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Switch S shown in the figure below has been closed for a long time, and the electric circuit carries a constant current. Take

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Solution :- Given, C,= 3pF = 3x10 of C = 6uf = 6x100F R, = 4 kr = 4x10² R2 = 7 ks = 7x10?? Power in R₂, P2=2w when switch s iLet the current flowing in R, & R2 he current lowing in is So I. Power in R₂ , P2 = 1/R => IR, = 2 = I x7X10% = 2 => I= 20(a) Charge on C, is given by, - Q, = C, v Now voltage across C, is equal to Ve, when S is closed. = Q, = C, Va, = 3x106x 67.6(b) Alor many milliseconds Clong time) 1 voltage across Cz is equal to applied voltage v which is equal to sum of VR, & Vaz =

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