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
To find the energy stored on capacitor C1, you can use the formula for the energy stored in a capacitor:
E = (1/2) * C * V^2
where E is the energy stored, C is the capacitance, and V is the voltage across the capacitor.
In this case, you know the capacitance of C1 (1 * 10^-6 F) and the voltage across it. However, you need to determine the voltage across C1 first.
To do that, you can analyze the circuit and use the concept of charge conservation. Since C1 is in series with C23, they share the same charge. The charge on C23 can be found using the equation:
Q = Ceq * V
where Q is the charge, Ceq is the equivalent capacitance (1.5 * 10^-6 F), and V is the voltage (15V).
Substituting the values:
Q = (1.5 * 10^-6 F) * 15V
Q = 2.25 * 10^-5 C
Since C23 and C1 have the same charge, the voltage across C1 can be found using the equation:
V1 = Q / C1
Substituting the values:
V1 = (2.25 * 10^-5 C) / (1 * 10^-6 F)
V1 = 22.5 V
Now that you know the voltage across C1 (22.5V), you can plug it into the energy formula to find the energy stored on C1:
E1 = (1/2) * (1 * 10^-6 F) * (22.5V)^2
E1 = 0.253 J (Joules)
Therefore, the energy stored on capacitor C1 is 0.253 Joules.