Answer to Question #159605 in Electric Circuits for Samychou

Question

A 2.5 x 10^(-6)F capacitor is charged to a potential difference if 100V and is disconnected from the supply. Its terminals are then connected to those of an uncharged 10 x10^(-6)F capacitor. Find
(a) the resulting potential difference across the two capacitors and
(b) the total energy stored in them.
Compare the result in (b) with the energy originally stored in the 2.5 x 10^(-6)F capacitor and comment on the difference.

Answer 1

Given quantites:

$C_1 = 2.5*10^{-6}F$ $U_1=100V$ $C_2=10*10^{-6}F$ $q_2 = 0$

Those capacitors are onnected in series the charge flow from one two another up to their chaarges to equal

$q_1 = C_1U = 2.5*10^{-4}C$

$q = q_1'=q_2' = \large\frac{q_1+q_2}{2}$ $=1.25*10^{-4}C$

(a) the resulting potential difference across the two capacitors and

$\Delta\varphi=\large\frac{q}{C_t}= \frac{q(C_1+C_2)}{C_1*C_2} = \frac{1.25*10^{-4}*12.5*10^{-6}}{2.5*10^{-6}*10^{-5}}$ $=62.5V$

(b) the total energy stored in them.

$W = \large\frac{qU}{2}=\frac{1.25*10^{-4}*62.5}{2}$ $=3.9*10^{-4}J$