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  • Paragraph-1 When a capacitor is connected to a battery through a resistancen

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When a capacitor \mathrm{C} is connected to a battery through a resistance \mathrm{R}, the plates of the capacitor will acquire equal and opposite charge and potential difference across it become equal to the emf of battery. The process called charging takes sometime and during this time there is an electric current through the resistance. If at any time t, q is the charge on capacitor, then from Kirchoff's law.
\mathrm{R \frac{d q}{d t}+\frac{q}{c}=E }
In the circuit shown, the battery is ideal with emf \mathrm{V}. The capacitor is initially uncharged. The switch \mathrm{S} is closed at \mathrm{t}=0

Question:-

The charge on the capacitor at time is
 

Option: 1

\mathrm{\frac{V C}{2}\left(1-e^{-2 t / 3 R C}\right)}


Option: 2

\mathrm{\frac{V C}{2}(1- \left.e^{-t / 3 R C}\right)}


Option: 3

\mathrm{\frac{V C}{3}\left(1-e^{-2 t / 3 R C}\right)}


Option: 4

\mathrm{\frac{V C}{3}(1- \left.e^{-t / 3 R C}\right)}


Answers (1)

best_answer

The charge on the capacitor at time t in charging C-R circuit is given as

\mathrm{q=q_0\left[1-e^{\frac{-t}{\rho}}\right] \quad \quad \quad \quad \quad (1)}
where \mathrm{q_0}= maximum charge on the capacitor.
and \rho = time constant

Here, \mathrm{q_0=\mathrm{C} \mathrm{V}_{\max }. At \mathrm{t} \longrightarrow \infty}

          \mathrm{V}_{\max }=\mathrm{V}_{\mathrm{CD}}=\mathrm{V}_{\mathrm{BE}}=\frac{V}{2}

Therefore\mathrm{ q_0=\frac{c V}{2}}.

Time constant (\rho)=\mathrm{CR}_{\mathrm{CD}}

\mathrm{=C\left(\frac{3}{2} R\right)=\frac{3 C R}{2} }
Putting the value of \mathrm{q_0} and \mathrm{\rho} in equation (1) We get,

\mathrm{q=\frac{V C}{2}\left(1-e^{\frac{-2 t}{2 R C}}\right) }

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Rishi

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