A capacitor having capacitance C is initially provided with a charge Q . At t = 0 , it is linked to an ideal battery of emf E via a resistance R . Determine the charge on the capacitor at time t .

Question

Quantrex Academy · Accepted Answer

Correct answer: D. C E (1 - e^ -(t)/(CR) ) + Q e^ -(t)/(CR)

Applying Kirchhoff's loop rule for the circuit at time

t

:

\mathcal{E} - iR - \dfrac{q}{C} = 0

Substituting

i = \dfrac{dq}{dt}

:

R \dfrac{dq}{dt} = \mathcal{E} - \dfrac{q}{C} = \dfrac{C\mathcal{E} - q}{C}

Rearranging and integrating with the initial condition

q(0) = Q

:

\int_Q^{q} \dfrac{dq}{C\mathcal{E} - q} = \int_{0}^{t} \dfrac{dt}{RC}

[-\ln(C\mathcal{E} - q)]_Q^{q} = \dfrac{t}{RC}

\ln\left(\dfrac{C\mathcal{E} - q}{C\mathcal{E} - Q}\right) = -\dfrac{t}{RC}

C\mathcal{E} - q = (C\mathcal{E} - Q) e^{-\dfrac{t}{RC}}

q = C\mathcal{E}\left(1 - e^{-\dfrac{t}{RC}}\right) + Q e^{-\dfrac{t}{RC}}

A capacitor having capacitance C is initially provided with a...

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