Discharging capacitor

Electricity and Magnetism Level 3

A capacitor, of capacitance $4F$ , is discharging through a $3$ ohm resistor. If the capacitor initially had $20C$ of charge, how much energy (in $J$ to 2 d.p.) will the capacitor dissipate in the first $5$ seconds?

The answer is 28.27.

1 solution

Hosam Hajjir
Aug 6, 2019

The charge $q(t)$ is given by ,

$q(t) = Q_0 e^{ -\dfrac{t}{RC} }$

And the energy stored in the capacitor is

$E(t) = \dfrac{1}{2} C v^2 = \dfrac{1}{2C} q^2(t) = \dfrac{Q_0^2}{2C} e^{-\dfrac{2 t}{RC}}$

Therefore the energy dissipated is

$\Delta E = E(0) - E(5) = \dfrac{400}{8} (1 - e^{-\dfrac{5}{6} } ) = 28.27 J$

Discharging capacitor

The answer is 28.27.

1 solution

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