1.

When n, identical capacitors are connected in series then equivalent capacitance becomes x, and when the same are connected in parallel then equivalent capacitance becomes y. Calculate y/x.

Answer»

Solution :`(1)/(C_(eq))=(1)/(C )+(1)/(C )+…`N times
`RARR C_(eq)=(C )/(n)=x "" …(i)`
Equivalent capacitanceof n identical CAPACITORS connected in PARALLEL is
`C_(eq)=C+C+C….` n times
`rArr C_(eq)=NC=y "" …(ii)`
Dividing (ii) by (i),
`(nC)/(C//n)=(y)/(x)`
`rArr n^(2)=y//x`
`or (y)/(x)=n^(2)`


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