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Two capacitors `C_(1) = 2 muF` and `C_(2) = 6 muF` in series, are connected in parallel to a third capacitor `C_(3) = 4 muF`. This arrangement is then connected to a battery of e.m.f `= 2V`, as shown in the figure. How much energy is lost by the battery in charging the capacitors A. `22 xx 10^(-6) J`B. `11 xx 10^(-6) J`C. `((32)/(3)) xx 10^(-6) J`D. `((16)/(3)) xx 10^(-6) J` |
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Answer» Correct Answer - B `C_(eq) = (C_(1)C_(2))/(C_(1) + C_(2)) + C_(3) = (2 xx 6)/(2 + 6) + 4 = 5.5 muF` Energy supplied `(E) = QV = CV^(2) = 22 xx 10^(-6) J` `P.E.` stored `(U) = (1)/(2) C_(eq)V^(2) = (1)/(2) xx 5.5 xx (2)^(2) = 11 xx 10^(-6) J` `rArr` Energy lost `= E - U = 11 xx 10^(-6) J` |
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