The frequency of oscillation of a certain LCcircuit is$\mathbf{200}\mathbf{}\mathbf{\text{kHz}}$. At time $\mathit{t}\mathbf{=}\mathbf{0}$, plate Aof the capacitor has maximum positive charge. At what earliest time$\mathit{t}\mathbf{>}\mathbf{}\mathbf{0}$ will

(a) plate Aagain have maximum positive charge,(b) the other plate of the capacitor have maximum positive charge, and (c) the inductor have maximum magnetic field?

1. Frequency of oscillation, $f=200\text{kHz or}200\times {10}^3\text{Hz}$
2. At $t=0$, plate A has maximum positive charge.

Using the concept, that the period is the reciprocal of the frequency, and applying the given conditions, we can find the earliest time when plate A will again have the maximum positive charge, and the other plate of the capacitor will have the maximum positive charge, and the inductor will have the maximum magnetic field.

Formula:

The period of oscillation of a LC circuit, $\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$ (i)

The value of when plate A will again have maximum positive charge is a multiple of the period.

Thus, it can be stated as: $t_A=nT$where, $n=1,2,3,4,\dots \dots \dots$

Using the above relation of time in equation (i), we get

$$\begin{gathered} t_A=\frac{n}{f} \\ =\frac{n}{\left(200\times {10}^3\text{Hz}\right)} \end{gathered}$$

For $n=1$, the earliest time is given as:

$$\begin{gathered} t_A=5.00\times {10}^{-6}\mu s \\ =5.00\mu s \end{gathered}$$

Hence, the value of the earliest time is $5.00\mu s$.

Comparing figure$t_A=n\left(5.00\times {10}^{-6}\text{s}\right)$ .31-1(a) and Fig.31-1(e), we see that it takes time$t=\frac{1}{2}T$for the charge on the other plate to reach its maximum positive value for the first time. And this is when plate A acquires its most negative charge. From that time onwards, it will repeat once every period.

Thus, the time at which the other plate has maximum charge is given as:

$$\begin{gathered} t=\left[\frac{1}{2}+\left(n-1\right)\right]T \\ =\left(n-\frac{1}{2}\right)T \\ =\left[\frac{\left(2n-1\right)}{2}\right]T \end{gathered}$$

where, $n=1,2,3,\dots \dots .$

For $n=1$, the earliest time is given by using equation (i) as follows:

$$\begin{gathered} t=\frac{\left(2n-1\right)}{2f} \\ =\frac{\left(2-1\right)}{2f} \\ =\frac{1}{2f} \\ =\frac{1}{2\left(200\times {10}^3\text{Hz}\right)} \\ =2.50\times {10}^{-6}\text{s} \\ =2.50\mu s \end{gathered}$$

Hence, the value of the earliest time is $2.50\mu s$.

Comparing steps Fig.31-1(a) and Fig.31-1(c), we see that it takes time$t=\frac{1}{4}T$for the current and the magnetic field in the inductor to reach the maximum value for the first time. Comparing steps Fig.31-1(c) and Fig.31-1(g), we see that this will repeat later every half-period.

Thus, the time at which the inductor has maximum magnetic field is given as:

$$\begin{gathered} t_L=\frac{T}{4}+\frac{\left(n-1\right)T}{2},where,n=1,2,3,....... \\ =\left[\frac{1}{4}+\frac{\left(n-1\right)}{2}\right]T \\ =\left[\frac{2+4\left(n-1\right)}{8}\right]T \\ =\left[\frac{2+\left(4n-4\right)}{8}\right]T \\ =\left[\frac{\left(4n-2\right)}{8}\right]T \\ =\frac{\left(4n-2\right)}{8f}\text{(}\because \text{from equation (i))} \end{gathered}$$

For$n=1$ , the earliest time is given by:

$$\begin{gathered} t_L=\frac{\left(4-2\right)}{8f} \\ =\frac{1}{4f} \\ =\frac{1}{4\left(200\times {10}^3\text{Hz}\right)} \\ =1.25\times {10}^{-6}\text{s} \\ =1.25\mu s \end{gathered}$$

Hence, the value of the earliest time for the inductor isrole="math" localid="1662754138564" $1.25\mu s$ .