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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorR-C CircuitEasy2 minQB

PhysicsEasysingle choice

The capacitor shown in the figure is initially uncharged, the battery is ideal. The switch $S$ is closed at time $t = 0$ , then the time after which the energy stored in the capacitor becomes one-fourth of the energy stored in it in steady-state is:

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Answer:

Solution:

$U = \frac{U_{m a x}}{4} \Rightarrow \frac{q^{2}}{2 C} = \frac{1}{4} \frac{Q_{m a x}^{2}}{2 C} \Rightarrow q = \frac{Q_{m a x}}{2}$

$q = \frac{C E}{2} \Rightarrow q = C E (1 - e^{- t / R C})$

$\frac{C E}{2} = C E (1 - e^{- t / R C}) \Rightarrow \frac{1}{2} = 1 - e^{- t / R C}$

$e^{- t / R C} = \frac{1}{2}$

$\Rightarrow \frac{t}{R C} = \ln 2$

$\Rightarrow t = R C \ln 2$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:R-C Circuit

⏱ 2mℹ️ Source: QB

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