Two capacitors C_1 = 8.0 mu F, C_2 = 16.0 mu F are charged individually to V_1 =
ID: 1414276 • Letter: T
Question
Two capacitors C_1 = 8.0 mu F, C_2 = 16.0 mu F are charged individually to V_1 = 14.0 V, V_2 = 3.3 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together. Calculate the final potential difference across the plates of the capacitors once they are connected. Remember capacitance in parallel: C_T = C_1 + C_2. The total charge is the sum of the individual charges. The voltage drop across both capacitors is the same. Submit Answer Tries 4/20 Calculate the amount of charge that flows from one capacitor to the other when the capacitors are connected together.______________Submit Answer Tries 0/20 By how much is the total stored energy reduced when the two capacitors are connected? Submit Answer Tries 0/20Explanation / Answer
in capacitor charqe Q = C V
Part A -
Q = C1 V1 + C2 V2 = ( C1 + C2 ) V
( 8 * 14 +16 * 3.3) * 10^(-6) = ( 8+ 16 )* 10^(-6) * V
V = 6.87 volt ......................................................................Ans
Part B
Q1 (initial) = C1 V1 = 112 * 10^(-6) coloumb Q1(finial) = C1 V = 54.96 * 10^(-6) coloumb
Q2 (initial) = C2V2 = 52.8 * 10^(-6) coloumb Q2(finial) = C2 V = 109.92 * 10^(-6) coloumb
charge flow from C1 to C2 = 59.2 * 10^(-6) coloumb .....................................Ans
Part B
total energy reduced = 1/2 * C1 * V1^2 + 1/2 *C2 * V2^2 -1/2 (C1+C2) V^2
= ( 0.5 * 10^(-6)) [ 8*14*14 +16*3.3*3.3 - 24 * 6.87 *6.87 ]
= 304.75 * 10^(-6) joule ............................................Ans
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