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

b. Will it be 1/4 of the maximum current?

Revelant knowledge

Charging of a capacitor: When capacitors are charged, the charge on it rises exponentially, and the formular is Q = Q_{max}(1−e^{t/}^{τ})

The capacitor’s current drops exponentially as: I = I_{max}e^{−t/}^{τ}

Ohm’s law describes the concepts that are required to solve the problem.

Calculate the voltage across resistor first, then calculate the magnitude of the current when the capacitor has attained one-fourth its maximum charge.

Ohm’s law-This states that the electric current flowing through a metal wire is directly proportional the potential differenceVacross its ends, provided its temperature does not change.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]

Vr indicates the voltage across resistor when the capacitor has absorbed 1/4 of its charge, andVthe 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]

Irefers to the current magnitude,Rthe resistance, andVthe 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]

Vdenotes the voltage, andRthe 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.