Two capacitors C1 = 5.8 F, C2 = 17.5 F are charged individually to v1 = 19.1 v,
ID: 1581524 • Letter: T
Question
Two capacitors C1 = 5.8 F, C2 = 17.5 F are charged individually to v1 = 19.1 v, v2 = 3.0 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. 7 V Submit Answer Submission not graded. Use more digits. Tries 0/10 Previous Tries Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. 70.2MC Pick one capacitor, you know the charge on this capacitor before they were connected Now that you know the potential difference after they are connected - remember the otential drop is the same for both of them-you can calculate the charge: O=CV. You ave to calculate the difference of Q before and after. Submit Answer Incorrect. Tries 3/10 Previous Tries By how much (absolute value) is the total stored energy reduced when the two capacitors are connected 70.2C Incompatible units. No conversion found between "c" and the required units. > Previous Tries Submit Answer Tries 0/10Explanation / Answer
initially
Q1 = charge stored by capacitor C1 = C1 V1 = 5.8 x 19.1 = 110.8 uC
Q2 = charge stored by capacitor C2 = C2 V2 = 17.5 x 3 = 52.5 uC
after connecting in parallel , the charges redistribute to Q1' and Q2' repectively so as to have equal potential on each capacitor as Voltage remain same in parallel
using conservation of charge
Q1' + Q2' = Q1 + Q2 = 110.8 + 52.5 = 163.3 uC
Q1' = 163.3 - Q2' eq-1
also , Q1'/C1 = Q2'/C2
(163.3 - Q2')/5.8 = Q2'/17.5
Q2' = 122.7 uC
using eq-1
Q1' = 163.3 - Q2' = 163.3 - 122.7 = 40.6 uC
V = common potential = Q1'/C1 = 40.6/5.8 = 7 Volts
amount of charge flowed = Q1 - Q1' = 110.8 - 40.6 = 70.2 uC
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