Question

A 1.50-$\mathit{\mu }$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 0.625A

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 time taken to charge the capacitor also increases.

Calculating the current

We know that the current in the circuit when the capacitor is ¼ maximum, we can calculate the exponential term:

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

We can finally calculate the current:

$$\begin{gathered} l=l_{max}{e}^{\frac{-t}{RC}} \\ =l_{max}\left(\frac{3}{4}\right) \\ =\frac{V}{R}\frac{3}{4} \\ =\frac{10.0V}{12.0\Omega }\frac{3}{4} \\ =0.625A \end{gathered}$$

Therefor the current is 0.625 A .