Electric Circuit Analysis/Resistors in Parallel

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Wikiversity Electrical Engineering School
The Lessons in
ELECTRIC CIRCUITS ANALYSIS COURSE

Lesson 3 : Review

What you need to remember from Resistors in Series. If you ever feel lost, do not be shy to go back to the previous lesson & go through it again. You can learn by repetition.

  • Total Series Resistance: (R_E = R_1 + R_2\!)
The total Resistance of Resistors in series is the sum of all resistors in series.
  • Voltage Divider Equation 2.3: V_1 = \frac{R_1 \times V_s}{R_1 + R_2}
  • Current through Resistors connected in Series is the same for all resistors.

Lesson 4: Preview

This Lesson is about Resistors in Parallel. The student/User is expected to understand the following at the end of the lesson.

  • two resistors connected in Parallel: R_{eq}=\frac{R_1 \times R_2}{R_1 + R_2}
  • Current Divider Principle: I_1 = I_s \times \frac{R_{eq}}{R1}
Lessons in Electric Circuit Analysis
Lesson #1:
Passive Sign Convention Isimple system icons app edit.png
Lesson #2:
Simple Resistive Circuits Isimple system icons app edit.png
Lesson #3:
Resistors in Series Isimple system icons app edit.png
Lesson #4:
Resistors in Parallel← You are here Sweden road sign C2.svg Isimple system icons app edit.png
Quiz Test:
Circuit Analysis Quiz 1 Crystal Clear app help index.svg
Lesson #5:
Kirchhoff's Voltage Law Isimple system icons app edit.png
Lesson #6:
Kirchhoff's Current Law Isimple system icons app edit.png
Lesson #7:
Nodal Analysis Isimple system icons app edit.png
Lesson #8:
Mesh Analysis Isimple system icons app edit.png
Quiz Test:
Circuit Analysis Quiz 2 Crystal Clear app help index.svg
Home Laboratory:
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Part 1

Introduction

The best way to understand Parallel circuits is to start with the definition. A circuit is parallel to another circuit or several circuits if and only if they share common terminals; i.e. if both the branches touch each other's endpoints. Here is an example:


Circuit1.JPG

Figure 4.1: A Parallel circuit


R1, R2, and the voltage source are all in parallel. To prove this fact consider the top and bottom parts of the circuit.


Circuit2.PNG

Figure 4.2: Components in parallel share a common nodes


The areas in yellow are all connected together, as well as the areas in blue. So all the branches have the same terminals, which means that R1, R2, and the source are all in parallel.


If we take this discussion of the water flow analogy. Electric current can be seen as water and the conductors as water pipes.

Something interesting happens as the current reaches the common node of resistors that are connected in parallel: the total current is divided into the parallel branches.


Part 2

Voltage Rule

If two or more branches are parallel then the voltage across them is equal. So based on this we can conclude that VR1=VR2=5volts. However unlike series resistors, the current across the branches is not necessarily equal.

Equivalent resistance

For series resistors to find the total resistance we simply add them together. For parallel resistors its a little more complicated. Instead we use the following equation:

R_{eq} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}

However for the case of only two resistors we can use the following simplified form:

Equation 4.2: Total Parallel Resistance

R_{eq}=\frac{R_1R_2}{R_1 + R_2}

It is well to note at this point that the total resistance of parallel-connected resistors will always be less than the resistance of smallest of the individual resistors.

Current Rule

In series connection, we deduced that voltage is divided amongst resistors. For parallel-connected resistors, however, current is divided. So, as we did with the voltage division principle, here is the mathematical formula:

Equation 4.3: Current Divider Formula

I_1 = I_s \times \frac{R_{eq}}{R1}

Using this formula you can work out the currents flowing through individual resistors.

Part 3

Application

We have spent three lectures hacking on about what and why resistors and& resistive circuits in two connection schemes are used, (i.e. series and parallel connections). The question now is, where & how in real life do these connections happen?


One simple application of these connection schemes is the Shunt Application. In the electric measurement industry, most often enough, we wish to measure currents and voltages of very high magnitudes (in the range of 500kV upwards). The problem is that metering devices have delicate electronic components and usually have small voltage and current ratings.


The solution to the above problem is to have a metering device connected in parallel to a resistor,with the resistor thus called a "shunt" resistor since it is there to protect (shunt) the metering device as shown in Part 4.

Part 4

EE-102-L03-Fig3.3.png

Figure 4.3: Application of Parallel Resistive circuits. Shunt connection

If we know the current rating of a device and the total current in the system, we can then work out the shunt current and, thus, the shunt resistance.

Part 5: Examples


EE-102-L03-Fig3.4.png

Figure 4.4: Example 3.1

Figure 3.4 shows a parallel resistive circuit with the following parameters.
V_s=10Volts ; R_1=3\Omega; R_2=7\Omega; Find R_{eq}; I_1 and I_2.

Solution: from Equation 4.2 we see that.

R_{eq}=\frac{R_1R_2}{R_1 + R_2}


Here are the solutions to the above problem:

\begin{matrix}\ R_{eq}&=&\frac{R_1 \times R_2}{R_1 + R_2} \\ \ \\ \ &=& \frac{(3\Omega \times 7\Omega)}{(3\Omega + 7\Omega)} \\ \ \\ \ & = & \frac{21}{10}\Omega \\ \ \\ \ &=&2.1\Omega \end{matrix}.

\begin{matrix}\ I_1 &=& \frac{V_s}{R_1} \\ \ \\ \ &=& \frac{10V}{3\Omega} \\ \ \\ \ & = & 3.33A \end{matrix}.

\begin{matrix}\ I_2 &=& \frac{V_s}{R_2} \\ \ \\ \ &=& \frac{10V}{7\Omega} \\ \ \\ \ & = & 1.43A \end{matrix}.


Thus it can be said that the supply current has been divided between R1 and R2.


We know that when solving these problems, we look at the data given and thus we can see how we need to manipulate our equations in order to achieve our objective.The following example highlights this point. See to it that you follow the method used and the reasoning behind it.

Part 6: Examples


EE-102-L03-Fig3.4.png

Figure 4.5: Example 3.1

Figure 4.5 shows a parallel resistive circuit with the following parameters.
I_s=5Amps ; R_1=2\Omega; R_2=3\Omega; Find: I_1 ; I_2 and V_s.

Solution: from Equation 3.2 we see that.

R_{eq}=\frac{R_1R_2}{R_1 + R_2}
Here are the solutions to the above problem:

First Find: R_{eq}: \begin{matrix}\ R_{eq}&=&\frac{(R_1 \times R_2)}{(R_1 + R_2)}\\ \ \\ \ &=& \frac{(2 \times 3)}{(2 + 3)} \\ \ \\ \ R_{eq} &=& 1.2\Omega \end{matrix}.

Then;

\begin{matrix}\ I_1&=&I_s\frac{R_{eq}}{R_1} \\ \ \\ \ &=& (5A)\frac{1.2\Omega}{2\Omega} \\ \ \\ \ & = & 3A \end{matrix}.

\begin{matrix}\ I_2&=&I_s\frac{R_{eq}}{R_2} \\ \ \\ \ &=& (5A)\frac{1.2\Omega}{3\Omega} \\ \ \\ \ & = & 2A \end{matrix}.

\begin{matrix}\ V_s = V_1 = V_2 \\ \ \\ \ V_2=I_2 \times R_2 \\ \ \\ \ =2 \times 3 \\ \ \\ \ = 6Volts \end{matrix}.


Part 7

Do you Remember?

Let's take some time to Reflect on Material covered thus far. We have learned a great deal about simple resistive circuits and the possible connections they afford us. Here I think you'll want to remember:

  • Voltage: (V or v - Volts)The electrical potential between two points in a circuit.
  • Current: (I or i - Amperes)The amount of charge flowing through a part of a circuit.
  • Power: (W - Watts)Simply P = IV. It is the current times the voltage.
  • Source: A voltage or current source is the supplier for the circuit.
  • Resistor: (R measured in Ω - Ohms)A circuit element that "constricts" current flow.
  • Total Series Resistance: (R_E = R_1 + R_2\!)
  • Voltage Divider : V_1 = \frac{R_1 \times V_s}{R_1 + R_2}
  • Current through resistors connected in series is the same for all resistors.
  • Two resistors connected in parallel: R_{eq}=\frac{R_1 \times R_2}{R_1 + R_2}
  • The Current Divider Principle: I_1 = I_s \times \frac{R_{eq}}{R1}


Do Exercise 4 in Part 8. After being completely satisfied with your work, you can go on to the next page - for the quiz! Good luck :-)

Related Topic(s) in Wikiversity

Please visit the following page to supplement material covered in this lesson.


Part 8: Exercise 4

  1. Given 2 Resistors R_1 = R_2 = 5\Omega in parallel find the R_{eq}.
  2. Given 3 Resistors R1 = 2\Omega; R2 = 3\Omega and R3 = 7\Omega in parallel and supply current as 15A, find R_{eq} ; I_1; I_2 & I_3 and the supply voltage across these resistors.
  3. Given 4 Resistors: R1 = 2\Omega connected in series to a parallel branch of 3 resistors: R2 = 3\Omega ; R3 = 7\Omega and R4 = 4\Omega Find: total resistance as seen by the voltage source.
  4. Is it possible to effectively connect voltage sources in parallel? If so, what conditions must be met?


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Completion list

Once you finish your Exercises you can post your score here! To post your score just e-mail your course co-ordinator your name and score *Click HereCrystal Clear app xfmail.png.

    1. Ozzimotosan -- 100%
    2. Doldham -- 75% & Corrected
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