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Four identical capacitors are connected as shown in diagram. When a battery of `6 V` is connected between `A` and `B`, the charges stored is found to be `1.5 muC`. The value of `C_(1)` is A. `2.5 muF`B. `15 muF`C. `1.5 muF`D. `0.1 muF` |
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Answer» Correct Answer - D The capacitance across `A` and `B` `= (C_(1))/(2) + C_(1) + C_(1) = (5)/(2)C_(1)` As `Q = CV`, `1.5 muC = (5)/(2)C_(1) xx 6` `rArrC_(1) = (1.5)/(15) xx 10^(-6) = 0.1 xx 10^(-6) F = 0.1muF` |
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