2. Electric Potential and Capacitance
medium

Two capacitors, each having capacitance $40\,\mu F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\,\mu F$. The value of $K$ will be.

A

$1.5$

B

$2.5$

C

$1.2$

D

$3$

(JEE MAIN-2022)

Solution

$C _{ eq }=\frac{ C ( KC )}{ C + KC }=\frac{ KC }{ K +1}$

$24=\frac{K 40}{K+1}$

${[K=1 \cdot 5] }$

Standard 12
Physics

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