Two capacitors, one that has a capacitance of 5 F and one that has a capacitance

Question

Two capacitors, one that has a capacitance of 5 F and one that has a capacitance of 15 F are first discharged and then are connected in series. The series combination is then connected across the terminals of a 11-V battery. Next, they are carefully disconnected so that they are not discharged and they are then reconnected to each other--positive plate to positive plate and negative plate to negative plate (a) Find the potential difference across each capacitor after they are reconnected 3.75 x v (5 F capacitor) V (15 F capacitor) (b) Find the energy stored in the capacitor before they are disconnected from the battery, and find the energy stored after they are reconnected (before disconnected) H (reconnected)

Explanation / Answer

(A) when connected in series.

Ceq = C1 C2 / (C1 + C2) = (5 x 15)/(15+ 5)

Ceq = 3.75 uF  

total charge , Q = Ceq V = (3.75 uF) (11)

= 41.25 uC  

now voltage cross them is V then applying charge conservation,

41.25 = (15 + 5)V

V = 2.06 Volt

(A) cross 5uF = 2.06 Volt

across 15uF = 2.06 Volt

(B) Ui = (3.75 x 10^-6) (11^2 ) /2  

Ui = 227 x 10^-6 J Or 227 uJ

Uf = (20 x 10^-6) (2.06^2)2/

Uf = 42.4 x 10^-6 J OR 42.4 uJ

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