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A capacitor of capacitance `10 m F` is charged up a potential difference of `2V` and then the cell is removed. Now it is connected to a cell of `emf 4V` and is charged fully. In both cases the polarities of the two cells are in the same directions. Total heat produced in the second charging process is :A. `10 mJ`B. `20 muJ`C. `40 muJ`D. `80 muJ` |
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Answer» Correct Answer - B b. Energy stored in capacitor when it is charged up to 2 V is `1/2 xx 10 xx 2^2 = 20 muJ = u_1` (suppose) Energy stored in capacitor when it is charged up to 4 V is `1/2 xx 10 xx 4^2 = 80 muJ = u_2` (suppose) Increase in charge = 40 -20 = `20muC`. Energy drawn from cell `= 20 xx 4 = 80 muC` = u (suppose) heat produced ` = u_1 + u-u_2` `= 20 + 80 - 80 = 20 muJ` . |
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