Two capacitors, C1 = 23.0 µF and C2 = 41.0 µF, are connected in series, and a 21

Question

Two capacitors, C1 = 23.0 µF and C2 = 41.0 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= 14.734F
total energy stored= 3.25e-3 J

(b) Find the energy stored in each individual capacitor.
energy stored in C1= .00208 J
energy stored in C2= .001167 J

(c) If the same capacitors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy as in part _________V

I have everything that I believe you need to answere this question. Parts A and B are both correct. I am just looking for some help with C.

Explanation / Answer

a) equivalent capacitance C= c1c2/c1+c2 = 14.734 µF V = 21 V energy stored = .5CV^2 = 3.25e-3 J b) voltage across c1 = c2V/(c1+c2) = 13.45 V energy across c1 = .5c1(13.45)^2 = .00208 J voltage across c2 = 21-13.45 = 7.55 energy across c2 - .5c2(7.55)^2 = .001167 J c) C = c1 + c2 = 23+41 = 64 µF energy = .5*(64*10^-6)*v^2 = 3.25e-3 v = 10.1 volts

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