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Two capacitors of `3 pF` and `6pF` are connected in series and a potential difference of `5000 V` is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates isA. `2250 V`B. `2222 V`C. `2.25 xx 10^(6)V`D. `1.1 xx 10^(6 V` |
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Answer» Correct Answer - B `(1)/(C ) = (1)/(3) + (1)/(6) rArr C = 2pE` Total charge `= 2 xx 10^(-12) xx 5000 = 10^(-8) C` The new potential when the capacitors are connected in parallel is `V = (2 xx 10^(-8))/((3 + 6) xx 10^(-12)) = 2222 V` |
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