Question

In an oscillating LCcircuit, when 75% of the total energy is stored in the inductor’s magnetic field,(a) What multiple of the maximum charge is on the capacitor? (b) What multiple of the maximum current is in the inductor?

Short answer

1. The multiple of maximum charge on the capacitor is 0.500Q.
2. The multiple of the maximum current in the inductor is$I\left(0.866\right)$ .

Step-by-step solution

The given data

The energy stored in the magnetic field is 75% of the total energy.

Understanding the concept of energy of LC circuit

In an oscillating LCcircuit, energy is shuttled periodically between the electric field of the capacitor and the magnetic field of the inductor; instantaneous values of the two forms of energy are given by equations 31-1 and equation 31-2. So we can take the ratio of that energy to find out the multiple of the maximum current in the inductor.

Formulae:

The electric energy stored by the capacitor in the LC circuit, $U_e=\frac{Q^2}{2C}$ (i)

The magnetic energy stored by the inductor in the LC circuit, $U_B=\frac{L{I}^2}{2}$ (ii)

a) Calculation of multiple of maximum charge on the capacitor

The energy stored in the magnetic field is 75 %, so the energy stored in the electric field is $\left(1-75\%\right)=25\%$.

Let the charge stored by the capacitor be q, and the total charge be Q.

Then the ratio of the electric energy stored by the capacitor to that the total energy of the system can be given using equation (i) as follows:

$\begin{array}{rcl}\frac{U_E}{U}& =& \frac{q^2/2C}{Q^2/2C}\\ 0.25& =& \frac{q^2}{Q^2}\\ \frac{q}{Q}& =& \sqrt{0.25}\\ & =& 0.500\\ & & \end{array}$

Hence, the multiple of maximum charge on the capacitor is 0.500Q.

b) Calculation of multiple of maximum current in the inductor

Let i be the current in the inductor, and the maximum current be I.

Now,the ratio of the magnetic energy stored by the inductor to that the total energy of the system can be given using equation (i) as follows:

$\begin{array}{rcl}\frac{U_B}{U}& =& \frac{L{i}^2/2}{L{I}^2/2}\\ 0.75& =& \frac{L{i}^2/2}{L{I}^2/2}\\ \frac{i^2}{I^2}& =& 0.75\\ \frac{i}{I}& =& \sqrt{0.75}\\ & =& 0.866\\ & & \end{array}$

Therefore, the multiple value of the current is $I\left(0.866\right)$.