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Association of resistors in parallel | Footer |
Series and Parallel Resistor Associations :
In the electric circuits, the elements which constitute them can be connected between them different manners according to needs ; we will examine the various types of connections and their particular properties, which they are resistances or the piles.
Let us return one moment to the examination of the circuit of figure 1. In this one, outgoing current I of the terminal " + " of the pile, crosses total resistance R and returns in the pile by its terminal " - " and, to distinguish these two resistances, we will call them R1 and R2.
Current I provided by the pile must cross R1 then R2 successively to be able to return on the terminal "-" pile.
When two or several elements of a circuit (in this case two resistances) are crossed successively by the same current, it is said that they are connected in series, or more simply than they are in series.
The fact that the current circulating in these elements is the same one for all is a specific characteristic of the connections in series, therefore several resistances in series all are crossed by the same current. (This is obvious and easy to include / understand).
The addition of R2 resistance makes the value resistive total of the circuit larger than if there were only R1 resistance, because the current, in addition to the obstacle caused by R1 with its passage, must also cross R2. We can say that the total resistance of the circuit of figure 1 above which is opposed in the passing of the current is given by the sum of the resistive values of each resistance. You point out that :
The equivalent resistance presented by several resistances connected in series is obtained by adding the resistive value with each resistance.
Now let us look at what it occurs of the tension delivered by the pile. At the boundaries of each resistance, it appears a tension and this in accordance with the law of Ohm.
For figure 1, the tension V of the pile is divided between two resistances R1 and R2 present in the circuit. At the boundaries of R1 appears a V1 tension (given by the values of I and R1) and at the boundaries of R2 appears a V2 tension (given by the values of I and R2). The sum of these two tensions is equal to the total tension of the pile : V1 + V2 = V.
Let us illustrate for an example what has just been marked.
Figure 2 is deferred the same circuit but certain electric quantities are decorated of a value.
In this circuit, we must determine the intensity of the current I which circulates in R1 resistances and R2, as well as the tensions V1 and V2 present at their terminals.
Two resistances being connected in series, both are crossed by the same current, therefore the total resistance offered to the circulation of this current is determined by the sum of two resistances is :
Equivalent resistance = R1 + R2 = 20 Ohms + 40 Ohms = 60 Ohms
The application of Ohm's law in the form I = V / R allows us to calculate I :
I = 6V / 60 Ohms = 0,1 A = 100 mA
100 mA is the intensity of the current which crosses R1 and R2. To calculate the tensions V1 and V2 present at the terminals of R1 and R2, the law of Ohm will be applied in form V = RI.
V1 = R1 x I = 20 Ohms x 100 mA = 20 Ohms x 0,1 A = 2 V
V2 = R2 x I = 40 Ohms x 100 mA = 40 Ohms x 0,1 A = 4 V
These found results, we note from the start that the tension V of the pile was divided in two parts and we carried out a circuit called tension divider.
In the electronic circuits, one often has recourse to the association of two resistances in series with an aim of obtaining a tension weaker than that provided by the food of the circuit.
For example, let us suppose duty to supply a lamp functioning under 6 V and absorbing a maximum current of 0,05 A (50 mA) starting from a pile of 9 V.
Under penalty of destroying the lamp, it is impossible to connect this one directly to the pile since the too important tension of this one would make circulate a too intense current in the lamp, current which " would roast " (like one says usually) the lamp.
To avoid this disadvantage, we can lay out in the circuit a resistance chutrice in series with the lamp, like illustrated figure 3. On this figure, you will also make knowledge with the graphic symbol of a lamp.
The value of resistance R must be calculated in an adequate way so that on its terminals, the tension is of 3 V (surplus provided by the pile). This value can be calculated by the law of Ohm because the current I which circulates in the circuit is imposed by the lamp L is 50 mA and tension VR on its terminals of 3 V.
R = VR / I
Let us replace VR by V - VL (VL : terminal voltage of the lamp L).
R = V - VL / I = (9V - 6V) / 0,05 A = 3 V / 0,05 = 60 Ohms
In this case, resistance R connected in series with the lamp L forms with this one a tension divider which reduces the tension applied to the lamp, so as to allow its lighting under good conditions.
It is said that resistance R thus "fell" part of the power provided by the pile. Resistances are largely used in the circuits to produce voltage drops, and to thus produce tension dividers.
There is a second type of connection used for resistances is association in parallel, illustrated figure 4.
ASSOCIATION OF RESISTORS IN PARALLEL
In this type of assembly, each of two R1 resistances and R2 have one their terminals connected to "+" of the pile and the other with " - ". Both are thus seen applying the same tension, that provided by the pile. This established fact is a specific characteristic of the connections in parallel ; you point out that :
At the boundaries of several associated elements in parallel, there is always the same tension.
In this type of connection, it is thus necessary primarily to analyze the behavior of the current. Figure 4, let us note for the current (I) which leaves the positive pole of the pile is divided at the point C in two currents called I1 and I2 ; each one of its currents crosses a resistance (I1 crosses R1 and I2 crosses R2) then meet in the point D to reform the initial current (I) which then joined the negative pole of the pile.
Current I provided by the pile is thus equal to the sum of the currents which cross each resistance.
To determine the equivalent resistance (Req) of such an assembly, we should use the law of Ohm. Figure 5 is deferred the electric circuit of figure 4 as well as the equivalent diagram in which Req appears.
Appear 5-a, we can determine the value of current I according to R1 and R2.
I = I_{1} + I_{2}
I_{1} = V_{R1} / R_{1}
I_{2} = V_{R2} / R_{2}
We know that in such an assembly, the terminal voltage of each resistance is equal to the power provided by the pile :
V = V_{R1} = V_{R2}
(1) from where I_{1} = V / R_{1}
(1) I_{2} = V / R_{2}
(1) and I = V / R_{1} + V / R_{2}
Figure 5-b, we deduce that : (2) I = V / R_{eq}
The two equalities (1) and (2) give same current I and are thus equal :
(1) = (2) ----------------------) V / R_{1} + V / R_{2} = V / R_{eq}
Let us multiply the two terms of the equality by 1 / V :
I / V . (V / R_{1} + V / R_{2}) = 1 / V . V / Req
Let us simplify the two terms of the equality.
We thus determined Req according to R1 and of R2. Extended to the general case several resistances in parallel, this formula becomes :
During our demonstration, we passed by the following intermediate result :
1 / R_{eq} = 1 / R_{1} + 1 / R_{2}
In other words, the reverse of equivalent resistance is equal to the sum of the opposite of resistances of the circuit, or to be more precise than the equivalent conductance is equal to the sum of the conductances of each resistance.
G_{eq} = G_{1} + G_{2}
Thus, to calculate equivalent resistance Req of two or several resistances in parallel, one can make the three following operations :
to determine the conductance of each resistance : G = 1 / R
to carry out the sum of the found conductances : G_{eq} = G_{1} + G_{2} + G_{3} + ...
to take the reverse of the sum obtained : R_{eq} = 1 / G_{eq}
When two resistances only are in parallel, one adopts the following formula which derives from the general formula :
R_{eq} = R_{1} x R_{2} / R_{1} + R_{2}
For a quantified practical example, let us apply what we have just seen :
That is to say to calculate resistance equivalent to the circuit represented figure 6.
To calculate Req, let us carry out the three necessary operations :
calculation of the conductance of each resistance :
G_{1} = 1 / R_{1} = 1 / 20 Ohms = 0,05 S
G_{2} = 1 / R_{2} = 1 / 40 Ohms = 0,025 S
summon conductances :
G_{eq} = G_{1} + G_{2} = 0,05 + 0,025 = 0,075 S
Calculation of equivalent resistance :
R_{eq} = 1 / G_{eq} = 1 / 0,075 = environ 13,3 Ohms
R1 resistances and R2 in parallel are thus equivalent to a single resistance of approximately 13,3 Ohms.
To compare the two types of associations of resistances, we can note that in the case of resistances in series, the value of equivalent resistance is always higher than the value of each resistance while in the case of a parallel association the value of equivalent resistance is in all the cases lower than the value of each resistance and even better, it is lower than smallest resistances.
The formulas presented are also used for more complex calculations of circuits born from the combination of the two types of associations.
Let us take for example the circuit of figure 7 and suppose duty to calculate its equivalent resistance.
One first of all calculates resistance equivalent (R2 - R3) to resistances R2 and R3 in parallel is:
For two resistances R2 and R3, one can substitute a single resistance of 20 Ohms (R2 and R3), as in the figure 7-b.
From this figure, one calculates resistance (Req) equivalent (figure 7-c) to R1 and R2-3 in series :
From this practical example, it comes out that in the presence of a complex circuit, it is necessary to treat the two types of associations separately so as to simplify the circuit gradually until obtaining a single resistance.
It is also interesting to see the behavior of the tensions and the currents in such a circuit.
In figure 8 is deferred the same circuit but is supplemented by the representation of the various currents and tensions.
We should now determine the parameters accompanied by a question mark in figure 8 is I, V1, V2, V3, I2 and I3.
Calculation of I :
I is the current total circulating in the circuit, we obtain it by dividing the power provided by the pile by the equivalent resistance of the circuit, which like is calculated previously of 25 Ohms :
I = V / R_{eq} = 9 / 25 = 0,36 A = 360 mA
Calculation of V_{1} :
V1, terminal voltage of >R1 resistance is obtained by multiplying R1 by the current which crosses it, but this current is not other than I :
V_{1} = R_{1} x I = 5 x 0,36 = 1,8 V
Calculation of V_{2} et V_{3} :
V_{2-3}, terminal voltage of unit is equal to the difference between the tension V of the pile and the V1 tension dropped by R1 :
V_{2-3} = V - V_{1} = 9 - 1,8 = 7,2 V
Calculation of I_{2} :
I_{2}, current circulating in R_{2} is obtained by dividing the terminal voltage of R_{2} is V_{2-3} by R_{2} :
I_{2} = V_{2-3} / R_{2} = 7,2 / 100 = 0,072 A = 72 mA
Calculation of I_{3} :
The current circulating in R_{3}, can be obtained in two ways :
I_{3 }= I - I_{2} = 360 mA - 72 mA = 288 mA
or I_{3} = V_{2-3} / R_{3} = 7,2 / 25 = 0,288 A = 288 mA
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