a circuit with three capacitors and a battery. The battery supplies 15V. The battery is in series with C3. C3 is in series with ( C23).
C3= 3*10^-6 F
C2= 2 *10^-6 F
C1= 1*10^-6 F
(respectively C2 +C3 are is parallel withe ach other).
I know the equivalent capacitance which is 1.5 *10^-6 F
But what I need to know is how to find the energy stored on Capacitor 1 (C1)
I meant C3 is in series with C2 and C1
which means that C2 and C1 are in parallel
Ceq(c1,c2):3uF. This combination is in series with the 3uF capacitor and both are connected to the 15v battery. The voltage will get divided equally since they are both of the same value(7.5v). Hence energy would be. Both c1 and c2 have 7.5v across them since they are in parallel and hence energy can be calculated as:0.5*1*7.5^2=28.125uJ