Test Index

VITEEE 2016 Paper

Section: Physics

Question : 4 of 125

Marks: +1, -0

C_1

C_2

C_3

\frac{3}{2}

C_2

The equivalent capacitance become

Solution:

When a dielectric slab of dielectric constant

K =

\frac{3}{2}

is inserted between the plates of

C_2

, its new capacitance

(C_2')

become

C_2'

\frac{3}{2}

Equivalent capacitance of

C_2'

and

C_3

C_{eq}

C_2' + C_3

\frac{3}{2}C

+ C =

\frac{5C}{2}

Now,

C_{eq}

and

C_1

are in series. Therefore, their equivalent capacitance is

C_{eq}

\frac{C_{eq} \times C_1}{C_{eq}+C_1}

\frac{2 \times \frac{5C}{2} \times C}{\frac{5C}{2}+C}

\frac{5C^2}{7C}

\frac{5C}{7}

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