Please help me answer this question with detailed explanantion. I now that the c

Question

Please help me answer this question with detailed explanantion. I now that the capacitors will be in parellel and the charges before they are disconnected will be Q1 = 150 micro C and Q2 = 350 microC

1) (6 points) Capacitor Ci = 3.0 F is charged to 50 V and capacitor C2-5.0 F is charged These capacitors are then disconnected from their batteries. Now the positive to 70 V. plate of Ci is connected to the negative plate of C2 and the negative plate of Ci is connected the positive plate of C2. Find (Show your work) a) The new potential difference across each capacitor. b) The new charge on each capacitor c) The total energy stored on capacitors before and after they are connected to each other. Sov

Explanation / Answer

When the switch ‘S’ is closed both the capacitors A and B are at same potential V. Since they have same capacities also, so charge qA and qB on both of them is

qA = qB = CV

Total energy of the system E1

Or, E1 = ½ CV2 + ½ CV2 = CV2   …... (1)

E1 =CV2 =3 x502 =7.5 KJ

E2 =CV2 =5 x702 =2.45 KJ

Total energy=

E1 + E2 =7.5+2.45=9.95 KJ

.

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