JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(3\)
Answer & Solution
Correct Answer
(A) \(1.5\)
Step-by-step Solution
Detailed explanation
\(C _{ eq }=\frac{ C ( KC )}{ C + KC }=\frac{ KC }{ K +1}\) \(24=\frac{K 40}{K+1}\) \({[K=1 \cdot 5] }\)
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