## Saturday, November 7, 2020

### Ohm's Law for R- and RC-Circuits: UNIZOR.COM - Physics4Teens - Electromagnetism - AC O...

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

R and RC Circuit Ohm's Law

In this chapter we will examine different aspects of the Ohm's Law as they occur in different alternating current (AC) circuits.
Four different types of AC circuits will be considered in this and subsequent lectures:
(a) R-circuit that contains only a resistor;
(b) RC-circuit that contains a resistor and a capacitor;
(c) RL-circuit that contains a resistor and an inductor;
(d) RLC-circuit that contains a resistor, an inductor and a capacitor.
The first two are subject to this lecture.

R-circuit

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 there is only a resistor with resistance R(ohm).

During any infinitesimal time interval the electric current going through a circuit depends only on the generated EMF and a resistance of a circuit components. Since the only resistive component is a resistance R, the electric current at any moment of time will obey the Ohm's Law for direct current.

Therefore the electric current I(t) in this R-circuit is
I(t) = E(t)/R = E0·sin(ωt)/R =
= (E0/R)·sin(ωt) = I0·sin(ωt)

where I0 = E0/R is a peak current.
Now we can express the Ohm's Law for peak voltage and peak amperage in the R-circuit as
I0 = E0 /R

Since in many cases we are interested in effective voltage and effective electric current, and knowing that
Eeff = E0 /2 and Ieff = I0 /2,
the Ohm's Law for effective voltage and effective amperage in the R-circuit can be expressed in a form
Ieff = Eeff /R

RC-circuit

Let's add a capacitor with capacity C (farad) to R-circuit, connecting it in series with a resistor. This is RC-circuit.
As we know, alternating current goes through a capacitor, so we have a closed circuit of resistor R and capacitor C connected in a series with EMF generator that produces sinusoidal voltage E(t)=E0·sin(ωt).
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 is VR(t) and the voltage drop on a capacitor is VC(t). Since a resistor 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) + VC(t)

As explained above 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 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 capacity, are in a relationship
C = QC(t)/VC(t) or
VC(t) = QC(t)/C

At the same time, by definition of the electric current, the electric current going through a capacitor is just a rate of change of electrical charge on its plates, that is, a derivative of an accumulated electric charge by time:
I(t) = Q'C(t)

This is the same electric current that goes through a resistor, since the circuit is closed. Therefore, we can substitute this expression of an electric current into a formula for a voltage drop on a resistor
VR(t) = Q'C(t)·R

We have expressed both voltage drops VR(t) and VC(t) in terms of an electric charge QC(t) accumulated on the plates of a capacitor:
VR(t) = Q'C(t)·R
VC(t) = QC(t)/C

Now we can substitute them into a formula for their sum being equal to a generated EMF
E(t) = VR(t) + VC(t) =
= Q'C(t)·R + QC(t)/C

Since E(t)=E0·sin(ωt) we should solve the following differential equation to obtain function QC(t)
E0·sin(ωt) =
= Q'C(t)·R + QC(t)/C

Let's divide this equation by R and use simplified notation for brevity:
y(t) = QC(t)
a = 1/(R·C)
b = E0/R
Then our equation looks simpler
y'(t)+a·y(t) = b·sin(ωt)

Let's discuss how to solve (integrate) this differential equation.
Recall the formula for a derivative of a product of two functions
(x(t)·y(t))' =
= x(t)·y'(t) + x'(t)·y(t)

Let's find such function x(t) that, if we use it as a multiplier of our differential equation, the left side will look like a complete derivative of x(t)·y(t).

Multiplied by this function x(t), our equation looks like this:
x(t)·y'(t)+a·x(t)·y(t) =
= b·x(t)·sin(ωt)

Let's find function x(t) such that a·x(t) is x'(t). Then the left side of the equation will be equal to (x(t)·y(t))'.

Obvious choice for such function is ea·t. It's an intelligent guess, but it can be determined analytically.
Indeed,
a·x(t) = x'(t)
a·x(t) = dx/dt
dt = dx/x(t)
d(a·t) = dln(x(t))
a·t = ln(x(t)) + K1
x(t) = K2·ea·t
Here K1 and K2 are any constants, so let's use K2=1.
Then x(t)=ea·t.

Now the equation takes the following form
ea·t·y'(t)+a·ea·t·y(t) =
= b·ea·t·sin(ωt)

which is equivalent to
[ea·t·y(t)]' = b·ea·t·sin(ωt)

The above equation should be integrated to get ea·t·y(t) and, finally y(t).

The indefinite integral of the left side is ea·t·y(t).

The indefinite integral of the right side can be found straight forward using Euler formula
cos(φ)+i·sin(φ) = ei·φ
and following from it expressions for trigonometric functions
cos(φ) = (ei·φ+e−i·φ)/2
sin(φ) = (ei·φ−e−i·φ)/(2i)

The result of integration of the right side (without multiplier b) is
ea·t·sin(ωt)·dt =
=
[ea·t/(a²+ω²)]·
·
[a·sin(ωt)−ω·cos(ωt)] + K3
(K3 is any constant)

Therefore,
ea·t·y(t) = b·[ea·t/(a²+ω²)]·
·
[a·sin(ωt)−ω·cos(ωt)] + K3

Reducing by ea·t both sides, the expression for y(t)=QC(t) looks like
y(t) = [b/(a²+ω²)]·
·
[a·sin(ωt)−ω·cos(ωt)] + K4

Consider two constants a/√(a²+ω²) and ω/√(a²+ω²). It is convenient to represent them as sin(φ) and cos(φ) correspondingly for some angle φ=arctan(a/ω).
Then
a·sin(ωt)−ω·cos(ωt) =
= √(a²+ω²)·
·
[sin(φ)·sin(ωt)−cos(φ)·cos(ωt)]

The last expression equals to
−√(a²+ω²)·cos(ωt+φ)
and our equation for y(t) looks like
y(t) = −b·cos(ωt+φ)/√(a²+ω²) +
+ K4

Since y(t) is the electric charge QC(t) accumulated on the plates of a capacitor, its derivative is an electric current in the circuit:
I(t) = b·ω·sin(ωt+φ)/√(a²+ω²)

Let's restore this equation to original constants
a = 1/(R·C)
b = E0/R
The factor at sin(ωt+φ) in the equation above then is
b·ω/√(a²+ω²) =
= E0·ω/(R·√1/(R·C)²+ω²) =
= E0/(√1/(ωC)²+R²) =
= E0/(√XC²+R²)

where XC=1/(ωC) was defined in one of the previous lectures as reactance of a capacitor (or capacitive reactance).

The angle φ=arctan(a/ω) in terms of original constants is
φ=arctan(1/(R·C·ω)) =
= arctan (XC / R)

This angle is called phase shift of the current from the voltage.

Now we have a formula for an electric current in the RC-cirucit that connects generated EMF, resistance of a resistor and capacity of a capacitor
I(t) = E0·sin(ωt+φ)/√XC²+R² =
= I0·sin(ωt+φ)

where I0=E0/√XC²+R² is the peak amperage of the electric current.
This is the Ohm's Law for RC-circuit.

The expression XC²+R² plays in this case the same role as a resistance in case of direct current circuits.

It is important that there is a phase shift φ of the oscillations of an electric current relatively to oscillation of generated EMF.

The last issue is to analyze the effective current in this RC-circuit. Since for sinusoidal oscillations the effective current is by √2 less than peak amperage, the effective current is
Ieff = I0 /2 =
= E0 /(√XC²+R²·√2) =
= Eeff /XC²+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/XC and phase shift φ=arctan(∞)=π/2, which corresponds to results obtained in a lecture "AC Capacitors" of the "Alternating Current Induction" chapter of this course dedicated to only a capacitor in the AC circuit.