Parallel DC Circuits
Two network configurations, series and parallel, form the framework for some of the most
complex network structures. A clear understanding of each will pay enormous dividends as
more complex methods and networks are examined. The series connection was discussed in
detail in the last chapter. We will now examine the parallel circuit and all the methods and
laws associated with this important configuration.
The term parallel is used so often to describe a physical arrangement between two elements
that most individuals are aware of its general characteristics.
In general,
Lets go for a deep calculation of the total resistance obtained from the parallel combination of resistors as given in eq.(1):
Looking to Fig. 2(a), applied voltage source $E$ is connected directly to all of the resisters in parallel. So voltage drops across $R_1$, $R_2$, and $R_2$ are same. But the currents in the parallel resistors are different due to different values of resistance. $I_1$ is the current passing through $R_1$, $I_2$ through $R_2$, and $I_3$ through $R_3$. Current $I$ is the overall current extracted from the battery, which divides further into three currents $I_1$, $I_2$, and $I_3$. In equation form, we can get:
where $I_1 = E/R_1$, $I_2 = E/R_2$, and $I_3 = E/R_3$. By inserting these values in the above equation, we can get:
But Fig. 2(b), which is the resulted form of parallel resistors, can gives us the value of current $I = E/R_T$. Now we get;
Since conductance G = 1/R, the equation can also be written in terms of conductance levels as follows:
Fig. 1:
Parallel resisters: three different configurations.- Fig. 1(a): the two resistors are in parallel because they are connected at points a and b.
- Fig. 1(b): Resistors $R_1$ and $R_2$ are in parallel because they again have points a and b in common. $R_1$ is not in parallel with $R_3$ because they are connected at only one point (b). Further, $R_1$ and $R_3$ are not in series because a third connection appears at point b. The same can be said for resistors $R_2$ and $R_3$.
- Fig. 1(c): Resistors $R_1$ and $R_2$ are in series because they have only one point in common that is not connected elsewhere in the network. while ($R_1 + R_2$) are in parallel with $R_3$, because their combined effect goes parallel with $R_3$. Resistors $R_1$and $R_3$ are not in parallel because they have only point a in common. In addition, they are not in series because of the third connection to point a.
Parallel Resistors in a circuit
For resistors in parallel as shown in [Fig. 2], the total resistance is determined from the following equation:$$\bbox[5px,border:1px solid grey] {{1 \over R_T} = {1 \over R_1} + {1 \over R_2} +{1 \over R_3}} \tag{1}$$
(a)
(b)
Fig. 2: Parallel combination of resistors.
$$ I = I_1 + I_2 + I_3$$
$$ I = {E \over R_1} + {E \over R_2} + {E \over R_3}$$
$$ I = E({1 \over R_1} + {1 \over R_2} + {1 \over R_3})$$
$$ {E \over R_T}= E({1 \over R_1} + {1 \over R_2} + {1 \over R_3})$$
$$\bbox[5px,border:1px solid grey] { {1 \over R_T}= ({1 \over R_1} + {1 \over R_2} + {1 \over R_3})}$$
$$\bbox[5px,border:1px solid grey] { G_T= G_1 + G_2+G_3}$$
Example 1: a: Find the total conductance of the parallel network in Fig. 3.
b. Find the total resistance of the same network using the results of part (a).
Solution:
a: $G_1 = 1/R_1 = 1/3Ω = 0.333 S$
$ G_2 = 1/R_2 = 1/6Ω = 0.167 S.$ $ G_T = G_1 + G_2 = 0.333 S + 0.167 S = 0.5 S$ b: $R_T = 1/G_T = 1/0.5S = 2Ω$
If you review the examples above, you will find that the total resistance
is less than the smallest parallel resistor. That is, in Example
1, 2 Ω is less than 3 Ω or 6 Ω. In general, therefore,
This is particularly important when you want a quick estimate of
the total resistance of a parallel combination. Simply find the smallest
value, and you know that the total resistance will be less than that
value. It is also a great check on your calculations. In addition, you
will find that
b. Find the total resistance of the same network using the results of part (a).
Fig. 3: Parallel resistors for example 1.
a: $G_1 = 1/R_1 = 1/3Ω = 0.333 S$
$ G_2 = 1/R_2 = 1/6Ω = 0.167 S.$ $ G_T = G_1 + G_2 = 0.333 S + 0.167 S = 0.5 S$ b: $R_T = 1/G_T = 1/0.5S = 2Ω$
Special Case: Equal Parallel Resistors
For equal resistors in parallel, the equation for the total resistance becomes significantly easier to apply. For N equal resistors in parallel,$${1 \over R_T}= {1 \over R} + {1 \over R} + {1 \over R} + . . .+ {1 \over R_N}$$
$$R_T = { 1 \over {{1 \over R} + {1 \over R} + {1 \over R} + . . .+ {1 \over R_N}}}$$
$$R_T = { 1 \over {{N \over R}}}$$
$$\bbox[5px,border:1px solid grey] {R_T = { R \over N }} \tag{2}$$
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