*Notes to a video lecture on http://www.unizor.com*

__RLC Circuit Ohm's Law__

This lecture combines the material of the previous two ones dedicated to the Ohm's Law in alternating current circuits.

The first one was analyzing a circuit with a resistor and a capacitor.

The second one analyzed a circuit with a resistor and an inductor.

This lecture analyzes a circuit with all three elements - resistor, inductor and capacitor.

Consider a closed AC circuit with a generator with no internal resistance producing a sinusoidal electromotive force (

*EMF*or

*voltage*)

**E(t)=E**(volt)_{0}·sin(ωt)where

*is a peak EMF produced by an AC generator,*

**E**_{0}*is angular speed of EMF oscillations and*

**ω***is time.*

**t**In this circuit we have a resistor with resistance

*, an inductor with inductance*

**R***and a capacitor with capacitance*

**L***connected in series with EMF generator. This is RLC-circuit.*

**C**Since the circuit is closed, the electric current

*going through a circuit is the same for all components of a circuit.*

**I(t)**Assume that the voltage drop on a resistor (caused by its

*resistance*) is

**R***, the voltage drop on an inductor (caused by its*

**V**_{R}(t)*inductive reactance*) is

**X**_{L}=ω·L*and the voltage drop on a capacitor (caused by its*

**V**_{L}(t)*capacitive reactance*) is

**X**_{C}=1/(ω·C)*.*

**V**_{L}(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) = V**_{R}(t) + V_{L}(t) + V_{C}(t)As explained in the previous lectures the Ohm's Law localized around a resistor states that

*or*

**I(t) = V**_{R}(t)/R

**V**_{R}(t) = I(t)·RThe inductance of an inductor

*, voltage drop on this inductor caused by a self-induction effect*

**L***, electromagnetic flux going through an inductor*

**V**_{L}(t)*and electric current going through it*

**Φ(t)***, as explained in the lecture "*

**I(t)***AC Inductors*" of the chapter "

*Electromagnetism - Alternating Current Induction*", are in a relationship

**V**d_{L}=**Φ/**d**t = L·**d**I(t)/**d**t**The capacity of a capacitor

*, voltage drop on this capacitor*

**C***and electric charge accumulated on its plates*

**V**_{C}(t)*, according to a definition of a*

**Q**_{C}(t)*capacitance*, are in a relationship

**C***or*

**C = Q**_{C}(t)/V_{C}(t)

**V**_{C}(t) = Q_{C}(t)/CWe have expressed both voltage drops

*and*

**V**_{R}(t)*in terms of an electric current*

**V**_{L}(t)*:*

**I(t)**

**V**_{R}(t) = I(t)·R

**V**_{L}(t) = L·I'(t)Since an electric charge

*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*

**Q**_{C}(t)*and*

**I(t) = Q**_{C}'(t)

**I'(t) = Q**_{C}"(t)Now all three voltage drops can be expressed as functions of

*and, equating their sun to a generated EMF*

**Q**_{C}(t)*, we can the following differential equation*

**E(t)**

**E**

+ L·Q

= L·Q

+ (1/C)·Q_{0}·sin(ωt) = Q_{C}'(t)·R ++ L·Q

_{C}"(t) + Q_{C}(t)/C == L·Q

_{C}"(t) + R·Q_{C}'(t) ++ (1/C)·Q

_{C}(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

*and inductance*

**C***, to use corresponding reactance*

**L***and*

**X**_{C}=1/(ω·C)*.*

**X**_{L}=ω·LSo, we will substitute

**C = 1/(ω·X**_{C})

**L = X**_{L}/ωSubstitute Q

_{C}(t)=y(t) for brevity. Then our equation looks like

**(X**

+ ω·X_{L}/ω)·y"(t) + R·y'(t) ++ ω·X

_{C}·y(t) = E_{0}·sin(ωt)Solving this equation for

*and differentiating it by time*

**y(t)=Q**_{C}(t)*, we will obtain the expression for an electric current*

**t***in this circuit as a function of all given parameters and time.*

**I(t)**First of all, let's find a particular solution of this differential equation.

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)**Then

*[*

**y'(t) =**

= ω·= ω·

*]*

**F·cos(ωt)−G·sin(ωt)***[*

**y"(t) =**

= −ω²·= −ω²·

*]*

**F·sin(ωt)+G·cos(ωt)**

**=**

= −ω²·y(t)= −ω²·y(t)

Using this trigonometric representation of potential solution

*, the left side of our equation is*

**y(t)***[*

**(X**

+ ω·X

= −(X

+ ω·X

= −ω·X

+ ω·X

= R·y'(t) + ω·(X

= R·ω·_{L}/ω)·y"(t) + R·y'(t) ++ ω·X

_{C}·y(t) == −(X

_{L}/ω)·ω²·y(t) + R·y'(t) ++ ω·X

_{C}·y(t) == −ω·X

_{L}·y(t) + R·y'(t) ++ ω·X

_{C}·y(t) == R·y'(t) + ω·(X

_{C}−X_{L})·y(t) == R·ω·

*]*

**F·cos(ωt)−G·sin(ωt)***[*

**+**

+ ω·(X+ ω·(X

_{C}−X_{L})·*]*

**F·sin(ωt)+**

+G·cos(ωt)+G·cos(ωt)

*[*

**=**

= ω·= ω·

*]*

**(X**_{C}−X_{L})·F−R·G*[*

**·sin(ωt)+**

+ω·+ω·

*]*

**R·F+(X**_{C}−X_{L})·G

**·cos(ωt)**Since this is supposed to be equal to

*, we have the following system of two linear equations with two variables*

**E**_{0}·sin(ωt)*and*

**F**

**G***[*

**ω·***]*

**(X**_{C}−X_{L})·F−R·G

**= E**_{0}*[*

**ω·***]*

**R·F+(X**_{C}−X_{L})·G

**= 0**or

**(X**_{C}−X_{L})·F−R·G = E_{0}/ω

**R·F+(X**_{C}−X_{L})·G = 0Determinant of the matrix that defines this system is

*.*

**(X**_{C}−X_{L})²+R²It's always positive and usually is denoted as

**Z² = (X**_{C}−X_{L})²+R²The value

*is called the*

**Z****impedance**of a circuit and, as we will see, plays the role of a resistance for an entire RLC-circuit.

This system of two linear equations with two variables has a solution:

**F = (E**_{0}/ω)·(X_{C}−X_{L})/Z²

**G = −(E**_{0}/ω)·R/Z²Consider two expressions that participate in the above solutions

*and*

**F***:*

**G***and*

**(X**_{C}−X_{L})/Z²

**R/Z²**Since

*, we can always find an angle*

**Z² = (X**_{C}−X_{L})²+R²*such that*

**φ***and*

**(X**_{C}−X_{L})/Z = sin(φ)

**R/Z = cos(φ)**Using this, we express the solutions to the above system as

**F = (E**_{0}/(ω·Z))·sin(φ)

**G = −(E**_{0}/(ω·Z))·cos(φ)Now the solution of the differential equation for

*that we were looking for in a format*

**y(t)=Q**_{C}(t)

**y(t) = F·sin(ωt) + G·cos(ωt)**looks like

**y(t)=(E**

− (E

= −(E_{0}/(ω·Z))·sin(φ)sin(ωt) −− (E

_{0}/(ω·Z))·cos(φ)cos(ωt) == −(E

_{0}/(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) =**

= (E

= (E

= I= (E

_{0}/(ω·Z))·ω·sin(ωt+φ) == (E

_{0}/Z)·sin(ωt+φ) == I

_{0}·sin(ωt+φ)where

*is an equivalent of the*

**I**_{0}=E_{0}/Z**Ohm's Law for an RLC-circuit**.

Similar relationship exists between effective voltage and effective current

**I**

= E_{eff}= I_{0}/√2 == E

_{0}/(√2·Z) = E_{eff}/ZHere

**impedance**

*, defined by resistance*

**Z***, inductive reactance*

**R***and capacitive reactance*

**X**_{L }*as*

**X**_{C }

**Z = √(X**_{C}−X_{L})²+R²plays a role of a resistance in the RLC-circuit.

There is a phase shift

*of the electric current oscillations relative to EMF. It is also defined by the same characteristics of an RLC-circuit:*

**φ**

**(X**_{C}−X_{L})/Z = sin(φ)

**R/Z = cos(φ)**Hence

**tan(φ) = (X**_{C}−X_{L})/RThis makes phase shift positive or negative depending on the values of capacitive and inductive reactance.

If

*is greater than*

**X**_{C}*, the phase shift is positive, if the reverse is true, the phase shift is negative.*

**X**_{L}If the values of reactance are the same, that is

*, there is no phase shift. From definition of reactance, it happens when*

**X**_{C}=X_{L}*or*

**1/(ω·C) = ω·L***or*

**1/(L·C) = ω²**

**L·C = 1/ω²**This relationship between inductance, capacitance and angular speed of EMF oscillation is call

*resonance*.

Expressions for electric current

**I(t) = (E**

= I_{0}/Z)·sin(ωt+φ) == I

_{0}·sin(ωt+φ)in RLC-circuit, where impedance

*and*

**Z=√(X**_{C}−X_{L})²+R²*, correspond to analogous formulas for R-, RC- and RL-circuits presented in the previous lectures. All it takes is to set the appropriate value of capacitance or inductance to zero.*

**tan(φ)=(X**_{C}−X_{L})/R
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