Question

In an oscillating LCcircuit with$\mathit{L}\mathbf{=}\mathbf{}\mathbf{50}\mathbf{}\mathbf{\text{mH}}$and$\mathit{C}\mathbf{=}\mathbf{40}\mathbf{}\mathbf{\mu F}$, the current is initially a maximum. How long will it take before the capacitor is fully charged for the first time?

Short answer

It takes$7.0\times {10}^{-4}\text{sec}$ long before the capacitor is fully charged for the first time.

Step-by-step solution

The given data

1. The inductance of the LC circuit,$L=50\text{mH or}50\times {10}^{-3}\text{H}$
2. The capacitance of the LC circuit,$C=40\mathrm{\mu F}=40\times {10}^{-6}\mathrm{F}$
3. Initially, the current is at maximum.

Understanding the concept of a period of oscillation

Angular frequency is also known as radial frequency, measured as the angular displacement per unit time. By using the formula for angular frequency in terms of the period, we will find the time taken for charging the capacitor fully.

Formulae:

The resonance frequency of an LC circuit, $\omega =\frac{1}{\sqrt{LC}}$ (i)

The angular frequency of an oscillation, $\omega =\frac{2\pi }{T}$ (ii)

Calculation of the time taken by the capacitor to fully charge 

From equations (i) and (ii), we can get the period of oscillations as follows:

$\begin{array}{rcl}T& =& 2\pi \sqrt{LC}\\ & =& 2\times 3.14\sqrt{50\times {10}^{-3}\times 4\times {10}^{-6}}\\ & =& 0.0028\text{sec}\\ & & \end{array}$

The time required for the charge to rise from zero to its first maximum value is one-quarter of a period. Thus, the required time for the capacitor being fully charged can be given as:

$$\begin{gathered} t=\frac{T}{4} \\ =\frac{0.0028}{4} \\ =7.0\times {10}^{-4}\text{sec} \end{gathered}$$

Hence, the time required to fully charge the capacitor for the first time is $7.0\times {10}^{-4}\text{sec}$.