Two capacitors, C1 = 27.0 µF and C2 = 35.0 µF, are connected in series, and a 21.0 V battery is connected across them.
(a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor.
equivalent capacitance=
µF
total energy stored=
(b) Find the energy stored in each individual capacitor.
energy stored in C1 J
energy stored in C2
The charge is the same on both
Q = C V
so C1V1 = C2 V2
V2 = (C1/C2)V1 and V1 + V2 = 21
let C1 = 27*10^-6
let C2 = 35*10^-6
then C1/C2 = .77
then V2 = .77 V1
1.77 V1 = 21
V1 = 11.9 volts
V2 = 9.1 volts
I think you can take it from there
To find the equivalent capacitance when capacitors are connected in series, you can use the formula:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...
In this case, there are only two capacitors connected in series, so the formula simplifies to:
1/Ceq = 1/C1 + 1/C2
Let's substitute the given values:
1/Ceq = 1/27 µF + 1/35 µF
To combine the fractions, you need to find a common denominator. The least common multiple of 27 and 35 is 945. So the equation becomes:
1/Ceq = (35/945 + 27/945) µF
Simplifying further:
1/Ceq = (62/945) µF
Now you can take the reciprocal of both sides to get the equivalent capacitance (Ceq):
Ceq = 945/62 µF ≈ 15.24 µF
So, the equivalent capacitance is approximately 15.24 µF.
To find the energy contained in the equivalent capacitor, you can use the formula:
E = (1/2) * C * V^2
where E is the energy, C is the capacitance, and V is the voltage.
Substituting the values:
E = (1/2) * 15.24 µF * (21.0 V)^2
E ≈ 3220.74 µJ
Therefore, the total energy stored in the equivalent capacitor is approximately 3220.74 µJ.
To find the energy stored in each individual capacitor, you can use the formula:
E = (1/2) * C * V^2
For C1:
E1 = (1/2) * 27.0 µF * (21.0 V)^2
Simplifying:
E1 ≈ 6172.5 µJ
Therefore, the energy stored in C1 is approximately 6172.5 µJ.
For C2:
E2 = (1/2) * 35.0 µF * (21.0 V)^2
Simplifying:
E2 ≈ 8082.75 µJ
Therefore, the energy stored in C2 is approximately 8082.75 µJ.