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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium2 minQB

PhysicsMediummultiple choice

In the circuit shown in figure, each capacitor has a capacitance C. The emf of the cell is E and circuit is already in steady state. If the switch S is closed :-

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Answer:

Solution:

Ans (A,D)

$C_{e ff} = (\frac{1}{C} + \frac{1}{2 C})^{- 1} = \frac{2 C}{3}$ , charge $= \frac{2 Cε}{3}$

$C_{e ff} = 2 C \Rightarrow$ charge $= 2 CE$

$2 CE - \frac{2 CE}{3} = \frac{4 CE}{3}$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2mℹ️ Source: QB

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