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PhysicsCapacitorBasic capacitor analysis with dielectricMedium2 minPYQ_2023
PhysicsMediumnumerical

Two parallel plate capacitors C1 and C2 each having capacitance of 10 μF are individually charged by a 100 V D.C. source. Capacitor C1 is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor C2 is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor C1 is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be _____ V.

(Assuming Dielectric constant =10)

Answer:
55.00
Solution:

As source is connected in the process therefore, additional charge will be provided. Therefore, charge on C1=KCV.

Since source is disconnected, so its charge will remain the same. Then, charge on C2=CV.

New capacitance in both cases will become, KC.

When they are connected in parallel charge will be equally divided so charge on one capacitor is q=K+12CV.

So common potential,

V'=qKC=K+12KV=55 V.

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Basic capacitor analysis with dielectric
2mℹ️ Source: PYQ_2023

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