Answer this question??????

Dear Student
Let C1 = 2 μF     C2 = 8  μF    When the switch is connected to 1 then energy dissipated isU = 12C V2                       U = 12 ×2μF V2     When the switch is connected to 2, then energy dissipated is U = 12  C1C2C1+C2 V22  U = 12  2μF×8μF 10μFV2= 12 ×1.6μF V2 Now the % of stored charge =   U Ux 100= 12 ×1.6μF V2 12 ×2μF V2 ×100 = 80%                          

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