Question

An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
     (i) completely electrical (i.e., stored in the capacitor)?
     (ii) completely magnetic (i.e. stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Answer 1

Hint: We will be using the formula of frequency of oscillator, energy stored in capacitor, time period, magnetic energy and electrical energy. Substituting the value of inductance, capacitance and charge on the formulas we can determine different values.

Formula used:
 \[\begin{align}
  & E=\dfrac{{{Q}^{2}}}{2C} \\
 & v=\dfrac{1}{2\pi \sqrt{LC}} \\
 & T=\dfrac{1}{f} \\
\end{align}\]

Complete step-by-step answer:
Given-
\[\begin{align}
  & Q=10mC \\
 & L=20mH \\
 & C=50\mu F \\
\end{align}\]

(a) Total energy stored initially
Let us consider Q be the charge flowing through the circuit and C be the capacitance in a capacitor. So energy stored E in the capacitor can be given by- \[E=\dfrac{{{Q}^{2}}}{2C}\]
Substituting the values \[=\dfrac{{{(10\times {{10}^{-3}})}^{2}}}{2\times 5\times {{10}^{-6}}}=1J\]
Thus energy is being conserved. It only gets converted from one form to another i.e. from electrical to magnetic energy and vice versa.

(b) Natural frequency of the circuit
Let L be the inductance of the circuit then the natural frequency can be calculated as \[v=\dfrac{1}{2\pi \sqrt{LC}}\]
Substituting the given values
\[\begin{align}
  & v=\dfrac{1}{2\pi \sqrt{20\times {{10}^{-3}}\times 50\times {{10}^{-6}}}} \\
 & =159.24Hz
\end{align}\]
Thus frequency of oscillation is found to be \[159.24Hz\].

(c) Energy stored
i. Completely electrical
First we need to calculate the time period. If f is the frequency then time period T can be given as \[T=\dfrac{1}{f}=6.28s\]
Let \[Q'\] be the total charge on capacitor at time t
\[t=Q'=Q\cos \left( \dfrac{2\pi }{T} \right)t\]
Electrical energy stored can be given as \[Q'=\pm Q\]
From cosine series we can infer that energy stored is completely electrical as \[t=0,\dfrac{T}{2},\dfrac{3T}{2},.......\]
ii. Completely magnetic
Magnetic energy will be maximum when the electrical energy will be zero. Therefore \[t=\dfrac{T}{4},\dfrac{3T}{4},\dfrac{5T}{4}.......\]

(d) Total energy stored between inductor and capacitor
Let us assume \[{{Q}_{1}}\] be the charge when total energy is shared equally
\[\begin{align}
  & \dfrac{1}{2}\dfrac{Q_{1}^{2}}{C}=\dfrac{1}{2}\left( \dfrac{1}{2}\dfrac{{{Q}^{2}}}{C} \right) \\
 & {{Q}_{1}}=\pm \dfrac{Q}{\sqrt{2}}
\end{align}\]
But \[{{Q}_{1}}=\dfrac{Q\cos 2\pi t}{T}\]
\[\cos \dfrac{2\pi t}{T}=\pm \dfrac{1}{\sqrt{2}}\]
\[t=(2n+1)\dfrac{T}{8}\]
Therefore \[t=\dfrac{T}{8},\dfrac{3T}{8},\dfrac{5T}{8}\]

(e) If resistor is inserted, loss of energy
Energy calculated initially in section (a) which is 1J.
Therefore 1J of energy is dissipated as heat.

Note: LC is also known as resonant circuit. LC circuit is a closed loop that contains only two components: inductor and capacitor connected together. These circuits are set at a certain resonance for that particular carrier frequency.