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## Homework Statement

So I came by this question yesterday which was in Solved examples part of my Textbook . I looked at the Solution and was not able to figure out how ?

The Question says .

A 5uF capacitor is first charged by applying a Potential Difference of 24 V . a different Capacitor of Capacitance 6uF is charged by applying a Potential Difference of 12 V . These capacitors are now connected such that the Positive plate of the first capacitor is connected to the Negative Plate of Second Capacitor and vice versa(does this mean they are connected in series or is there a different name for such configuration?) . Find the new Potential Difference and the New charge on the capacitors .

## Homework Equations

The only formula used would be Q = CV

## The Attempt at a Solution

The solution had this C1 = 5uF V1 = 24uF therefore = Q1 = 120uC

C2 = 6uF V1 = 12uF therefore = Q2 = 72uC

Then they found out what is called effective potential (thats what confusing me ) . They used this equation

V{effective}(C1+C2) = V1C1 - V2C2

V{eff} = 4.36 V

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then they Calculated the V{effective} and the new charges . what I am not able to understand is how did they came by this formula ??

Also when the configuration is like Positive plate is connected to positive and negative to negative(does this mean they are connected in parallelor is there a different name for such configuration?) the effective Potential difference comes out to be 17.45 V . How ? Please explain I just cant get it .

Thanks in advance