Thursday, April 28, 2011

Algebra: What is mean Proportion and How to solve it?

A proportion such as 1:2 =2:4 or a:b = b:c  is called a mean proportion because the second and the third terms are equal. In the latter b is the mean proportional between the first and the last terms a and c, respectively and c is the third proportional to a and b.

Solving for the mean proportional
 Given
 in proportion, the product of the extremes is equal to the product of the means.
  extracting the square roots of booth members of the equation

Note; like roots and like powers of equal numbers are equal

It follows that:
The mean proportional between two quantities is equal to the square root of their product.

example 1 :
Solve for x in






example 2 :
Solve for x in



Algebra: How to solve the Equations containing decimals

Equations containing decimals are equations with fractions. hence the solution of decimal equations is similar to that of fractional equations with this difference-in decimal equation the multiplier of both members is that power
of ten that will clear the equations of its decimal.

Example1.
Solve for x:                                 .3x + 5 = .2x
Multiplying both members          
of the equation by 10,                 3x + 50 = 2x
                                                  x = -50

Check                                     .3x(-50) + 5 = .2(-50)
                                              -15+5  = -10
                                               -10 = -10

Example no 2.                                      Solve for y in the equation
                                                         .5y/3 - .18/4 = .1y/6
Clear the equation of decimals by multiplying both members by 100
50y/3 - 18/4 = 10y/6
Clear the equation of fractions by multiplying both members by 12, the LCD
200y - 54 = 20y
180y = 54
y = 54/180 = .3

check:
.5y/3 - .18/4 = .1y/6
(.5(.3))/3 -.18/4  = (.1(.3))/6
.05-.045 = 0.005
.005 = .005



                                            

Sunday, February 27, 2011

Algebra : How to solve a Literal equation

Equations where the known quantities are expressed as letters are called literal equations. Formulas are literal equations. Generally, the last letter of the alphabet are used to represent the unknown quantities, and the first letters, the known quantities.

Examples:

Solve for the unknown and check

1. 3x + 2a = 2x+5a     transposing
    3x - 2x = 5a - 2a      combining like terms
           x = 3a  


check
3.3a + 2a = 2.3a + 5a
9a + 2a = 6a + 5a
11a = 11a


2. 8z + 6m = 5z + 9m
    8z - 5z = 9m - 6m        transposing
    3z = 3m                   combining like terms
    z = m                       axiom of division

check
   8.m + 6m = 5.m + 9.m
   8m + 6m= 5m + 9m
   14m = 14m  

Friday, December 10, 2010

Power Factor for a more efficient circuit and loading

Power factor is an important aspect to consider in AC circuit, because any factor less than 1 means that the circuit wiring has to carry more current than what would be necessary with zero reactance in the circuit to deliver the same true power to the resistive load.
    When expressed as fraction, the ratio between true power and  apparent power is called the power factor of the circuit.
                                                     Kw
The power factor is given cosÆŸ = --------
                                                    KVA
in the case of single phase supply

                     VI                    KVAx1000
KVA=     ----------  or I=    -----------  
                     1000                    V

In case of three-phase supply
            sqrt(3)VL IL                 KVAx 1000
KVA= --------------     or IL=-------------
                1000                           sqrt(3)VL
In the case of three phase supply
   All current will cause losses in the supply and distribution system. A load with a power of 1.0 result in the most efficient loading of the supply and a load with a power of 0.5 will result in much higher losses in the supply such as an induction motor, power transformer, lighting ballast,welder or variable speed drive, switched mode power supply, discharge lighting or other electronic load.
  The heating and lighting loads supplied from three-phase supply have power factors ranging from 0.95 to unity. But motor loads have usually low lagging power factor ranging from 0.5 to 0.9. Single phase motors may have a power factor of as low as 0.4 and electric welding units have even lower power factor of 0.2 or 0.3
 A poor power factor due to an inductive load can be improved by the addition of power factor correction,but a poor power factor due to a distorted current waveform requires a change in equipment design or expensive harmonic filter s to gain an appreciable improvement. many inverters are quoted as having a power factor of better than 0.95 when in reality, true power factor is between 0.5 and 0.75. the Figure of 0.95 is based on the cosine of the angle between the voltage and current but does not take into account that the current waveform is discontinues and therefore contributes to increased losses on the supply
   in each  equation above, the KVA is directly proportional to the current. The chief disadvantage of a low power factor is that the current requires for a given power is very high. this fact leads to the following undesirable results.

1. Large KVA for a given amount of power.
    All electric machines such as alternators, transformers, cables are limited in their current-carrying capacity by the permissible temperature rise which is proportional to I^2 .hence, they may all be fully loaded with respect to their rated KVA without delivering their full power.

2. Poor voltage regulation . When loading a low lagging power factor is switched on,there is a large voltage drop in the supply lines and transformers. this drop in voltage adversely affects the starting torque of motors and needs expensive voltage stabilizing equipment for keeping the consumers voltage fluctuations within the statutory limits

RLC parallel circuit formula and Phasor diagram

RLC in parallel


RCL parallel circuit can be considered as two reactance's XC and XL where the currents are in phase opposition in parallel with the resistor.
 parallel RLC circuit connection
Through the use of  the phasor diagram the effective total resistance can be found  . It can be observed that IR is at the base of the right triangle and IL minus IC minus IL forming the perpendicular.
The amplitude of the resultant current is the hypotenuse.
 parallel RLC phasor diagram

To calculate the resultant we use the Pythagorean theorem . Ic and IL  are 180 degree out of phase  therefore cancel out. The resultant current is the difference between the two.

RLC parallel connection formula current calculation
              








Impedance  phase angle  is calculated from the difference between IL and IC divided by IR