Resistors in Parallel

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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 repitition.

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 connested 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

Lesson #2:

Simple Resistive Circuits

Lesson #3:

Resistors in Series

Lesson #4:

Resistors in Parallel← You are here

Quiz Test:

Circuit Analysis Quiz 1

Lesson #5:

Kirchhoff's Voltage Law

Lesson #6:

Kirchhoff's Current Law

Lesson #6:

Nodal analysis

Lesson #6:

Mesh Analysis

Quiz Test:

Circuit Analysis Quiz 2

Home Laboratory:

Circuit Analysis - Lab1

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 it share common terminals. That is if both of the branches touch each other endpoints they are in parallel. Here is an example:

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.

Figure 4.2: Components in parallel share a common nodes

The areas in yellow all are 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 and only two resistors we can use this 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 smallest of the individual Resistors.

Current Rule

In Series Connection we deduced that Voltage is divide amongst resistors. For Parallel connected Resistors, Current is divided. So here is a mathematical formula as we did with voltage division principle.

Equation 4.3: Current Divider Formula

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

Using this formula you can workout the currents flowing through individual Resistors.

Part 3

Application

We have spent three lectures hacking on about What & Why Resistors & 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 Electric Measurement industry, most often enough, we wish to measure Currents and Voltages of Very High Magnitudes ( e.g some ranges of 500kV and upwards or 1000kA and upwards ). The problem is that metering devices have delicate electronic components and usually have small Voltage and Ampere operating ratings.

Solution to the above is to have a metering device connected in parallel to a resistor, this resistor is thus called a "shunt" resistor since it is there to protect the metering device as shown in the next figure in part 4.

Part 4

Figure 4.3: Application of Parallel Resistive circuits. Shunt connection

If we know what the ampere rating of a device and what the total current is then we can work out the shunt current and thus the Shunt Resistor.

Part 5: Examples

Figure 4.4: Example 3.1

Figure 3.4 shows a Parallel resistive circuit with the following parameters.
$V s = 10 V o l t s$ ; $R 1 = 3Ω$ ; $R 2 = 7Ω$ ; Find $R e q$ ; $I 1 a n d I 2 .$

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 that you can follow the Method used and the reasoning behind.

Part 6: Examples

Figure 4.5: Example 3.1

Figure 4.5 shows a Parallel resistive circuit with the following parameters.
$I s = 5 A m p s$ ; $R 1 = 2Ω$ ; $R 2 = 3Ω$ ; 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: $R e q$ : $\begin{matrix}\ R_{eq}&=&\frac{(R_1 \times R_2)}{(R_1 + R_2)}\\ \ \\ \ &=& \frac{(2 \times 3)}{(2 + 3)} \\ \ \\ \ R_{eq} &=& 1.2\Omega \end{matrix}$ .

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: $R_{eq}=\frac{R_1 \times R_2}{R_1 + R_2}$

Current Divider Principle: $I_1 = I_s \times \frac{R_{eq}}{R1}$

Do Exercise 4 in part 8. After being completely satisfied of your work, you can go on and try The next Page which is a quick quiz test. Good luck :-) !

Part 8: Exercise 4

Here are some questions to test yourself with.

Given 2 Resistors: $R 1 = R 2 = 5Ω$ in parallel find The $R e q$ .
Given 3 Resistors: $R 1 = 2Ω$ ; R2 = 3 $Ω$ and R3 = 7 $Ω$ in parallel and Supply Current is 15Amps. Find: $R e q$ ; $I 1$ ; $I 2$ & $I 3$ and Supply Voltage across these Resistors.
Given 4 Resistors: R1 = 2 $Ω$ is connected in series to a parallel branch consisting of R2 = 3 $Ω$ ; R3 = 7 $Ω$ and R4 = 4 $Ω$ Find: Total Resistance as seen by the Voltage source.
Is it possible to connect Voltage sources in Parallel, If so what conditions must be met?

Answers to Exercise 4

Completion list

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