RLC Circuit Ohm Law: UNIZOR.COM - Physics4Teens - Electromagnetism - AC Ohm Law
Channel: Zor Shekhtman on UNIZOR Education
Date: 16/11/2020

UNIZOR.COM - Physics through a Looking Glass of Mathematician


RLC Circuit Ohm's Law


Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (EMF or voltage)
E(t)=E0·sin(ωt)
where E0 is a peak EMF produced by an AC generator,
ω is angular speed of EMF oscillations and
t is time.
In this circuit we have a resistor with resistance R, an inductor with inductance L and a capacitor with capacitance C connected in series with EMF generator. This is RLC-circuit.
Since the circuit is closed, the electric current I(t) going through a circuit is the same for all components of a circuit.

Assume that the voltage drop on a resistor (caused by its resistance R) is VR(t), the voltage drop on an inductor (caused by its inductive reactance XL=ω·L) is VL(t) and the voltage drop on a capacitor (caused by its capacitive reactance XC=1/(ω·C)) is VL(t).

Since a resistor, an inductor and a capacitor are connected in a series, the sum of these voltage drops should be equal to a generated EMF E(t):
E(t) = VR(t) + VL(t) + VC(t)

As explained in the previous lectures the Ohm's Law localized around a resistor states that
I(t) = VR(t)/R or
VR(t) = I(t)·R

The inductance of an inductor L, voltage drop on this inductor caused by a self-induction effect VL(t), electromagnetic flux going through an inductor Φ(t) and electric current going through it I(t), as explained in the lecture "AC Inductors" of the chapter "Electromagnetism - Alternating Current Induction", are in a relationship
VL = dΦ/dt = L·dI(t)/dt

The capacity of a capacitor C, voltage drop on this capacitor VC(t) and electric charge accumulated on its plates QC(t), according to a definition of a capacitance C, are in a relationship
C = QC(t)/VC(t) or
VC(t) = QC(t)/C

We have expressed both voltage drops VR(t) and VL(t) in terms of an electric current I(t):
VR(t) = I(t)·R
VL(t) = L·I'(t)
Since an electric charge QC(t) is also involved to express the voltage drop on a capacitor, we will use this electric charge as a main variable, using the definition of an electric current as a rate of change of electric charge
I(t) = QC'(t) and
I'(t) = QC"(t)

Now all three voltage drops can be expressed as functions of QC(t) and, equating their sun to a generated EMF E(t), we can the following differential equation
E0·sin(ωt) = QC'(t)·R + L·QC"(t) + QC(t)/C =
= L·QC"(t) + R·QC'(t) + (1/C)·QC(t)


As we know, capacitive and inductive reactance are functionally equivalent to resistance, and they all have the same unit of measurement - ohm. Therefore, it's convenient, instead of capacitance C and inductance L, to use corresponding reactance XC=1/(ω·C) and XL=ω·L.
So, we will substitute
C = 1/(ω·XC)
L = XL/ω

Substitute QC(t)=y(t) for brevity. Then our equation looks like
(XL/ω)·y"(t) + R·y'(t) + ω·XC·y(t) = E0·sin(ωt)

Solving this equation for y(t)=QC(t) and differentiating it by time t, we will obtain the expression for an electric current I(t) in this circuit as a function of all given parameters and time.
The form of the right side of this equation prompts to look for a solution in trigonometric form
y(t) = F·sin(ωt) + G·cos(ωt)


Using this trigonometric representation of potential solution y(t), the left side of our equation is
(XL/ω)·y"(t) + R·y'(t) + ω·XC·y(t) =
= ω·[(XC−XL)·F−R·G]·sin(ωt)+ω·[R·F+(XC−XL)·G]·cos(ωt)


Since this is supposed to be equal to E0·sin(ωt), we have the following system of two linear equations with two variables F and G
ω·[(XC−XL)·F−R·G] = E0
ω·[R·F+(XC−XL)·G] = 0
or
(XC−XL)·F−R·G = E0/ω
R·F+(XC−XL)·G = 0


This system of two linear equations with two variables has a solution:
F = (E0/ω)·(XC−XL)/Z²
G = −(E0/ω)·R/Z²
where 
Z² = (XC−XL)²+R²
The value Z is called the impedance of a circuit and, as we will see, plays the role of a resistance for an entire RLC-circuit.


We can always find an angle φ such that
(XC−XL)/Z = sin(φ) and
R/Z = cos(φ)

Using this, we express the solutions to the above system as
F = (E0/(ω·Z))·sin(φ)
G = −(E0/(ω·Z))·cos(φ)
Now the solution of the differential equation for y(t)=QC(t) that we were looking for in a format
y(t) = F·sin(ωt) + G·cos(ωt)
looks like
y(t)=(E0/(ω·Z))·sin(φ)sin(ωt) − (E0/(ω·Z))·cos(φ)cos(ωt) =
= −(E0/(ω·Z))·cos(ωt+φ)

As we noted before, differentiation of this function gives the electric current in the circuit I(t):
I(t) = y'(t) =
= (E0 /(ω·Z))·ω·sin(ωt+φ) =
= (E0 /Z)·sin(ωt+φ) =
= I0·sin(ωt+φ)
where I0=E0 /Z is an equivalent of the Ohm's Law for an RLC-circuit.

Similar relationship exists between effective voltage and effective current
Ieff = I0 /√2 = E0 /(√2·Z) = Eeff /Z