A 21.0-mu F capacitor and a 48.0-mu F capacitor are charged by being connected a

Question

A 21.0-mu F capacitor and a 48.0-mu F capacitor are charged by being connected across separate 55.0-V batteries. Determine the resulting charge on each capacitor. (Give your answer to at least three significant figures.) 21.0- mu F capacitor 48.0- mu F capacitor The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor? 21.0- mu F capacitor 48.0-mu F capacitor What is the final potential difference across the 48.0- mu F capacitor?

Explanation / Answer

(a) We know that Q = CV

So Q = 21 * 10^-6 * 55 = 1155 *10^-6 = 1155 microcouloumb

In case of 48 microfarad capacitor

Q = 48 *10^-6 * 55 = 2640 microcoulumb

(b) When they are both connected in reverse fashion i.e positive plate to negative and negative plate to positive,

Let x is the amount of charge moves from one capacitor to another to neutralize.

So We have Charge restructuring as

Q1 -x   -Q1 +x    +q1 -x and -q1 +x

Now finally the potential should be equal.

So (Q1-x)/C1 = (x -q1)/C2

So Solving gives x = 1607 micro coulumb

So Final charge on 48 is (2640-1607) = 1033 and on 21 capacitor is 1607 - 1155 = 452

Final potential is 1033 / 48 = 21.5 V

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