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A capacitor `C_(1)=4mu` F is connected is series with another capacitor `C_(2)=1mu`F the combination is connected across DC source of 200 V the ratio of potential across `C_(2)` to `C_(1)` isA. `2:1`B. `4:1`C. `8:1`D. `16:1` |
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Answer» Correct Answer - a In series combination `(1)/(C_(s))+(1)/(C_(2))=1/4+1/1=5/4` `C_(s)=4/5 mu F` now total charge in series combination is given by `q=c_(s)xxV=4/5 xx200 =600 V` `therefore V_(1)` potential across `R_(1)=(160)/(C_(1))=160/4=40 V` so `(V_(2))/(V_(1))=160/40=4/1 rarr V_(2) L: V_(1)=4:1` |
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