a)

The resistor [katex]10.0{\rm{ }}\Omega [/katex] is the voltage across it

[katex]{V_r} = V – \left( {\frac{1}{4}} \right)V[/katex]

V r indicates the voltage across resistor when the capacitor has absorbed 1/4 of its charge, and V the battery voltage.

In the following equation, substitute 10.0 V for :

[katex]\begin{array}{c}\\{V_r} = V – \left( {\frac{1}{4}} \right)V\\\\ = \left( {10.0{\rm{ V}}} \right) – \frac{1}{4}\left( {10.0{\rm{ V}}} \right)\\\\ = 7.5{\rm{ V}}\\\end{array}[/katex]

Calculate the current at the capacitor’s maximum charge.

This is how ohm’s laws calculates the magnitude of current:

[katex]I = \frac{{{V_r}}}{R}[/katex]

I refers to the current magnitude, R the resistance, and V the voltage at which the capacitor has accumulated one-fourth the charge.

Substitute 7.5 V to V

_r

[katex]10.0{\rm{ }}\Omega [/katex]

[katex]\begin{array}{c}\\I = \frac{{{V_r}}}{R}\\\\ = \frac{{7.5{\rm{ V}}}}{{10.0{\rm{ }}\Omega }}\\\\ = 0.75{\rm{ A}}\\\end{array}[/katex]

b)

The maximum current is

[katex]I = \frac{V}{R}[/katex]

V denotes the voltage, and R the resistance.

Substitute V for V or [katex]10.0{\rm{ }}\Omega [/katex]

[katex]\begin{array}{c}\\I = \frac{{10{\rm{ V}}}}{{10{\rm{ }}\Omega }}\\\\ = 1{\rm{ A}}\\\end{array}[/katex]Ans Part a

The current magnitude is 0.75 A.

Part b

The maximum current is not 1/4 when the charge exceeds one-fourth of its maximum.