Question

A 1.50- $\mathbf{\mu }$F capacitor is charging through a 12.0-Ω resistor using a 10.0-V battery. What will be the current when the capacitor has acquired$\frac{\mathbf{1}}{\mathbf{4}}$of its maximum charge? Will it be$\frac{\mathbf{1}}{\mathbf{4}}$of the maximum current?

Short answer

When the capacitor has acquired$\frac{1}{4}$ of its maximum charge then the current will be 0.625 A

Step-by-step solution

Relation between Resistance and capacitance

Resistance and capacitance are related in the manner that resistance affects directly the time taken to charge a capacitor. Or we can also say that as resistance increases the time taken to charge the capacitor also increases.

Calculating the current

We know that the current in the circuit when the capacitor is $\frac{1}{4}$of the maximum, we can calculate the exponential term:

$$\begin{gathered} \mathrm{Q}=\frac{{\mathrm{Q}}_{\max }}{4} \\ \mathrm{Q}=\frac{{\mathrm{Q}}_{\max }}{4} \\ \mathrm{Q}={\mathrm{Q}}_{\max }\left(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}-}}\right) \\ {\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}}=\frac{3}{4} \end{gathered}$$

We can finally calculate the current:

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_{\max }{\mathrm{e}}^{\frac{-1}{\mathrm{RC}}} \\ ={\mathrm{I}}_{\max }\left(\frac{3}{4}\right) \\ =\frac{\mathrm{V}}{\mathrm{R}}\times \left(\frac{3}{4}\right) \\ =\frac{10.0\mathrm{V}}{12.0\mathrm{\Omega }}\times \left(\frac{3}{4}\right) \\ =0.625\mathrm{A} \end{gathered}$$

Therefore the current is 0.625 A.