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Three capacitors of `2 muF, 3 muF` and `6 muF` are joined in series and the combination is charged by means of a `24` volt battery. The potential difference between the plates of the `6 muF` capacitor is
A. `4` volt
B. `6` volt
C. `8` volt
D. `10` volt

1 Answer

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Best answer
Correct Answer - A
`(1)/(C_(eq)) = (1)/(2) + (1)/(3) + (1)/(6) rArr C_(eq) = 1muF`
Total charge `Q = C_(eq).V = 1 xx 24 = 24 muC`
Potential diff. across `6muF` capacitor `= (24)/(6) = 4` volt

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