Series LC Circuit: UNIZOR.COM - Physics4Teens - Electromagnetism - Alternating Current Induction
Channel: Zor Shekhtman on UNIZOR Education
Date: 07/10/2020

UNIZOR.COM - Physics through a Looking Glass of Mathematician


Series LC Circuit

Consider a circuit that contains an AC generator, an inductor of inductance L and a capacitor of capacity C in a series.


The current IL(t) going through an inductor is the same as the current IC(t) going through a capacitor. So, we will use an expression I(t) for both.

The electromotive force (EMF) generated by an AC generator depends only on its own properties and can be described as a function of time t
E(t) = E0·sin(ωt)
where
E0 is a peak voltage on the terminals of a generator,
ω is an angular velocity of a rotor in radians per second.

Inductance L of an inductor and capacity C of a capacitor produce voltage drops VL(t) and VC(t) correspondingly.

As we know, the voltage drop on an inductor is causes by self-induction and depends on the rate of change (that is, the first derivative by time) of a magnetic flux Φ(t) going through it
VL(t) = dΦ(t)/dt
Magnetic flux, in turn, depends on a current going through a wire of an inductor I(t) and the inductor's inductance L
Φ(t) = L·I(t)
Therefore, the voltage drop on an inductor equals to
VL(t) = L·dI(t)/dt

As we know, the amount of electricity Q(t) accumulated in a capacitor is proportional to voltage VC(t) applied to its plates (that is, voltage drop on a capacitor). The constant capacity of a capacitor C is the coefficient of proportionality (see lecture "Electric Fields" - "Capacitors" in this course) that depends on a type of a capacitor
C = Q(t)/VC(t)
Therefore,
Q(t)=C·VC(t)

Knowing the amount of electricity Q(t) accumulated in a capacitor as a function of time t, we can determine the electric current I(t) in a circuit, which is a rate of change (that is, the first derivative by time) of the amount of electricity
I(t) = dQ(t)/dt = C·dVC(t)/dt

The sum of voltage drops on an inductor and a capacitor is supposed to be equal to EMF produced by an AC generator E(t)=E0·sin(ωt), which is the final equation in our system:
VL(t) = L·dI(t)/dt
I(t) = C·dVC(t)/dt
E0·sin(ωt) = VL(t) + VC(t)

To solve this system of three equations, including two differential ones, let's resolve the third equation for VC(t) and substitute it in the second one.
VC(t) = E0·sin(ωt) − VL(t)
I(t)=C·d[E0·sin(ωt)−VL(t)]/dt

In the last equation we can differentiate each component and, using symbol ' for a derivative, obtain
I(t)=CωE0·cos(ωt)−C·V'L(t)

Together with the first equation from the original system of three equations above we have reduced the system to two equations (again, we use symbol ' for brevity to signify differentiation)
VL(t) = L·I'(t)
I(t)=CωE0·cos(ωt)−C·V'L(t)
Therefore,
I(t)=CωE0·cos(ωt)−CL·I"(t)
or in a more traditional for differential equation form
a·I"(t) + b·I(t) = E0·cos(ωt)
where
a = L/ω
b = 1/(Cω)

Let's try to find such coefficient K that function I(t)=K·cos(ωt) satisfies our equation.
I'(t) = −ωK·sin(ωt)
I"(t) = −ω²K·cos(ωt)

Now our differential equation for I(t) is
−a·ω²K·cos(ωt) + b·K·cos(ωt) = E0·cos(ωt)

From this we can easily find a coefficient K
K = E0/(b−aω²)

Since a=L/ω and b=1/(Cω)
K = E0/{[1/(Cω)] − Lω}

In the previous lectures we have introduced concepts of capacitive reactance XC=1/(Cω) and inductive reactance XL=Lω. Using these variables, the expression for coefficient K is K = E0/(XC−XL)
Therefore,
I(t) = E0·cos(ωt)/(XC−XL)

The last equation brings us to a concept of a reactance of the LC circuit
XC−XL
that is similar to resistance of regular resistors.
Using a concept of reactance, the last equation resembles the Ohm's Law.

Let's determine the voltage drops on a capacitor VC(t) and an inductor VL(t) using the expression for the current I(t).

Since Q(t)=C·VC(t) and I(t)=Q'(t), we can find VC(t) by integrating I(t)/C.
VC(t) = XC·E0·sin(ωt)/(XC−XL)

VL(t) = L·I'(t) = −XL·E0·sin(ωt)/(XC−XL)

Summary

EDF generated by a source of electricity
E(t) = E0·sin(ωt)
where
E0 is a peak voltage on the terminals of a generator,
ω is an angular velocity of a rotor in radians per second.

Alternating electric current in the circuit
I(t) = I0·cos(ωt)
where
I0 = E0/(XC−XL)
XC = 1/(ω·C)
XL = ω·L

Voltage drop on a capacitor
VC(t) =
= XC·E0·sin(ωt)/(XC−XL)

Voltage drop on an inductor
VL(t) =
= −XL·E0·sin(ωt)/(XC−XL)

Phase Shift

Notice that cos(x)=sin(x+π/2). Graph of function y=sin(x+π/2) is shifted to the left by π/2 relative to graph of y=sin(x).
Therefore, oscillations of the current I(t) in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity E(t) by a phase shift of π/2.

Oscillations of the voltage drop on a capacitor VC(t) in the LC circuit are synchronized (in phase) with generated EMF.

Notice that −sin(x)=sin(x+π). Graph of function y=sin(x+π) is shifted to the left by π relative to graph of y=sin(x).
Therefore, oscillations of the voltage drop on an inductor VL(t) in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity E(t) by π (this is called in antiphase).