Many real-life circuits are more complicated, with combinations of both series-connected and parallel-connected components. In this article, we’ll study such series-parallel circuits. You’ll find that the basic rules of the game are the ones that you’ve already learned for series circuits and parallel circuits, but the job of applying those rules can get pretty tricky.
The circuit in which the resistance is connected in series, the current flowing through each resistance is the same and the voltage drop across each resistance depends on its resistance value. Consider an electrical network shown in the figure.1, let three resistance R1, R2, and R3 be connected in series across a battery source of V volts. The current I flowing through R1, R2, and R3 is the same.
By applying Ohm’s law to the above circuit,
By applying Kirchhoff’s Voltage Law to the above circuit,
i.e. supply voltage = sum of the voltage drop across the elements
Hence, the equivalent resistance
Multiplying I on both sides of equation 3. We get
Current flowing through the elements is the same
Supply voltage is equal to sum of voltage drop. The voltage drop in the element is depends on the value of that element
The equivalent resistance is equal to sum of the resistance of the elements
Powers are additive
Determine the voltage drop and power dissipated in each resistor of the circuit shown below
Step.1: Find the equivalent resistance
Total current delivered by the d.c source
Voltage drop across 1 Ω =
Voltage drop across 2 Ω =
Voltage drop across 3 Ω =
Total voltage drop = 1+2+3 = 6 V = Supply voltage
Step.2: Power dissipated in each resistor
Power dissipated in 1 Ω =
Power dissipated in 2 Ω =
Power dissipated in 3 Ω =
Total Power dissipated = 1+2+3 =6 watts = Power delivered by the source = V×I = 6×1 =6 V
Resistance connected in parallel
If three resistances connected in parallel, the voltage drop across each resistor are same and the current through the resistors are different. The total current supplied by the system is equal to sum of voltage drop across the resistor.
Figure.3. shows three resistors are connected in parallel across a D.C source of voltage V.
Applying Ohm’s law to the network, the current through each resistor is,
Applying KVL to the network,
Figure.3 shows a sample network in which three resistors are connected in parallel across a battery of 6 V. Determine,
Power deviled by the source = V × I = 6 × 11 = 66 = Sum of the power consumed by each resistor.
Series and Parallel network
n a series-parallel circuit, the components are connected in series-connected and parallel-connected combinations. The below circuit shows an example of a series-parallel combination circuit. In this circuit, the Resistors R2, R3, and R4 are connected in series. This combination is in parallel with resistor R1. The series combination (i.e R2, R3, R4) and the parallel resistor R1 are connected to a common point called a node or Junction at which the current flowing to two paths. One is through the series combination and the other through parallel resistor R1. The amount of current division is based on the value of resistance in each path.
Notation for Series-Parallel Relationships
As a short-hand way of describing series-parallel relationships in a circuit, we’ll use two symbols:
we use the symbol + to indicate a series connection between components
we use the symbol || to indicate a parallel connection
So for the example circuit shown above, we would write R2 + R3 + R4 to indicate that R2, R3, and R4 form a series path. Then, to show that R1 is in parallel with this series path, we would write R1 || (R2 + R3 + R4)
Figure.5 shows the Series-Parallel network, in which three resistors such as 1ohm, 2 ohm and 3 ohm connected in parallel, and this combination is connected in series with a 6-ohm resistor. To solve the given network the following steps are to be followed,
Step.1. Reduce the parallel network by finding the equivalent resistance of parallel combination, (Ref. Figure.6)
The reduction of parallel network and the equivalent circuit is shown in Figure.5. With reference to figure.5, the reduced network has two resistors 0.5455Ω and 6Ω are connected in series. Now the problem treated similar to the resistance connected in series network.
Step.2. Find the equivalent resistance of the reduced network
Step.3. Find the current supplied by the source
Step.4. Find the voltage drop across each resistor
Voltage drop across 0.5455Ω resistor = Source voltage – Voltage drop across 6Ω resistor =6V – 5.5002V = 0.5V
Step.5. Find the current through each resistor
I1Ω = V0.5455Ω /R1Ω = 0.5/1 = 0.5 Amps
I2Ω = V0.5455Ω /R2Ω = 0.5/2 = 0.25 Amps
I3Ω = V0.5455Ω /R3Ω = 0.5/1 = 0.1667 Amps
the readers must remember the following notations while analyzing the series- parallel circuits
As you know, the three forms of Ohm’s law are:
Kirchhoff’s Voltage Law in Series-Parallel Circuits
Recall that KVL says that the sum of the voltage drops around any closed loop in a circuit equals the sum of the voltage rises around that loop.
We originally studied KVL in the context of series circuits, but KVL holds for all circuits.
Kirchhoff’s Current Law in Series-Parallel Circuits
Recall that KCL says that the sum of all currents entering a point is equal to the sum of all currents leaving that point.
We originally studied KCL in the context of parallel circuits, but KCL holds for all circuits
Components connected in series must have the same current (but usually they won’t have the same voltage).
Components connected in parallel must have the same voltage (but usually they won’t have the same current)
Power in a Series-Parallel Circuit
The following fomulas that can be used to find the power dissipated in a resistor in a series-parallel circuit,
P = V × I
P = I2 × R
P = V2 ÷ R
R is the resistor’s resistance,
V is the voltage across the resistor, and
I is the current through the resistor.
Total Circuit Power
Just as with series circuits and parallel circuits, there are two ways to compute total power dissipated in a series-parallel circuit. You’ll get the same answer either way:
Either find the power for each resistor, and then add these powers:
PT = P1 + P2 + P3 + … + Pn
Or apply any one of the power formulas to the entire circuit:
PT = VS × IT
PT = IT2 × RT
PT = VS2 ÷ RT
These are the same power formulas from above, except that now we’re applying them to the entire circuit, instead of to a single resistor.
Voltage-Divider Rule in Series-Parallel Circuits
The series- parallel circuits is analysed by using Ohm’s law, KVL, and KCL In some cases, the voltage-divider rule and the current-divider rule are also useful. However, it’s easy to make mistakes when you apply the voltage-divider rule and current-divider rule to series-parallel circuits. Let us consider a series-parallel circuit as shown below figure
The Current-Divider Rule in Series-Parallel Circuits
In discussing parallel circuits, we stated the current-divider rule as follows: for branches in parallel, the current Ix through any branch equals the ratio of the total parallel resistance RT to the branch’s resistance Rx, multiplied by the total current IT entering the parallel combination. In equation form:
Ix = (RT ÷ Rx) × IT
The same rule applies to parallel branches in a series-parallel circuit. But you have to be careful to apply it correctly. In particular, note that Rx is the total resistance of the branch whose current you are trying to find. Depending on the circuit you’re analyzing, Rx may be a single resistor’s resistance, or it may be the combined resistance of several resistors in series and/or parallel. Also, note that in this equation RT means the total resistance of the parallel branches, which may or may not be the same as the entire circut’s total resistance. Similarly, IT in this equation means the total current entering the parallel branches, which may or may not be the same as the entire circut’s total current.
In our example circuit, for instance, we have two parallel branches. One branch contains just R1, and the other branch is a series path containing R2, R3, R4. If we know the total current entering both branches, then the current-divider rule lets us find the current into either branch. For instance, the current into the series path is given by:
Ipath = (RT ÷ Rpath) × IT
But the following would not be a proper application of the current-divider rule, since R2 does not by itself form a complete branch:
I2 = (RT ÷ R2) × ITIncorrect!
Because students often apply the current-divider rule incorrectly in series-parallel circuits, I recommend that you rely more heavily on Ohm’s law, KVL, and KCL, and try to avoid using the current-divider rule.
We have now reviewed the use of Ohm’s law, Kirchhoff’s Laws, and the divider rules in series-parallel circuits. Analysis of any particular series-parallel circuit will involve the use of one or more of these rules and laws.
Usually there will be more than one way to approach a particular problem or circuit. In finding a particular voltage, for instance, one student might use Ohm’s law and KVL, while another student uses KVL and the voltage-divider rule, and a third student uses the current-divider rule and Ohm’s law. All three approaches will give the same answer, as long as the students apply the rules correctly.
As mentioned earlier, a good first step is usually to find the circuit’s total equivalent resistance by combining pairs of series-connected or parallel-connected resistors until you’ve combined all the resistors into a single resistance. Then find the circuit’s total current by using Ohm’s Law in the form:
IT = VS ÷ RT
Ground (or Common) Symbol
Recall from Unit 4 that instead of drawing lines to show connections in a schematic diagram, we often use a ground symbol, also called a common symbol, to designate terminals that are connected together. The symbol looks like this:
The main reason for using this symbol is that it let you eliminate lines that clutter a schematic diagram and make it difficult to visualize current flow between components.
Example: Take a look again at this circuit diagram:
Notice that the voltage source, R1, and R4 all come together at a common point. Using the ground symbol to show that these three components are connected together, we could redraw the diagram as shown below.
This is not a very good example, because the redrawn diagram with the ground symbol is no easier to read than the original diagram. But in a complicated schematic diagram with many resistors, the ground symbol can eliminate a lot of lines and therefore make the diagram much easier to read. The important point in this example is to see that the two diagrams are equivalent.
Bubble Symbol for Voltage Sources
In schematic diagrams where the common symbol is used, we often omit that symbol from the side of any voltage source that is connected to common.
The terminal that isnot connected to common is shown by a small circle (or “bubble”) and is labeled + or – the voltage of the voltage source.
Example: Continuing the same example from above, we could redraw the schematic diagram once more using the bubble symbol to arrive at the following diagram:
Unit 9 Review
This e-Lesson has covered several important topics, including:
identifying series and parallel relationships
analyzing series-parallel circuits
power in series-parallel circuits
ground symbol and bubble symbol.
To finish the e-Lesson, take this self-test to check your understanding of these topics.
Congratulations! You’ve completed the e-Lesson for this unit.