Question

26.44) A 1.70 $-\,\mu F$ capacitor is charging through a 10.0 $-\, \Omega$ resistor using a 10.0 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?

Answer 1

Concepts and reason

The concepts required to solve the given problem is Ohm’s law.

First calculate the voltage across the resistor and then the magnitude of current, when the capacitor has acquired one-fourth of its maximum charge will be calculated.

Fundamentals

Ohm’s law- It states that electric current flowing through a metallic wire is directly proportional to the potential difference V, across its ends provided its temperature remains the same.

a)

The voltage across the resistor $10.0{\rm{ }}\Omega$ is calculated when the capacitor has acquired ¼ of its maximum charge as follows:

${V_r} = V - \left( {\frac{1}{4}} \right)V$

Here, V_r is the voltage across the resistor, when the capacitor has acquired ¼ of charge and V is the battery voltage.

Substitute 10.0 V for V in the equation as follows:

$\begin{array}{c}\\{V_r} = V - \left( {\frac{1}{4}} \right)V\\\\ = \left( {10.0{\rm{ V}}} \right) - \frac{1}{4}\left( {10.0{\rm{ V}}} \right)\\\\ = 7.5{\rm{ V}}\\\end{array}$

Calculate the current when the capacitor has acquired one-fourth of its maximum charge.

The magnitude of current is calculated by using ohm’s law as follows:

$I = \frac{{{V_r}}}{R}$

Here, I is the magnitude of current, R is the resistance and V_r is the voltage when the capacitor has acquired one-fourth of charge.

Substitute 7.5 V for V_r and $10.0{\rm{ }}\Omega$ for R in equation as follows:

$\begin{array}{c}\\I = \frac{{{V_r}}}{R}\\\\ = \frac{{7.5{\rm{ V}}}}{{10.0{\rm{ }}\Omega }}\\\\ = 0.75{\rm{ A}}\\\end{array}$

b)

The maximum current is,

$I = \frac{V}{R}$

Here, V is the voltage and R is the resistance.

Substitute 10 V for V and $10.0{\rm{ }}\Omega$ for R,

$\begin{array}{c}\\I = \frac{{10{\rm{ V}}}}{{10{\rm{ }}\Omega }}\\\\ = 1{\rm{ A}}\\\end{array}$

Ans: Part a

The magnitude of current is 0.75 A.

Part b

The current when the charge reaches one-fourth of the maximum charge is not ¼ of the maximum current.