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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)/(2)) xx 10^(-6) J`D. `((16)/(3)) xx10^(-6) J` |
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Answer» Correct Answer - B `E_(lost)=(1)/(2)C_(eff)V^(2)`. |
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