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Consider the following. (Let C_1 = 27.20 muF and C

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Answer #1

A) c2||6uF =21.2 + 6 =27.2 uf

1/Ceq = 1/C1 + 1/27.2 + 1/ C1 = 3/27.2

Ceq= 9.07 uF

b) voltage will be divided equally among capacitors as their capacitance is same, each equal to 27.2uF

V across all capacitors will be 9/3 = 3V

Charge on right 27.2 uf capacitor= 27.2×3= 81.6uC

Charge on left 27.2 uf capacitor= 27.2×3= 81.6uC

Charge on 21.2 uf capacitor= 21.2×3= 63.6uC

Charge on right 6 uf capacitor= 6×3= 18 uC

C) potential difference is already found in previous part

potential difference across right 27.2 uF capacitor = 3V

  potential difference across left 27.2 uF capacitor = 3V

potential difference across 21.2 uF capacitor = 3V

potential difference across 6 uF capacitor = 3V

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