CONSTRUCTION AND READING OF THE CARTESIAN DIAGRAM "PART 2"
APPLICATION OF THE CARTESIAN DIAGRAM
Using the Cartesian axis system, we will now construct a graphical representation of Ohm's law, which we have already examined from the point of view of literal mathematical expressions (1 - The formulas).
Consider the case of a circuit formed by a battery of cells (B) and by a resistor (R) having the value of 1 ohm (Figure 3-a). It is admitted that drivers have negligible resistance.
We can determine for each voltage value (V), a well-defined value of intensity (I), which, according to Ohm's law, is even higher than the voltage of the battery itself important.
Initially, when the battery is not connected, no voltage is applied to the resistor and there is obviously no current.
This can be represented in the diagram of Figure 3-a, by marking from one point the origin 0 which corresponds to the zero value of the voltage (abscissa axis) and to the zero value of the intensity (axis of the ordinate).
By supplying a voltage of 2 volts, the intensity crossing the resistance has the value :
I = V / R
I = 2 / 1 = 2 amperes
In the diagram of Figure 3-a, we mark a point corresponding to the values : voltage = 2 volts, intensity = 2 amperes.
With a voltage of 4 volts, the intensity at value :
I = V / R
I = 4 / 1 = 4 amperes
In the diagram of Figure 3-a, we mark a new point corresponding to V = 4 volts and I = 4 amperes.
By this method, we can introduce into the diagram as many points as there are voltage values to be considered and corresponding intensity values. Let us limit ourselves to these three points which are sufficient to allow us to continue our development.
Let's move the diagram of Figure 3-a in Figure 3-b. If the plot of the axes and the positioning of the three points were made with the desired precision, we can see that three points are aligned on a line.
This inclined line, passing through the origin of the axis system, constitutes the representative curve of Ohm's law, in the particular case of a circuit having a resistance of 1 ohm.
The verification is very simple. Starting from a voltage value not yet taken into consideration but indicated on the abscissa axis, we draw the perpendicular to this same axis until we meet the right of the graph.
Remark: In mathematics, the word "curve" is synonymous with "line". A "curve" can therefore be a straight line as in the present case. We will revisit this terminology of the study of functions. From the intersection of these two lines, one leads perpendicular to the ordinate axis thus determining on this axis, the value of the intensity corresponding to the value of the chosen voltage. The two values, that of the voltage and that of the intensity, as well as the value of the resistance (1 ohm), must satisfy the equality established by Ohm's law.
Example:
Take the value of 1 volt for the voltage (x-axis). Let's draw the perpendicular that meets the line at point A (Figure 3-b).
From point A, draw the perpendicular to the ordinate axis ; it indicates on this axis the value 1, that is to say, the value of 1 ampere (intensity).
Let us now calculate from one of the forms of Ohm's law, the value of the resistance starting from a voltage of 1 volt and an intensity of 1 ampere:
R = V / I
1 = 1 / 1
1 = 1
which effectively corresponds to the value of the resistance of the circuit.
Take the value 3 volts. Let's draw the perpendicular. She meets the line at point B (Figure 3-b).
From point B, let's draw the perpendicular to the ordinate axis ; it indicates the value 3. Now replace the values 3 volts, 3 amps and 1 ohm in another form of Ohm's law.
I = V / R
3 = 3 / 1
3 = 3
In this case too, the equality between the two members of the formula is satisfied. We can therefore conclude that point B, like point A, does indeed belong to the graph of Ohm's Law.
Many other examples of this kind could be presented, choosing at will the value of the tension and using the right of the graph to determine the value of the intensity on the ordinate axis.
They would always show that the points of the inclined line represent the link between tension and intensity, just as the formulas of Ohm's Law represent this same link.
Thus, the inclined line passing through the origin of the diagram and the three formulas of Ohm's law can be considered equivalent to each other.
However, we have arrived at this result only by considering the particular case in which the resistance of the circuit is equal to 1 ohm. We will later extend our considerations to the more general case in which the resistance of the circuit has any value.
"The surface of the square, constructed from the hypotenuse, is equal to the sum of the surfaces of the squares constructed from the other two sides".
This theorem is illustrated in Figure 4.
The second form of the theorem that we meet most often :
In a rectangle triangle, the square of the hypotenuse is equal to the sum of the squares on both sides of the right angle.
Let us write the formula of this theorem (a and b are the lengths of the sides, and c the length of the hypotenuse).
c² = a² + b²
This relationship helps to find the hypotenuse. Let's write it as :
One can also find a side by writing the relation like this :
Suppose we want to calculate the hypotenuse of a right triangle whose sides would have : a = 3 m and b = 5 m ; let's replace the letters with their value in the formula :
Suppose we want to calculate the side (b) of a right triangle by knowing the hypotenuse c = 6 cm and the other side (a = 4 cm) ; Let's replace the letters by their value in the formula :
We thus finish our 2nd lesson of mathematics. In the next, we will discuss algebraic expressions.
Pythagore's theorem
Take the value of 1 volt for the voltage (x-axis). Let's draw the perpendicular that meets the line at point A (Figure 3-b).
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