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
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)` is
A. `2:1`
B. `4:1`
C. `8:1`
D. `16:1`
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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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