I have two questions. 2) Suppose after charging a 4.7 mu F capacitor to 5 V, you

Question

I have two questions.

2)

Suppose after charging a 4.7 mu F capacitor to 5 V, you connect it to the second uncharged capacitor of 14 mu F. What voltage will your voltmeter read across the second capacitor after repeating the charge sharing process two times? Express the answer with one decimal place. Five capacitors are connected across a potential difference V ab as shown in Fig. Because of the dielectrics used, each capacitor will break down if the potential across it is V=28 V. What should be the value of V ab (with one decimal place) if you reach the breakdown voltage one of the capacitor.

Explanation / Answer

You must use 4 principles in answering this: (a) since the caps are joined in parallel their final voltage (V) will be equal; (b) charge (Q) is conserved, so after having made the connection, the final charge wil be equal to the sum of the 2 initial charges; (c) Q = V*C in all cases (C = capacitance); (d) when caps are joined in parallel the total capacitance (Ctot) will be the sum of the individual capacitances.


Let initial charge on 4.7uF cap be Q, that on 14uf cap be zero. Total charge = Q.

Q = V*C so for initial conditions Q = 4.7*5uC=23.5uC

For conditions when caps are in parallel V = Q/Ctot = 23.5/18.7 = 1.25V.

you can refer this

http://answers.yahoo.com/question/index?qid=20120205195321AAzI6vO

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