 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.

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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 