Equivalent capacitance for two capacitors:

$\overline{){{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}\frac{{\mathbf{C}}_{\mathbf{1}}{\mathbf{C}}_{\mathbf{2}}}{{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}}}$

The time constant for RC circuits:

$\overline{){\mathbf{\tau}}{\mathbf{=}}{\mathbf{R}}{\mathbf{C}}}$

Discharging a capacitor in RC circuits:

$\overline{){\mathbf{Q}}{\mathbf{=}}{{\mathbf{Q}}}_{{\mathbf{0}}}{\mathbf{(}}{\mathbf{1}}{\mathbf{-}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{\tau}}{\mathbf{)}}}$

C_{eq} = (C_{1}C_{2})/(C_{1} + C_{2}) = (3C * 5C)/(3C + 5C) = (15/8)C = (15/8)0.13 = 0.24375 F

Two capacitors of capacitance 3C and 5C (where C = 0.13 F) are connected in series with a resistor of resistance R Randomized Variables R=5.5 Ω

How long will it take the amount of charge in the circuit to drop by 75% in seconds?

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