Alternating Current

LC Oscillations

Have you ever noticed that the small LED lights of some devices like chargers keep on glowing even after these devices are switched off? What causes the bulbs to glow? These circuits have capacitors and inductors. The energy keeps oscillating in the circuit even after the battery is disconnected. The combination of an inductor and a capacitor creates an LC oscillator circuit. Let us learn more.

Suggested Videos

previous arrow
next arrow
previous arrownext arrow

LC Oscillator

Whenever we connect a charged capacitor to an inductor the electric current and charge on the capacitor in the circuit undergoes LC Oscillations. The process continues at a definite frequency and if no resistance is present in the LC circuit, then the LC Oscillations will continue indefinitely.

This circuit is known as an LC oscillator. Let’s take a capacitor with capacitance C and an inductor with inductance L. The capacitor is fully charged with charge Qo. What we do here is to connect the capacitor and the inductor end to end.

LC oscillator

Assuming the inductor and capacitor to be ideal (meaning resistance will be zero in the overall circuit). Initially, the capacitor C of the LC circuit carries a charge Qo and current I in the Inductor is zero. Therefore at time T = 0, the charge on the capacitor will be:

$$q\left ( T=0 \right )=Q_{o}$$

Current Flowing: $$I=0$$

At time T = t, the capacitor now begins to discharge through the inductor. The current begins to flow in an anti-clockwise direction. Therefore the charge of the capacitor decreases, but the energy of the inductor increases. The energy gets transferred from the capacitor to the inductor.

At this stage, there is the maximum value of the current in the inductor. Then the relationship between the current and the charge will be:

$$ I=-\frac{\mathrm{d} q}{\mathrm{d} t} \quad …………..(1)$$

The negative sign is added because as the time passes from 0 to t the, charge on the plates of capacitor decreases i.e. charge decreases with respect to time and thus the dq/dt obtained will be negative and this is why we add a negative sign to make a current positive.

Browse more Topics under Alternating Current

Applying Energy Conservation

Since, Heat Loss = 0 and Kinetic Energy = 0. Total energy which is constant is given by:

$$\frac{1}{2}\times \frac{q^{2}}{C}+\frac{1}{2}\times{L}{I^{2}}=E_{o}$$

Now initially we had charge Qand current flowing was zero I=0. So, the total energy at T = 0, will only be the energy stored in the capacitor and energy stored in the inductor will be zero.

$$E_{o}= \frac{1}{2}\frac{Q_{o}^{2}}{C}$$

Now on equating the value of total energy we get:

$$\frac{1}{2}\times\frac{q^{2}}{C}+\frac{1}{2}\times{L}{I^{2}}= \frac{1}{2}\frac{Q_{o}^{2}}{C}$$

On differentiating with respect to t we get:

$$\frac{1}{2}\frac{2q}{C}\frac{\mathrm{d}q}{\mathrm{d}t}+\frac{1}{2}L\times 2I\frac{\mathrm{d} I}{\mathrm{d} t}=0$$

On further calculation we get the equation as:

$$L\frac{\mathrm{d} I}{\mathrm{d} t}=\frac{1}{C}q$$

Substituting the value of I from equation 1 we get:

$$\frac{\mathrm{d^{2}}q }{\mathrm{d} t^{2}}=-\frac{1}{LC}q \quad …………..(2)$$

And now double differentiating equation 1 with respect to t we get:

$$\frac{\mathrm{d^{2}}I }{\mathrm{d} t^{2}}=-\frac{1}{LC}I \quad …………..(3)$$

We can see that the above both the equation 2 and 3 of Charge and Current both represent the Simple Harmonic Motion. That is both charge and current are oscillating as a simple harmonic waves with respect to time.

Solution of Simple Harmonic Motion Equation

In general charge as a function of time in SHM will be given as:

$$q\left ( t \right )=Q_{o}\sin \left ( \omega t+\phi  \right )$$

At t = 0, q(0) = Qo. Therefore putting the value of t=0 and q= Qo in above equation we get:

$$ Q_{o}=Q_{o}\sin \left (0+\phi  \right )$$

$$ \sin \phi =1\\ \phi =\frac{\pi }{2}$$

So charge as a function of time will be:

$$ q\left ( t \right )=Q_{o}\sin \left ( \omega t+\frac{\pi }{2}\right )$$

$$ q\left ( t \right )=Q_{o}\cos \left ( \omega t\right )$$

Now differentiating charge with respect to time so we will get current as a function of time:

$$ i(t)=-\frac{\mathrm{d} q(t)}{\mathrm{d} t}$$

Solved Examples for You on LC Oscillator

Q1: In a LC circuit we have a inductor of L = 20mH and  a capacitor of capacitance 50F. Initially charge on the plate of capacitor is 10mC. What is the total electric field energy stored in the capacitor and also mention whether that energy in the capacitor is going to decrease with time?

Solution: Given, charge Q=10 mC=10 x 10-3 C and capacitance C = 50 x 10-6 F. The electric field energy stored in capacitor is :


$$ U_{E}=1J$$

Q2: In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is:

Solution: Let Q denote maximum charge on capacitor. Let q denote charge when energy is equally shared


$$ Q^{2}=2q^{2}$$

$$ q=\frac{Q}{\sqrt{2}}$$

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

2 responses to “Representation of AC Current and Voltage by Rotating Vectors – Phasors”

  1. Tanishk says:


  2. sridhar Dasari says:

    Awesome post…

Leave a Reply

Your email address will not be published. Required fields are marked *

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.