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QJEEM22Ph1782: JEE Main 2022 Physics - Class 12 JEEM22Ph1782 Solution: Two capacitors, each having capacitance are connected in…

Two capacitors, each having capacitance 40 μF40 \,\mu \mathrm{F} are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K\mathrm{K} such that the equivalence capacitance of the system became 24 μF24 \,\mu \mathrm{F} . The value of K\mathrm{K} will be :

Multiple Choice Options

  • A 1.5
  • B 2.5
  • C 1.2
  • D 3

Step-by-Step Verified Solution

JEE Main 2022 (Online) 28th July Morning Shift Physics - Capacitor Question 45 English Explanation

40K×4040K+40=24{{40K \times 40} \over {40K + 40}} = 24

40K=24(K+1)40K = 24(K + 1)

40K=24K+2440K = 24K + 24

16K=2416K = 24

K=2416=32=1.5K = {{24} \over {16}} = {3 \over 2} = 1.5

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