91影视

The charge of the capacitor varies as $q=Q\text{cos}\left(蝇t+蠒\right)$.

Figure 31-1 describes the oscillatory variation of energy in an LC circuit. We can determine various phases of the situations described in the question by considering the time t = 0.

Formula:

The charge of the capacitor varies as $\mathit{q}\mathbf{=}\mathit{Q}\mathbf{}\mathbf{\text{cos}}\left(蝇t+蠒\right)$ (i)

For $t=0$, we can say using equation (i) that the charge will vary as

$q=Q\text{cos}\left(蠒\right)............................\left(i\right)$

Situation a indicates that the charge on the capacitor is maximum. Thus, using equation (i), we get the maximum charge condition as follows:

$q=Q$

This implies that

$$\begin{gathered} \cos 蠒=1 \\ 蠒=0卤2n蟺 \end{gathered}$$

Hence, the value of the phase constant for situation a is $0卤2n蟺$.

Situation c indicates that the charge on the capacitor is zero. Thus, using equation (i), we can get the phase constant as follows:

$$\begin{gathered} dq=0 \\ \cos 蠒=0 \\ 蠒=\frac{蟺}{2}卤n蟺 \end{gathered}$$

Hence, the value of the phase constant is $\frac{蟺}{2}卤n蟺$.

Situation e indicates that the charge on the capacitor is maximum.

But capacitor is fully charged withtheopposite polarity. Thus, using equation (i), we can get the phase constant as follows:

$$\begin{gathered} dq=-Q \\ \cos 蠒=-1 \\ 蠒=蟺卤2n蟺 \end{gathered}$$

But the charge on the capacitor is with opposite polarity.

Hence, the value of the phase constant is $蟺卤2n蟺$.

Situation g indicates that the charge on the capacitor is zero and the current is flowing in the opposite direction. Thus, using equation (i), we can get the value of the phase constant as follows:

$$\begin{gathered} dq=0 \\ \cos 蠒=0 \\ 蠒=\frac{3蟺}{2}卤2n蟺 \end{gathered}$$

Hence, the value of the phase constant is $\frac{3蟺}{2}卤2n蟺$.