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Capacitor question

2022 · 28 Jul · Shift 1 · Q55
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  5. /2022 · 28 Jul · Shift 1 · Q55

Capacitor question

2022 · 28 Jul · Shift 1 · Q55

JEE MainPhysicsCapacitorMCQ+4 / −1
Two capacitors, each having capacitance 40 μF40 \,\mu \mathrm{F}40μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K\mathrm{K}K such that the equivalence capacitance of the system became 24 μF24 \,\mu \mathrm{F}24μF. The value of K\mathrm{K}K will be :
  1. A
    1.5
  2. B
    2.5
  3. C
    1.2
  4. D
    3
View written solutionFree

Correct answer: A

  1. Initial setup

Each capacitor has capacitance C=40 μFC = 40\,\mu FC=40μF

When a dielectric of constant KKK is inserted fully into one capacitor, its capacitance becomes C1=K⋅40 μFC_1 = K\cdot 40\,\mu FC1​=K⋅40μF

The other capacitor remains C2=40 μFC_2 = 40\,\mu FC2​=40μF

Since they are connected in series, the equivalent capacitance is Ceq=C1C2C1+C2C_{eq} = \frac{C_1 C_2}{C_1 + C_2}Ceq​=C1​+C2​C1​C2​​

Given: Ceq=24 μFC_{eq} = 24\,\mu FCeq​=24μF

  1. Substitute the values

24=(40K)(40)40K+4024 = \frac{(40K)(40)}{40K + 40}24=40K+40(40K)(40)​

Factor the denominator: 24=1600K40(K+1)24 = \frac{1600K}{40(K+1)}24=40(K+1)1600K​

24=40KK+124 = \frac{40K}{K+1}24=K+140K​

  1. Solve for KKK

24(K+1)=40K24(K+1) = 40K24(K+1)=40K

24K+24=40K24K + 24 = 40K24K+24=40K

24=16K24 = 16K24=16K

K=2416=1.5K = \frac{24}{16} = 1.5K=1624​=1.5

  1. Check the options

The correct option is: A: 1.5\boxed{A:~1.5}A: 1.5​

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