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Three series capacitors of capacitances 2.0, 3.0 and `6.0 mu F` are charged by a 60 V vattery. Find the total energy stored. |
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Answer» Here, `C_(1) = 2.0 mu F, C_(2) = 3.0 mu F`, `C_(3) = 6.0 mu F, V = 60 V`, As `(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (1)/(C_(2)) + (1)/(C_(3)) + (1)/(6) = 1` `:. C_(s) = 1 mu F = 10^(-6) F` `U = (1)/(2) C_(s) V^(2) = (1)/(2) xx 10^(-6) (60)^(2) = 1.8 xx 10^(-3) J` |
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