LC Circuit in series with an AC source

When the inductance L and capacitance C are connected in series to an alternating source of voltage, then the circuit is called an LC circuit or tank circuit. Colloquially, a tuning fork is the best analogy for an LR circuit. The LR circuit stores electrical energy generated from the oscillating resonant frequency of the circuit.

Let us quickly look over individual elements before jumping into combination circuits. 

• If only resistance is connected to an AC source, then it has a current in phase with the potential, represented by the formula I = Iosinwt, and voltage is V = Vosinwt. Hence the equation for current becomes I = Vo/R

• If the only inductor is connected to an AC source, then the current lags the potential by 900, represented by the formula I = I0sin(wt- π/2), and voltage is V = Vosinwt. Hence the equation for current becomes I = Vo/XL where XL is the inductive reactance calculated by 

XL = 2πfL and its unit is the ohm (Ω)

• If the only capacitor is connected to an AC source, the current leads the potential by 900, which is represented by the formula I = I0sin(wt+ π/2), and voltage is V = Vosinwt. Hence the equation for current becomes I = Vo/XC where XC is the capacitive reactance calculated by XC =1/ 2πfC, and its unit is the ohm (Ω)

LC Circuit Meaning 

In an LC circuit, a pure inductor L is connected in series to a pure capacitor C connected to an AC source. AC voltage is a sinusoidal curve that changes in magnitude and direction concerning time denoted by V = V0sin wt.

Here the inductive reactance XL = 2πfL and capacitive reactance XC is 1/2πfC.

The inductive reactance and capacitive reactance are opposite, as the current lags the potential by 90° in the inductor while the current leads the potential by 90° in the capacitor. 

Although there is no resistor in this circuit, the wires, the inductor coil, and the capacitor plates offer some resistance. But, let an ideal pure LC circuit is supposed to have a resistance of R = 0

Thus the current keeps going round and round in the circuit. The oscillating pendulum clock stores all the energy generated from the oscillations until the internal components exhaust the charge. 

Phasor diagram of the LC circuit

Here, in the capacitor, current leads the potential by 900, and in the inductor, the current lags the potential by 900.

In the phasor diagram, the potential EL of the inductor leads the current by 900, and the potential EC of the capacitor lags the current by 900. Hence, the vector follows the potential direction, which is more in magnitude.

For instance,

1. i) If EC > EL , Where EL = IXL and EC = IXC

E = EC – EL, where the voltage will lag the current by 900

1. ii) If EL > EC , Where EL = IXL and EC = IXC 

E = EL – EC, where the voltage will lag the current by 900

Hence collectively, we can write the value of E as 

         E = EL ~ EC 

            = IXL ~ IXC

The Impedance is the collective resistance of Capacitance XC and Inductance XL. Here the resistance offered by the capacitor and inductor is not the same. The capacitor absorbs or stores the energy, and the inductor tends to oppose the charge going through it. Hence the collective resistance or Impedance of the capacitor and inductor is XL ~ XC.

This can be proved by a simple formula of the Impedance of the circuit calculated by

Z = √ [R2 + (XL-XC)2] 

In an ideal LC circuit R = 0

Thus, Z = XL – XC

As the value of Z can never be negative, we will make a slight change to the formula. 

ZLC = E/I = XL ~ XC = wL – 1/wc or vice versa

Hence, ZLC = wL ~ 1/wC

Solved LC circuit example

Here an inductor of 300 mH and a capacitor of 50 μF are connected in series across a 200 V, 50 Hz supply. Find the value of the current. Also, find the voltage drop in the inductor and capacitor. 

Let’s dive right in,

L = 300 mH = 0.3 H, C = 50 μF = 50 x 10-6 F, V = 200, f= 50

XL = 2πfL = 2π50 x 0.3 = 94.2 Ω

XC = 1/ 2πfC = 1/ 2π50 x 10-6 = 63.66 Ω

XL > XC; hence the circuit is inductive. 

Net reactance is 94.2 -63.66 = 30.54 Ω

R = 0

Z = XL – XC = 30.54Ω

I = V/Z = 200/30.54 = 6.55 A

I = 6.55 A

Voltage drop in inductor = IXL = 6.55 x 94.2 = 617V

Voltage drop in capacitor = IXC = 6.55 x 63.66 = 416.97V  

Conclusion

Thus, with the help of simple formulas and definitions, we can solve an LC circuit quickly. The Impedance of the LC circuit plays a vital role in deciding the total lag of the current from the potential. Various aspects such as oscillating frequency, Impedance, magnitude of the potential together for the efficient functioning of an LC circuit.