Tuesday, November 10, 2020

Ohm's Law for RL_Circuit: UNIZOR.COM - Physics4Teens - Electromagnetism ...

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

RL 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)(volt)
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 and an inductor with inductance L connected in series with EMF generator. This is RL-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 is VR(t) and the voltage drop on an inductor caused by self-induction effect is VL(t). Since a resistor and an inductor 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)

As explained in the previous lecture for an R-circuit, 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

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)

Now we can substitute them into a formula for their sum being equal to a generated EMF
E(t) = VR(t) + VL(t) =
= I(t)·R + I'(t)·L

Since E(t)=E0·sin(ωt) we should solve the following differential equation to obtain function I(t)
E0·sin(ωt) = I(t)·R + I'(t)·L

Let's divide this equation by L and use simplified notation for brevity:
y(t) = I(t)
a = R/L
b = E0/L
Then our equation looks simpler
y'(t)+a·y(t) = b·sin(ωt)

This exact differential equation was solved in the previous lecture dedicated to RC-circuit, notes for that lecture contain detail analysis of its solution
y(t) = −b·cos(ωt+ψ)/√(a²+ω²) + K
where phase shift ψ=arctan(a/ω) and K is a constant that can be determined by initial condition.

Using original notation,
I(t) = −(E0/L)·cos(ωt+ψ)/
/(R/L)²+ω² + K =
= −E0·cos(ωt+ψ)/
/R²+(ω·L)² + K =
= −E0·cos(ωt+ψ)/R²+XL² + K

where XL=ω·L is inductive reactance of an inductor - a characteristic of an inductor functionally equivalent to a resistance for a resistor and measured in the same units - ohm and
tan(ψ) = a/ω = R/(L·ω) = R/XL

Let's apply some Trigonometry to simplify the above formula.
−cos(ωt+ψ) =
= −sin((π/2)−ωt−ψ) =
= −sin((π/2)−ψ−ωt) =
= sin(ωt−((π/2)−ψ)) =
= sin(ωt−φ)

where φ=(π/2)−ψ and, therefore, tan(φ)=1/tan(ψ)=XL/R.

Using phase shift φ, the equation for the current I(t) looks like
I(t) = E0·sin(ωt−φ)/R²+XL² + K
where tan(φ)=XL/R.

The value of a constant K can be defined by initial conditions. Since we don't really know these conditions (like what is the value of a current at some moment in time), traditionally this constant is assigned the value of zero, motivating it by the fact that, if the generated EMF is oscillating between its minimum and maximum of the same magnitude with different signs, the current also will be "symmetrical" relative to zero level, which requires the value K=0

The final version of the current in this RL-circuit is
I(t) = E0·sin(ωt−φ)/R²+XL² =
= I0·sin(ωt−φ)

where I0 = E0/R²+XL²

The last issue is to analyze the effective current in this RL-circuit. Since for sinusoidal oscillations the effective current is by √2 less than peak amperage, the effective current is
Ieff = I0 /2 =
= E0 /(√XL²+R²·√2) =
= Eeff /XL²+R²

This is the Ohm's Law for effective voltage and amperage in RC-circuit.

It's important to notice that in the absence of a resistor (that is, R=0) the formula for I(t) transforms into
I(t) = I0·sin(ωt−φ)
where peak amperage I0=E0/XL and phase shift φ=arctan(∞)=π/2, which corresponds to results obtained in a lecture "AC Inductors" of the "Alternating Current Induction" chapter of this course dedicated to only an inductor in the AC circuit, taking the current as given and deriving the EMF.