Electrical Theory: AC Series-Parallel Circuits

This tutorial assumes the reader has the following knowledge:

Ohm's Law

DC Series Circuits

DC Parallel Circuits

DC Series-Parallel Circuits

Polar Coordinates

Review

For a review of complex number math and polar coordinates, click on the following link

www.associatedcontent.com/article/1792048/complex_number_math_and_polar_coordinate.html

For a review of AC Circuit Analysis, click on the following link:

www.associatedcontent.com/article/1752554/ac_circuit_analysis_phase_relation.html

AC Series Parallel Circuits

Problem:

Find the total current flowing in the circuit in Figure One.

R1 = 10 ohms

R2 = 10 ohms

C1 = 1 microfarad

Frequency = 10000 Hz

Voltage Source = 10 volts @ 0 degrees

Solution:

Following the math can be a tedious job. Therefore I will provide an explanation of each step in the solution. The objective is to find the total impedance of the circuit and then divide the voltage by the impedance to find the current. The impedance will be associated with a negative angle because we are dealing with capacitance. The current will be associated with a positive angle.

CALCULATE CAPACITIVE REACTANCE

The capacitive reactance Xc is

Xc = 1/(2*pi*f *c)

where

pi = 3.14

f = 10000 hz = 10⁴ hz

c = 1 microfarad = 10^-6 farads

Hence Xc = 1/(2 * 3.14 * 10⁴ * 10^-6)

Xc = 1/(6.28*10⁴*10^-6)

Xc = 1/(6.28*10^-2)

Xc = 0.159*10² =15.9

SOLVE FOR R2 IN PARALLEL WITH C1

Now that we have the capacitive reactance, we can solve for the parallel combination of the capacitive reactance Xc and the resistance R2. The capacitive reactance Xc is associated with an angle of -90^o. (See figure Two.)

Z1 = R2//Xc

where // means "in parallel with."

Z1 = (R2/0 * Xc/-90) divided by (R2 - jXc)/-n^o Equation One

First we will solve for the complex expression R2 - jXc by using the Pythagorean theorem

R2 - jXc = ((R2)² + Xc²)^1/2

R2 = 10

Xc = 15.9

This impedance is associated with an angle defined by arctan (Xc/R2)

R2 - jXc = (10² + 15.9²)^1/2

R2 - jXc = (100 + 252.81)^1/2 =

R2 - jXc = (352.81)^1/2

The tangent of the angle is always the opposite side of the triangle
divided by the adjacent side of the triangle. In this case, the opposite side is the capacitive reactance and the adjacent side is the resistance.

R2 - jXc = 18.78 arctan (-Xc/R)

R2 - jXc = 18.78 arctan(-15.9/10)

R2 - jXc = 18,78 arctan(-1.59)

R2 - jXc = 18.78 /-57.8^o

Now we can turn back to equation one which gave the total parallel impedance of R2 and Xc . Returning to Equation one, we already know that R2 - jXc = 18.78/-57.8o

SOLVE FOR TOTAL IMPEDANCE

The total impedance of a parallel circuit is given by the product of the two impedances divided by the sum of the two impedances.

Z1 = (R2/0^o* Xc/-90^o) divided by (R2 - jXc)/-n^o Equation One

We already know that (R2 - j Xc) is equal to 18.78 /-57.8^o

Z1 = (R2/0^o * Xc/-90^o) divided by (18.78 /-57.8^o)

Z1 = 10*15/-90^o divided by (18.78 /-57.8^o)

Z1 = 150/-90^o divided by (18.78 /-57.8^o)

Z1 = 7.99/-32.2⁰

Now we can calculate the total impedance seen by the voltage source. See figure one.

Zt = R1 + Z1 Equation two

Zt = 10/0 + 7.99/-32.2^o

In any right triangle the length of the adjacent side is equal to the length of the hypotenuse multiplied by the cosine of the angle between the hypotenuse and the adjacent side. Likewise, the length of the opposite side is equal to the length of the hypotenuse times the sine of that angle.

Zt = 10 + 7.99*cos(32.2) - j*7.99*sin(32.2)

Zt = 10 + 7.99 * 0.846 - j*7.99 * 0.533

Zt = 10 + 6.76 - j*4.26

Zt = 16.76 - j*4.26

Now we can use Pythagorean's theorem.

Zt = (16.76² + 4.26²)^1/2

Zt = (280.90 + 18.15)^1/2

Zt = (299.15)^1/2

Zt = 17.29

arctan(4.26/16.76) = arctan(0.254)

arctan(0.254) = 6.78^o degrees

Zt = 17.29/-14.25^o

SOLVE FOR TOTAL CURRENT

Now we can use Ohm's Law to solve for the current.

It = V/Zt = (10/0^o) divided by (17.29/-14.25^o)

It = 0.578/14.25^o