Boldface symbols

indicate a complex

quantity. Later, we use that Re(u(t))

= Ua cos ?t, but here we

treat voltage and current as complex functions of time even though this is not physically possible. (In the real world,

voltages and currents are always scalar quantities.)

If we apply the complex voltage (10) to the

right-hand side of (1), we can find the forced response.

The current is now a complex function on the

format

(11)

Substitution of the complex

voltage (10) and ditto current (11) into (1) gives

(12)

Which after differentiation leads to

(13)

All

terms have a common factor

and dividing all terms with

this factor gives

(14)

Since

the modules of

is 1, we immediately get from (14) that

(15)

Now,

because

(16)

Substitution of (15) and (16) into (11), the complex-valued forced

response to the complex-valued

input

voltage becomes

(17)

which

may be reformulated as

(18)

By equating the real part on both sides of (17), the actual real current

in the circuit in Figure 1 is given by

(19)

which

corresponds to (9) derived earlier.

Concluding

remarks

In (18)

it is convenient to define

as the complex

impedance of the series connection of

the resistor and the inductor shown in Figure 1. Then, (18) is on the form

and

hence for ac circuit analysis (18) extends Ohm’s

Law

for resistive circuits to also include circuits

having inductors as well as resistors.

?

As illustrated by this small example, the

principle of using complex-valued voltage and current functions and complex impedance’s

is a very powerful tool for analysis of electric circuits. Basically, what

happens is that a time-domain differential equation is transformed to an

equivalent algebraic equation containing complex coefficients, but the latter is

in general much easier to solve. see 3 for further reference.

Matrix formulation of chemical problems

Balance

of chemical equation

One of the most frequent tasks, in both

laboratory and chemistry class, is obtaining a set of stoichiometric

coefficients to ‘balance’ a chemical equation. This task is a necessary first

step towards solving most school problems. As has been pointed out, the expression “balancing a chemical

equation” is a contradiction of terms. A chemical equation is

fundamentally a conservation statement about atomic species and, thus, it

should already be balanced. Usually, students obtain coefficients by a trial

and error method. This effort yields no profit and loses a lot of time. When a moderate number of chemical species (8

or more) are involved in a given

skeletal equation, the task is nearly impossible to solve by trial and error. Thus, some special methods are

necessary. Up to best of our knowledge, three main methods can be used which

are: the oxidation-number method; the ion-electron method; and the matrix

method. However, there are a lot of

limitations to use the first and second method to balance some systems, and due

to this matrix play a key role in chemical reactions.

For example, we have used matrices concept

to balance the following chemical

equation. see 4 for further reference.

a FeCl2 + b Na3(PO4) ? cFe3(PO4)2 +

d NaCl

where a, b, c, and d are the stoichiometric coefficients.

Writing the balance for each ion we get the

following homogeneous equations:

Fe:

Cl:

Na:

(PO4):

For this reaction, it is

better to treat the phosphate ion

as an element, and not break it

into two individual elements. The coefficient matrix

for the homogeneous linear equation system

above is

and a few row operations gives the reduced row

echelon form

Hence, the smallest positive integer solution

is a = 3, b = 2, c = 1, and d = 6.

Example 2:

A firm wishes to market bags of lawn fertilizer

which contain 25% nitrogen, 7% phosphoric acid, and 8% potash. The firm has

four chemical precursors C1, C2, C3, C4 which are to be combined to make the

fertilizer. The percentage of each ingredient in a pound of these chemicals is

given in Table 1. How much of each chemical should be mixed to obtain 100

pounds of fertilizer meeting

Nitrogen

C1

20

C2

25

C3

0

C4

3

Phosphoric

Acid

12

5

6

7

Potash

0

5

15

10

Table 1:

Percentage of chemicals

these criteria?

Solution :

Let

be the

number of pounds of chemical Ci used. Then since the total adds up to 100

pounds we have

Now

pounds

of chemical C1 contains 0.20

pounds

of nitrogen,

pounds

of C2 contains 0.25

pounds

of nitrogen,

pounds

of C3 contains 0 pounds of nitrogen, and

pounds

of C4 contains 0.30

pounds

of nitrogen.

Since there should be

pounds

of nitrogen in the mixture we obtain

0.20

+ 0.25

+ 0

+ 0.30

= 25.

Similar expressions can be derived for

phosphoric acid and potash giving us all together a system of linear equations

with augmented matrix