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What is the Formula for LC Oscillations?
In the circuit shown in Fig.(a), S1S2S3 is a two way switch. To charge the capacitor C first, one has to connect the switch S1S2 [Fig. (b)]. If now S1S3 is connected, the charged capacitor and the inductance L constitute an LC circuit [Fig.(c)].
As soon as the switch is disconnected from S2, no current flows through the circuit any more. In this condition if the amount of charge in the capacitor C is Q [Fig.(b)], the energy stored in the electric field of the capacitor is EC = \(\frac{1}{2} \frac{Q^2}{C}\), but the inductor possesses no energy, i.e., EL = 0. Due to accumulation of charge in the capacitor it acts as a battery. Hence immediately after connecting the switch S1S3, the capacitor starts to send current in the LC circuit thus formed [Fig.(c)]. As this discharging current gradually rises, a magnetic field starts develop-ing in the inductor. At this stage it may be said that the capacitor is being discharged gradually through the inductor L. After some time, the capacitor C becomes completely discharged and the current uf the LC circuit reaches a peak value I0 i.e., EC becomes Zero and EL = \(\frac{1}{2} L I_0^2\). It means that the electrical energy in the capacitor has been transferred fully into the magnetic energy in the inductor.
Again, due to flow of current in the LC circuit, the capacitor is being charged in the reverse direction i.e., the capacitor is now charged with a polarity opposite to its initial state [Fig.(d)].
Obviously with the increase of charge in the capacitor, the energy stored in its electric field begins to increase and the energy stored in the magnetic field of the inductance gradually decreases. Hence the magnitude of the current decreases and finally comes to zero. Thus after some time, EL becomes zero again and EC = \(\frac{1}{2} \frac{Q^2}{C}\). this process of discharging and charging of the capacitor occurs alternately. We know that a capacitor is able to store electric energy whereas an inductor can store magnetic energy in it. Now if a charged capacitor with an inductor is connected in an ac circuit, a periodical energy tranformation starts. The energy of the capacitor is converted to the energy of the inductor and back again. This phenomenon is called LC oscillations.
In the above discussion one half cycle of this oscillation has been discussed. At the end of next half cycle the circuit comes back to its initial condition, i.e., the polarity of the plates becomes equal to its initial state and the current in the circuit becomes zero again. Obviously the direction of current in the second half cycle is just opposite to that in the first half cycle. So the current in an LC circuit fluctuates periodically between the peak values and -I0.
The resistance of a pure inductor is zero and so there is no energy loss due to Joule effect. So as time elapses, no loss of total energy takes place, i.e., there is no damping of LC oscillations. So the peak value of the alternating current in IC circuit remains unchanged [Fig.(a)]. But practically, the resistance of any coil cannot be ignored. Thus, there is always sorne resistance in the circuit due to which sorne energy is lost in the form of heat. So the current remains osculatory, but is damped (Fig.(b)). To maintain the alternating current, the circuit must he
supplied with the same amount of energy as is being lost during each cycle, from some external source.
If damping is absent, the frequency of LC oscillations,
f = \(\frac{1}{2 \pi \sqrt{L C}}\)
When this frequency becomes equal to the frequency of the applied alternating emf then resonance occurs in that circuit.
Oscillators convert direct current (dc) from a power supply to an alternating current signal. LC circuit is used in many cases as an important component of an oscillator.
Numerical Examples
Example 1.
A 220 V, 50 Hz ac source is connected to an inductance of 0.2 H and a resistance of 20 Ω in series. What is the current in the circuit?
Solution:
Example 2.
An ac source of frequency 50 Hz is connected with a resistance (R = 36Ω) and L of 0.12 H in series. What is the phase difference between current and voltage?
Solution:
If θ be the phase difference, then
tanθ = \(\frac{\omega L}{R}\)
or θ = tan-1\(\frac{\omega I}{R}\) = tan-1(1.047) = 46.3°
[here ω = 2π × 50 = 2 × 3.14 × 50 = 314 Hz]
So, phase difference = 46.3°.
Example 3.
A current of 1 A flows in a coil when connected to a 100 V dc source. If the same coil is connected to a 100 V, 50 Hz ac source, a current of 0.5 A flows in the coil. Calculate the inductance of the coil.
Solution:
If R be the resistance of the coil in dc circuit \(\frac{E}{I}\) = R
Example 4.
A lamp in which 10 A current can flow at 15 V is connected with an alternating source of potentIal 220 V and frequency 50 Hz. What should be the inductance of choke coil required to light the bulb?
Solution:
Resistance of the lamp, R = \(\frac{15}{10}\) = 1.5 Ω
To send 10 A current through the lamp, required impedance of the ac circuit, L = \(\frac{220}{10}\) = 22 Ω
Now if L be the inductance of the choke coil and its resistance is negligible, then
Example 5.
What will be the peak value of the alternating current when a condenser of 1 µF is connected to an alternating voltage of 200 V, 60 Hz?
Solution:
C = 1µF = 10-6F; w = 22πF = 2 × 3.14 × 60Hz
Peak value of current,
I0 = Irms × \(\sqrt{2}\) = \(\frac{E_{\mathrm{rms}}}{\frac{1}{\omega C}} \cdot \sqrt{2}\) = Erms ᐧ ωC\(\sqrt{2}\)
= 200 × (2 × 3.14 × 60) × 10-6 × 1.414
= 0.106 A (approx.)