Two capacitors (C) = 6.3uF, Cy = 19.5uF) are charged individually to (V) = 16.2V, V5 = 3.6V). The two capacitors are then connected together in parallel with the positive plates together and the negative plates together. 2a Calculate the final potential difference (in 1) across the plates of the capacitors ence they are connected. 2b Following the connection in (a) above, calculate the amount of charge (absolute value, in ') that flows from one capacitor to the other when the capacitors are connected together, Please state your answer in Sl units and multiply the final answer by 10. 2 By how much (absolute value) is the total stored energy (in J) reduced when the two capacitors are connected? Please state your answer in Sl units and multiply the final answer by 10%. Given two Capacitors - @ 1 = 6:3MF C 2 = 19 . 5 MF V, = 16 . 2 V V2 = 3 . 6 V Charges of individual capacitors -> We know ( c = Q ) V So , 8. = c, V, = 6 3 X 10 x 1612 - 9, = 1 02 x10 0 19/2 = 2 12 = 3. 6 x 10 *1915 -> 8 = 7 02 x10 4 C Now let's connect the two capacitors parallely -> Equivalent capacitance Ceg= ", + 0 2 7 0 = ( 6 . 3 + 19, 5) x10 6 - = 25. 8x10 6 FFrom Conservation of charge , we say Initial Charge = Final Charge = => cv , + 2 Vz = cear ear -> (6. 3 x10 6 x 18.2) + (19,5 x10 6 x 3.6) = (25. 8 x18 ) x Veg 7 Vea = 6:6 8 V - This is the common potential difference across the parallely connected capacitors. ". Final potential after connection Near = 6.68VNow, Beg = 8,+ 82 = Cea lea ~ 172. 29 x10 'e After connection : ING ME here ; Vee = 6.6 8 V 2, = 6.3 X 10 OF 8 0 2 - 19.5 x10 6 F so , we can calculate the charges 9 , = e , veg = 6.3 x10 6 x 8.68 = 42, 08 x10 6 e 8 ) = 0 , v = 19.5 x10 x 6,68 - 130.26 */10 6 e Now ; if we add Q, f, we get 172.34 x10 6 e which is geg indeed Thi of charge - 1AB, = Q,'NO = 42.08 x10- ~ 102. 06 * 10 8 = 60 *10 6 C *AQ = Q2ND = 130,26 *10 N 70. 2x 10 = 60 * 10 -6 0 :. The flow of charge from one capacitor to another is 60x10 e = 6x10-5 = 6 -Multiplied T day 105 ( im SI )