In given circuit, C1=C2=C3=C initially. Now, a dielectric slab of dielectric constant K =(3/2) is inserted in C 2. The equivalent capacitance become

Question

In given circuit, C1=C2=C3=C initially. Now, a dielectric slab of dielectric constant K =(3/2) is inserted in C 2. The equivalent capacitance become (A) (5C/7) (B) (7C/5) (C) (2C/3) (D) (C/2)

Tardigrade Admin · Accepted Answer

When a dielectric slab of dielectric constant

K = \frac{3}{2}

is inserted between the plates of

C_{2}

, its new capacitance

(C_{2}^{'})

becomes

(C_{2}^{'}) = \frac{3}{2} C

Equivalent capacitance of

(C_{2}^{'})

and

C_{3}

is

C_{e q} = C_{2}^{'} + C_{3} = \frac{3}{2} C + C = \frac{5 C}{2}

Now,

C_{e q}

and

C_{1}

are in series.
Therefore, their equivalent capacitance is

C_{e q} = \frac{C _{e q} \times C _{1}}{C _{e q} + C _{1}} = \frac{\frac{5 C}{\frac{2}{5}} \times C}{\frac{5 C}{2} + C}

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

Q. In given circuit, $C_{1} = C_{2} = C_{3} = C$ initially. Now, a dielectric slab of dielectric constant $K = \frac{3}{2}$ is inserted in $C_{2}$ .
The equivalent capacitance become

$\frac{5 C}{7}$

$\frac{7 C}{5}$

$\frac{2 C}{3}$

$\frac{C}{2}$

Q. In given circuit, C1​=C2​=C3​=C initially. Now, a dielectric slab of dielectric constant K=23​ is inserted in C2​. The equivalent capacitance become

75C​

57C​

32C​

2C​

Q. In given circuit, $C_{1} = C_{2} = C_{3} = C$ initially. Now, a dielectric slab of dielectric constant $K = \frac{3}{2}$ is inserted in $C_{2}$ .
The equivalent capacitance become

$\frac{5 C}{7}$

$\frac{7 C}{5}$

$\frac{2 C}{3}$

$\frac{C}{2}$