A 10.0-µF capacitor is charged so that the potential difference between its plat

Question

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V. The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate.

Find the charch that flows from one capacitor to the other when the capacitors are connected (17 microcoulombs) and the energy that is dissipated when the two capacitors are connected together (42 microjoules)

i think that those are the answers, can you please show m how to get to that answer?

Explanation / Answer

initial charge on capacitor 1 Q1= C1V1 = 10*10 = 100 µC
initial charge on second capacitor = C2V2 = 25 µC

Here charge is conserved
Final total charge = Q1' + Q2' = 100+25 = 125 µC
since they are connected in parallel
potential difference across them will be same
so
Q1'/C1 = Q2'/C2
Q2' = Q1'(5/10) = Q1'/2

plug this in the first equation we got
Q1'+Q1'/2 = 125 µC
Q1' = 125*2/3 = 250/3

V1'=V2'=V = Q1'/C1 = 25/3
charge flown = 100 - 250/3 = 50/3 = 16.7 17 µC

Initial energy = .5C1V1^2 + .5C2V2^2 = 562.5 x 10^-6

Final energy = .5(C1+C2)V^2 = 520.83 x 10^-6

energy dissipated = 562.5 x 10^-6 - 520.83 x 10^-6 = 41.7 J = 42 micro joules

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