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2. Electric Potential and Capacitance
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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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