# Problem: A 1.70-μF capacitor is charging through a 10.0-Ω resistor using a 10.0-V battery.What will be the current when the capacitor has acquired 1/4 of its maximum charge? Will it be 1/4 of the maximum current?

###### FREE Expert Solution

Since the capacitor is charging:

$\overline{){\mathbf{Q}}{\mathbf{=}}{{\mathbf{Q}}}_{{\mathbf{0}}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{-}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}}{\mathbf{\right)}}}$

New current:

$\overline{){\mathbf{I}}{\mathbf{=}}{{\mathbf{I}}}_{{\mathbf{0}}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}}}$

In the given case, Q = Q0/4

Substituting:

$\begin{array}{rcl}\frac{\overline{){\mathbf{Q}}_{\mathbf{0}}}}{\mathbf{4}}& \mathbf{=}& \overline{){\mathbf{Q}}_{\mathbf{0}}}\mathbf{\left(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}}\mathbf{\right)}\\ \frac{\mathbf{1}}{\mathbf{4}}& \mathbf{=}& \mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}}\\ {\mathbf{e}}^{\mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}}& \mathbf{=}& \mathbf{1}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\\ {\mathbf{e}}^{\mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}}& \mathbf{=}& \frac{\mathbf{3}}{\mathbf{4}}\\ \mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}& \mathbf{=}& \mathbf{l}\mathbf{n}\mathbf{\left(}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{\right)}\\ \mathbf{-}\mathbf{t}}{\mathbf{R}\mathbf{C}}& \mathbf{=}& \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{28768}\\ \mathbf{t}& \mathbf{=}& \mathbf{\left(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{28768}\mathbf{\right)}\mathbf{\left(}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\right)}\end{array}$

t = 4.89056 × 10-6s ###### Problem Details

A 1.70-μF capacitor is charging through a 10.0-Ω resistor using a 10.0-V battery.

What will be the current when the capacitor has acquired 1/4 of its maximum charge? Will it be 1/4 of the maximum current?