Question

^{$\mathit{C}\mathbf{=}\mathbf{1}\mathbf{.4}\mathit{\mu }\mathit{F}$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?}

Short answer

a) The maximum charge on the capacitor during the oscillation is $1.27\times {10}^{-6}\text{\hspace{0.17em}C}$.

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

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

Step-by-step solution

Step 1: Given

• 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 .$I=12\text{mA}$

Determining the concept 

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

(a) Determining the maximum charge on the capacitor during the oscillation

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.${\text{1.27×10}}^{\text{-6}}\text{\hspace{0.17em}C}$

(b) Determining the earliest time, when the rate of change of energy in the capacitor is maximum

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 .$8.31\times {10}^{-5}\text{\hspace{0.17em}s}$

(c) Determining the maximum rate of change of energy

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 ,$Q\text{and}\omega$

$\begin{array}{l}{\left(\frac{d{U}_E}{dt}\right)}_{max}=\left(\frac{{\left(I\sqrt{LC}\right)}^2}{\text{2}C}\right)\left(\frac{I}{\sqrt{LC}}\right)\\ {\left(\frac{d{U}_E}{dt}\right)}_{max}=\frac{I^2}{\text{2}}\sqrt{\frac{L}{C}}\\ {\left(\frac{d{U}_E}{dt}\right)}_{max}=\frac{{\left(\text{0.012}\right)}^{\text{2}}}{\text{2}}\sqrt{\frac{\left(\text{0.008}\right)}{\left({\text{1.40×10}}^{\text{-6}}\right)}}\\ {\left(\frac{d{U}_E}{dt}\right)}_{max}={\text{5.44×10}}^{\text{-3}}\text{J/s}\end{array}$

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