^{$\mathit{C}\mathbf{=}\mathbf{1}\mathbf{.4}\mathit{\mu }\mathit{F}$In an oscillating LCcircuit, $\mathit{L}\mathbf{=}\mathbf{\text{8.0\hspace{0.17em}mH}}$and . At the time , $\mathbf{t}\mathbf{}\mathbf{=}\mathbf{\text{0}}$the current is maximum at $\mathbf{\text{12.0mA}}$. (a) What is the maximum charge on the capacitor during the oscillations? (b) At what earliest time $\mathbf{t}\mathbf{}\mathbf{>}\mathbf{\text{0}}$is the rate of change of energy in the capacitor maximum? (c) What is that maximum rate of change?}

• Inductance in the circuit, .$L=8.\text{0\hspace{0.17em}mH}$
• Capacitance,$\text{C=1.40\hspace{0.17em}}\mu \text{F}$
• At ,$t=0$ maximum current .$I=12\text{mA}$

By using the formulae for angular frequency and amplitude of current, find the maximum charge on the capacitor. By differentiating the total energy of the capacitor with respect to time, find the time corresponding to the maximum energy and the maximum rate of change of energy.

Formulae:

• Angular frequency of oscillation,.$\omega =\frac{1}{\sqrt{LC}}$
• The amplitude of varying current,.$I=\omega Q$
• The total energy in the capacitor,.$U_E=\frac{q^2}{\text{2}C}=\frac{Q^2}{\text{2}C}{\left(\text{sin\hspace{0.17em}}\omega t\right)}^2$

The angular frequency of oscillation is .$\omega =\frac{1}{\sqrt{LC}}$

Also, the amplitude of varying current, .$I=\omega Q$

Using these two equations, it can be written as,

$\begin{array}{c}Q=I\sqrt{LC}\\ Q=\left(\text{0.012}\right)\sqrt{\left({\text{1.40×10}}^{\text{-6}}\right)\left(\text{0.008}\right)}\\ Q={\text{1.27×10}}^{\text{-6}}\text{\hspace{0.17em}C}\end{array}$.

Hence, the maximum charge on the capacitor during the oscillation is.${\text{1.27×10}}^{\text{-6}}\text{\hspace{0.17em}C}$

The total energy in the capacitor is,

.$U_E=\frac{q^2}{\text{2}C}=\frac{Q^2}{\text{2}C}{\left(\text{sin\hspace{0.17em}}\omega t\right)}^2$

Differentiating it with time and use,$\sin \left(2\theta \right)=2\sin \theta \cos \theta$

$\frac{d{U}_E}{dt}=\frac{Q^2}{\text{2}C}\omega \text{sin}\left(\text{2}\omega t\right)$

The maximum value occurs when .$\text{sin}\left(\text{2}\omega t\right)\text{=1}$

Therefore,

.$\begin{array}{c}\text{2}\omega t=\frac{\pi }{\text{2}}\\ t=\frac{\text{1}}{\text{2}\omega }\left(\frac{\pi }{\text{2}}\right)\\ t=\frac{\pi }{\text{4}}\sqrt{LC}\\ t=\frac{\text{3.14}}{\text{4}}\sqrt{\left({\text{1.40×10}}^{\text{-6}}\right)\left(\text{0.008}\right)}\\ t={\text{8.31×10}}^{\text{-5}}\text{\hspace{0.17em}s}\end{array}$

Hence, the earliest time when the rate of change of energy in the capacitor is maximum is .$8.31\times {10}^{-5}\text{\hspace{0.17em}s}$

Since,

$\frac{d{U}_E}{dt}=\frac{Q^2}{\text{2}C}\omega \text{sin}\left(\text{2}\omega t\right)$.

The maximum value occurs when .$\text{sin}\left(\text{2}\omega t\right)=\text{1}$

${\left(\frac{d{U}_E}{dt}\right)}_{max}=\frac{Q^2}{\text{2}C}\omega$.

Substituting the values of ,$Q\text{and}\omega$

Hence,the maximum rate of change in energy is .$5.44\times {10}^{-3}\text{\hspace{0.17em}J/s}$