Question

A 27.0-uF capacitor and a 48.0-uF capacitor are charged by being connected across separate 20.0-V batteries.

A) determine the resulting charge on each capacitor (Give the answer in at least three sig figs)

27.0-uF capacitor _____mC

48.0-uF capacitor ____mC

B) 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?

27.0-uF capacitor _____uC

48.0-uF capacitor ____uC

C) What is the final potential difference across the 48.0-uF capacitor?

Answer 1

(a)

Charge on each capacitor is,

          q₁ = C₁V₁

               = (27.0 *10^-6 F) (20.0 V)

                = 540*10^-6 F or 540 μF

And q₂ = C₂V₂

               = (48.0 *10^-6 F) (20 V)

                = 960 *10^-6 F or 960 μF

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(b)

When the capacitors are connected in series, then the potential difference on the capacitors is,

                V₁ + V₂ = 40 V                       ...... (1)

   Since, charge on the two capacitors is same, V∞1/C

                    V₁ /V₂ = C₂ /C₁

                                 = 48 μF/ 27 μF

                                 = 1.78

                       V₁ = 1.78 V₂

Taking equation (1), we get

                      1.78V₂ + V₂ = 40 V

                                       V₂ = 14.4 V

             And,   V₁ = 1.78(14.4V)

                              = 25.63 V

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Charge on each capacitor is,

                     q₁ = C₁V₁

                          = (27.0 *10^-6 F) (25.63 V)

                          = 692.06*10^-6 F or 692.06 μF

And           q₂ = C₂V₂

                         = (48.0 *10^-6 F) (14.4 V)

                        = 691.2 *10^-6 F or 691.2 μF

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(c)

The potential difference across the 48.0-µF capacitor is 14.4 V