Discussion Forum



1)

 In given circuit, C1=C2=C3= C  initially Now a dielectric slab of dielectric constant   $K=\frac{3}{2}$  is inserted in C2

1652021554_p1.JPG

 

The equivalent capacitance become


A) $\frac{5C}{7}$

B) $\frac{7C}{5}$

C) $\frac{2C}{3}$

D) $\frac{C}{2}$

Answer:

Option A

Explanation:

 When dielectric slab of dielectric constant

 K= $\frac{3}{2} $ is  inserted between the plates of C2,

 Its new capacitance (C2') becomes

   C2'= $\frac{3}{2}$ C

 Equivalent capacitance of C2'  and C3 is

 $C_{eq}=C_2'+C_{3}=\frac{3}{2}C+C=\frac{5C}{2}$

 1652021634_pp.JPG

 Now , Ceq and C1 are in series . Therefore , their equivalent capacitancce is 

$C_{eq}=\frac{C_{eq}\times C_{1}}{C_{eq}+ C_{1}}=\frac{\frac{5C}{2}\times C}{\frac{5C}{2}+C}$

  $=\frac{5C^{2}}{7C}=\frac{5C}{7}$