The discharge of a capacitor in an RC circuit is given by
[tex] \frac{dQ}{dt} = - \frac{1}{RC} Q[/tex]
where
Q = charge,
R = resistance,
C = capacitance,
t = time.
Let Q(0) = Q₀.
Then
[tex]\int _{Q_{0}}^{Q} \frac{dQ}{Q} =- \frac{1}{RC} \int_{0}^{t} dt \\\\ ln \frac{Q}{Q_{0}} =- \frac{t}{RC} \\\\ Q(t) = Q_{0} e^{- \frac{t}{RC} } [/tex]
Because Q = 0.2Q₀ whent t= 2.1 ms, therefore
[tex]e^{- \frac{2.1 \times 10^{-3}}{RC}} = \, 0.2 \\\\ - \frac{2.1 \times 10^{-3}}{RC} = ln(0.2) \\\\ RC = \frac{-2.1 \times 10^{-3}}{ln(0.2)} = 1.3048 \times 10^{-3}[/tex]
Because C = 2.60 μF = 2.6 x 10⁻⁶ F, therefore
R = (1.3 x 10⁻³)/(2.6 x 10⁻⁶) = 501.85 Ω
Answer: 501.9 Ω
