Answer to Question #159591 in Electric Circuits for Ivo

Question

Two capacitors C1 =18.0 mF and C2 = 36.0 mF are connected in series and a 12.0V battery is connected across the two capacitors. Find
(i) the equivalent capacitance and the energy stored in this equivalent capacitance.
(ii) Find the energy stored in each individual capacitor

Answer 1

$C_1 = 18.0mkF$ $C_2 = 36.0mkF$ $U = 12.0V$

Two capacitors are connected in series.

(i) the equivalent capacitance and the energy stored in this equivalent capacitance.

the equivalent capacitance for connected in series is $C = \large\frac{C_1*C_2}{C_1+C_2}= \frac{18*36}{18+36}$ $=12mkF$

 the energy stored in this equivalent capacitance. $W = \large\frac{CU^2}{2}$ $= \large\frac{12*144}{2}$ $=864mkJ$

(ii) Find the energy stored in each individual capacitor

the charges in capacitors and total charge for system is equal to each other in connected series.

$q = q_1 = q_2 = CU= 12mkF*12V = 144mkC$

for the the first capacitor $W_1 = \large\frac{q_1^2}{2C_1}$ $= \large\frac{144*10^{-6}*144*10^{-6}}{2*18*10^{-6}}$ $= 576mkJ$

for the the second capacitor $W_2 = \large\frac{q_2^2}{2C_2}$ $= \large\frac{144*10^{-6}*144*10^{-6}}{2*36*10^{-6}} = 288mkJ$