In given circuit, C 1 = C 2 = C 3 = C initially. Now, a dielectric slab of dielectric constant K = 2 3 is inserted in C 2 . The equivalent capacitance become

When a dielectric slab of dielectric constant K = 2 3 is inserted between the plates of C 2 its new capacitance ( C 2 ′ ) becomes C 2 ′ = 2 3 C Equivalent capacitance of C 2 ′ and C 3 is C eq = C 2 ′ + C 3 = 2 3 C + C = 2 5 C Now, C eq and C 1 are in series. Therefore, their equivalent capacitance is C eq = C eq + C 1 C eq × C 1 = 2 5 C + C 2 5 C × C = 7 C 5 C 2 = 7 5 C

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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium2 minPYQ_2016

PhysicsMediumsingle choice

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

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Solution:

When a dielectric slab of dielectric constant $K = \frac{3}{2}$ is inserted between the plates of $C_{2}$ its new capacitance $(C_{2}^{'})$ becomes $C_{2}^{'} = \frac{3}{2} C$ Equivalent capacitance of $C_{2}^{'}$ and $C_{3}$ is $C_{eq} = C_{2}^{'} + C_{3} = \frac{3}{2} C + C = \frac{5 C}{2}$
Now, $C_{eq}$ and $C_{1}$ are in series. Therefore, their equivalent capacitance is $C_{eq} = \frac{C _{eq} \times C _{1}}{C _{eq} + C _{1}} = \frac{\frac{5 C}{2} \times C}{\frac{5 C}{2} + C} = \frac{5 C ^{2}}{7 C} = \frac{5 C}{7}$

Stream:VITEEESubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2016

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