Created it, 05/10/16
Update it, 05/10/16
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CONSTRUCTION AND READING OF THE CARTESIAN DIAGRAM “2nd PART”
While serving to us as the system of Cartesian axes, we now will build a chart of the law of Ohm, which we already examined from the point of view of the literal mathematical expressions (1 - formulas).
Let us consider the case of a circuit formed by a primary battery (B) and by a resistance (R) having the value of 1 ohm (figure 3-a). It is admitted that the drivers have a negligible resistance.
We can determine for each value of tension (V), a value of intensity (I) well defined, which, according to the law of Ohm, is all the more high as the tension of the battery is it even more important.
At the beginning, when the battery is not connected, no tension is applied to resistance and, there is obviously no current.
That can be represented in the diagram of the figure 3-a, by marking of a point the origin 0 which corresponds to zero value of the tension (x-axis) and to zero value of the intensity (y-axis).
By providing a power of 2 volts, intensity crossing resistance to for value:
On the diagram of the figure 3-a, we mark a point corresponding to the values: tension = 2 volts, intensity = 2 amps.
With a tension of 4 volts, intensity with for value:
On the diagram of the figure 3-a, we mark a new point corresponding to V = 4 volts and I = 4 amps.
By this process, we can as many introduce into the diagram points as there are values of tension to take into account and corresponding values of intensity. We limit to these three points which are sufficient to enable us to continue our development.
Let us defer the diagram of the figure 3-a on the figure 3-b. If the layout of the axes and the positioning of the three points were carried out with the wanted precision, one can see that three points are aligned on a line.
This inclined line, passing by the origin of the system of the axes, constitutes the curve representative of the law of Ohm, in the particular case of a circuit having a resistance of 1 ohm.
The checking is very simple. On the basis of a value of tension not yet taken into account but indicated on the x-axis, one traces the perpendicular with this same axis until meeting the line of the graph.
Note: In mathematics, the “curved” word is synonymous with “line”. A “curve” can thus be a line as in this case. We will re-examine this terminology of the study of the functions. From the intersection of these two lines, one leads a perpendicular to the y-axis thus determining on this axis, the value of the intensity corresponding to the value of the selected tension. The two values, that of the tension and that of the intensity, as well as the value of resistance (1 ohm), will have to satisfy the equality established by the law of Ohm.
Example:
Let us take the value of 1 volt for the
tension (x-axis). Let us trace the perpendicular which meets the straight lines
at point A (figure 3-b).
From point A, let us trace the perpendicular with the y-axis ; it indicates on this axis value 1, i.e., the value of 1 amp (intensity).
Let us calculate now on the basis of one of the forms of the law of Ohm, the value of resistance on the basis of a tension of 1 volt and an intensity of 1 amp:
what corresponds indeed to the value of the resistance of the circuit.
Let us take the value 3 volts. Let us trace
the perpendicular. It meets the straight lines at the point B
(figure 3-b).
From the point B, let us trace the perpendicular with the y-axis; it indicates value 3. Let us replace the values now 3 volts, 3 amps and 1 ohm in another form of the law of Ohm.
In this case also, the equality between the two members of the formula is satisfied. We can thus conclude that the point B, just like point A, belongs indeed to the graphics of the law of Ohm.
One could present many other examples of this kind, by choosing the value of the tension at will and by using the line of the graph to determine on the y-axis the value of the intensity.
They would always show that the tilted line points represent the existing bond between the tension and the intensity, just like the formulas of the law of Ohm represent this same bond.
Thus, the tilted line passing by the origin of the diagram and the three formulas of the law of Ohm can be regarded as equivalent between them.
However, we arrived to this result only by considering the particular case in which the resistance of the circuit is equal to 1 ohm. We will extend later on our considerations to the more general case in which the resistance of the circuit presents an unspecified value.
THEOREM
OF PYTHAGOREIt can be stated in the following form:
“The surface of the square, built starting from the hypotenuse, is equal to the sum of surfaces of the squares built starting from the two other sides”.
This theorem is illustrated on figure 4.
The second form of the theorem which one generally meets:
In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the two sides of the right angle.
Let us write the formula of this theorem (has and B are the lengths on the sides, and C the length of the hypotenuse).
This relation makes it possible to find the hypotenuse. Let us write it in the form:
One can also find a side by writing the relation like this:
Let us suppose that one wants to calculate the hypotenuse of a right-angled triangle whose sides would have: a = 3 m and b = 5 m ; let us replace the letters by their value in the formula:
Let us suppose that one wants to calculate the side (b) of a right-angled triangle by knowing the hypotenuse C = 6 cm and the other side (= 4 cm have); let us replace the letters by their value in the formula:
We thus finish our 2nd lesson of mathematics. In the next one, we will approach the algebraical expressions.
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