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Three capacitors `2, 3` and `6 mF` are joined in series with each other. What is the minimum effective capacitanceA. `(1)/(2)muF`B. `1 muF`C. `2 muF`D. `3 muF` |
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Answer» Correct Answer - B `(1)/(C_(eq)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3))` `= (1)/(2)+(1)/(3)+(1)/(6)` `= (3+2+1)/(6) = (6)/(6) = 1 muF` |
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