This article assumes the reader has knowledge in the following areas:

Ohm’s Law

Series Circuits

Parallel Circuits

Series Parallel Circuits

Alternating Current (AC)

Trigonometry

Inductive Reactance

Capacitive Reactance

Impedance

When we discussed DC circuits, we assumed a constant voltage output from a DC source. In this discussion our voltage source outputs a sine wave with equal positive and negative peak voltages.

The reader could review inductive reactance, capacitive reactance and impedance in my article at

http://www.associatedcontent.com/article/1594075/capacitive_and_inductive_reactance.html?cat=58

__The sine wave__

The typical sine wave output could be described by the following formula

v = Vp * sin (wt)

where

v is the instantaneous voltage

Vp is the peak voltage of the sine wave

w is equal to 2*pi*F

pi equals 3.14

F is the frequency of the sine wave

t is the time period

The unit of measurement of the product w*t is the radian.

Likewise the equation for the current is

i = Ip * sin (wt)

where

i is the instantaneous voltage

Ip is the peak current of the sine wave

__1) The resistor__

Consider an AC circuit containing one resistor.

The instantaneous voltage v is equal to

v = Vp * sin (wt)

Then the instantaneous current i would be equal to

i = Ip * sin (wt)

__2) The capacitor__

Consider an AC circuit containing one capacitor.

__
__The instantaneous voltage v is

v = Vp * sin (wt)

The instantaneous current i is

i = Ip * sin (wt + 90

^{o})

Let’s figure out what this means in terms of the relation between voltage and current.

We define the time t = 0 as the instant that the voltage is at zero volts.

Assume that

Vp = 1 volt

Ip = 1 ampere

If t = 0, then w*t = 0 and

v = Vp * sin 0 = 0 volts

The current I is

i = Ip * sin (wt + 90) = Ip * sin 90 = Ip * 1 = Ip.

Look at figure one. It shows an AC sine wave and relates it to a circle. One cycle of the sine wave represents 360 degrees of the circle.

Figure two shows the relation between the voltage and current for an AC circuit containing a capacitor. The current leads the voltage by 90 degrees.

Hence the instantaneous current is at its maximum value when the instantaneous voltage is zero volts.

__3) The inductor__

Consider an AC circuit containing one inductor.

The source voltage v is

v = Vp * sin (wt)

Assume that

Vp = 1 volt

Ip = 1 ampere

At t = 0, the current is 0 amps.

i = Ip * sin (wt) = Ip * sin 0 = 0 amps

v = Vp * sin(wt + 90) = Vp * sin 90 = Vp * 1 = Vp

Figure three shows the relation between the voltage and current for AC circuit containing an inductor. The voltage leads the current by 90 degrees.

__4) A resistor and a capacitor in series__

In a series circuit containing a resistor and a capacitor, we are dealing with impedance Zt

The total impedance Zt is

Zt = R – 1/(jwc)

and

arc tan (1/jwc/r) = arc tan (Xc/R)

Assume that

Vp = 1 volt

Ip = 1 ampere

Now if v = Vp sin (wt)

then

i = Ip sin (wt + arc tan (Xc/R) )

The current leads the voltage by the angle

O = arc tan (Xc/R)

__5) A resistor and an inductor in series__

In a series circuit containing a resistor and a inductor, we are dealing with impedance Zt

The total impedance Zt is

Zt = R + jwL

and

arc tan (jwL/r) = arc tan (XL/R)

Assume that

Vp = 1 volt

Ip = 1 ampere

Now if i = Ip* sin (wt)

then

v = Vp sin (wt + arc tan (XL/R) )

The voltage leads the current by the angle

O = arc tan (XL/R)

__5) A Resistor and a Capacitor in Parallel__

The total impedance is

Zt = R*jXc/(R-jXc)

The current leads the voltage by the angle

O = arc tan (Xc/R)

__6) A Resistor and an Inductor in Parallel__

The total impedance is

Zt = R*jXL/(R+jXL)

The voltage leads the current by the angle

O = arc tan (XL/R)

__7) A resistor, inductor and capacitor in series__

Consider a circuit with a resistor, a capacitor and an inductor in series. Assume that the inductive reactance is greater than the capacitive reactance.

The total impedance is

Zt = R + j(XL – Xc)

and

O = arc tan ((XL-Xc)/R)

Assume that

Vp = 1 volt

Ip = 1 ampere

Now if i = Ip* sin (wt)

then

v = Vp sin (wt + arc tan ((XL-Xc)/R) )

The voltage leads the current by the angle

O = arc tan ((XL-Xc)/R)

The voltage leads the current because the inductive reactance is greater than the capacitive reactance.

Now lets repeat the above exercise assuming that the capacitive reactance is greater than the inductive reactance.

The total impedance is

Zt = R + j(XL – Xc)

and

O = arc tan ((XL-Xc)/R)

Assume that

Vp = 1 volt

Ip = 1 ampere

Now if v = Vp sin (wt)

then

i = Ip sin (wt + arc tan ((XL-Xc)/R) )

The voltage lags the current by the angle

O = arc tan ((XL-Xc)/R)

The current leads the voltage because the capacitive reactance is greater than the inductive reactance.

__8) A resistor, inductor and capacitor in parallel__

Consider a parallel circuit containing a resistor, an inductor and a capacitor. Assume that the inductive reactance is greater than the capacitive reactance.

The total impedance is

1/Zt = 1/R + 1/jXL + 1/(-jXc)

In polar coordinates this equation is

1/(Zt__/____O__) = 1/R + 1/XL__/90__ + 1/Xc__/-90__

Since the inductive reactance is greater than the capacitive reactance, the net reactance is inductive and the phase angle is

O = arc tan ((XL-Xc)/R)

Therefore the voltage leads the current by the angle

O = arc tan ((XL-Xc)/R)

If the capacitive reactance is greater than the inductive reactance in the above exercise, the phase angle would be

O = arc tan ((XL-Xc)/R)

Therefore the voltage lags the current by the angle

O = arc tan ((XL-Xc)/R)

This concludes this article on AC analysis.

References:

I have a Bachelor of Science in Electrical Engineering

Introductory Circuit Analysis Third Edition

ISBN 0-675-8559-4