BILL ADONGO: REVIEW

LIFE TIME:

BILL ADONGO(1983-present), a Ghanaian mathematician, actuary, chemist and physicist. He defined modern science by his Least Whole Normal Theory which is used for fitting two or more quantitative products in sciences. He is a brilliant scientist and he formulated new laws in sciences and mathematics that have great impact in the life of present scientists. BILL ADONGO keeps diary recording all applications of his Least Whole Normal theory. Below is copy of his diary.

CAPACITOR, VOLTAGE AND CHARGE

A capacitor is a passive element designed to store energy in its electric field. Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics communications, computers, and power systems. For example, they are used in the tuning circuits at radio receivers and as dynamic memory element,

In many practical applications, the plates may be aluminum foil which the dielectric may be air, ceramic, paper, or mica. When a voltage source deposits a positive charge q on one plate and a negative charge   -q on the other, Capacitor is said to store the electric charge. The amount of charge stored, represented by q, is directly proportional to the applied voltage V so that

q=CV

Where C, the constant of proportionality, is known as the capacitance of the capacitor. Finding way of measuring unknown amount of charge stored q sand unknown capacitance of the capacitor C, with only known voltage V is impossible without employing my Least Whole Normal Measurement. From previous publications, I talk of theory of quantitative products where two or more quantitative products in science can be fitted by my Least Whole Normal Function and when applying this theory, the equation above will yield

T= µ_c*τ_v- ф^-1(γ%)√(S²τ_v)

E(C)=1/V_p[T+ф^-1(γ%)√(S²V_p)]

E(q)=E(C)V_p

where V_p is the independent voltage for predicting the expected of amount of charge stored E(q) and the expected capacitance of the capacitor E(C).

EXAMPLE

Calculate the capacitance of the capacitor C and amount of Charge stored q if the voltage is 8V based on the previous experiment made on six Capacitance of the capacitors and their known voltage.(take γ=95).

Capacitance of the capacitor(C)	Voltage(V)
60mF=60x10-3	5
50mf=50x10-3	6
40mF=40x10-3	7.5
30mF=30x10-3	10
20mF=20x10-3	15
10mF=10x10-3	30

SOLUTION

MEAN AND STANDARD DEVIATION

C	V	CV	Deviation of c from mean(d)	d2	Vd2
60mF=60x10-3	5	0.3	0.03551	0.001261	0.006305
50mf=50x10-3	6	0.3	0.02551	0.000651	0.003905
40mF=40x10-3	7.5	0.3	0.01551	0.000241	0.001804
30mF=30x10-3	10	0.3	0.00551	0.000030	0.003036
20mF=20x10-3	15	0.3	0.00449	0.000020	0.000302
10mF=10x10-3	30	0.3	-0.01449	0.000210	0.006299

Mean(µ_c)=∑CV/∑V=1.8/73.5=0.02449

Standard Deviation(S)=√(∑Vd2/∑V=√(0.021651/73.5=0.017147

From the data above we have;

T= µ_c*τ_v- ф^-1(γ%)√(S²τ_v)

But τ_v=∑V/N=73.5/6=12.25

T=0.02449x12.25- ф^-1(95%)√(0.017147²x12.25)

T=0.2012761475

E(C)=µ_c*V_p=1/V_p[T+ф^-1(γ%)√(S²V_p)]

E(C)= 1/8[0.2012761475+ф^-1(95%)√(0.017147²x8)]

E(C)=0.035=35x10^-3=35mF

q=E(C)V_p=35x10^-3x8=0.3

Hence, the capacitance of the capacitor and charge stored are 35mF and 0.3 respectively if the voltage is 8V.