In given circuit- C1-C2-C3-C initially- Now- a dielectric slab of dielectric constant K-3-2 is inserted in C2-The equivalent capacitance become

The correct option is B 5C7 When a dielectric slab of dielectric constant K=32 is inserted between the plates of C_2 , its new capacitance ...

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Byju's Answer

Standard XII

Physics

Parallel and Series Combination of Capacitors

In given circ...

Question

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}

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\frac{7 C}{5}

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\frac{2 C}{3}

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\frac{C}{2}

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Solution

The correct option is B $\frac{5 C}{7}$
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
$C_{e q} = \frac{C_{e q} \times C_{1}}{C_{e q} + C_{1}} = \frac{\frac{5 C}{2} \times C}{\frac{5 C}{2} + C}$
$= \frac{5 C^{2}}{7 C} = \frac{5 C}{7}$

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In given circuit, C1=C2=C3=C initially. Now, a dielectric slab of dielectric constant K=32 is inserted in C2.The equivalent capacitance become

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