Till now, we have learnt about the Resistor, basically how it works, what its applications are, how to use it in a circuit, etc. But we come across the circuits that involve some combinations of resistors. Our aim in this article is to simplify this network and obtain the equivalent resistance of the network.

1. Types of Resistors:

Fixed Value Resistor:

These resistors have their values fixed by the manufacturers themselves. But there can be a slight deflection from the value. We call this deflection ‘tolerance’.

Suppose we have been given a resistor of value $(100 \pm 5\%)\ \Omega$,

It means the expected value of the resistance is $100 \Omega$ and the Tolerance is 5%

This tells us that the value of the resistor (fixed) can vary from $96 \Omega – 100\Omega$

How to calculate the Value of the Resistance? – Color Coding

• The first band gives the first digit
• The second band gives the second digit
• Third band gives the multiplier (the power raised to base 10)
• The fourth band gives you the tolerance value

To clarify even further, we take an example of a fixed resistor having its value of resistance as:

$(35 \times 10^2) \pm 10\% \ \Omega$

For this, we need to predict the color coding of this resistor. This can be done with the help of the table below, which contains the colors corresponding to their digit, and also tells the tolerance value associated with that color.

So, for the above example $(100 \pm 5\%)\ \Omega$, the color coding will be:

• First band – Orange
• Second band – Green
• Third band – Red
• Fourth band – Silver

Here’s a tool to make your job of choosing the resistor based on color coding even more easy!

Variable Resistors:

• Resistors, whose value can be altered, are known as Variable Resistors. The manufacturer just prints the maximum value that the device can go up to. (For example, if 10K is printed on a potentiometer, it implies that you can adjust its resistance to any value from 0 to 10K ohms).
• The way you can vary the resistance changes from component to component

2. Power Rating of Resistors:

We have learnt about choosing the value of Resistance but whenever current passes through resistors, there is some heat generated as well. Now, we need to take care that this heat doesn’t damage the resistor. For this, we have power ratings assigned to resistors. Power is basically heat generated per second and is measured in Watts (W)

• There is a relation of power with Voltage and Current,

$P = V \times I$

Here, P is the Power (in watts)

V is the potential difference across the resistor (in volts)

I is the current flowing through the resistor (in amperes)

• There are resistors available of 1/8 watt, 1/4 watt, 1/2 watt, 1 watt, etc. The larger the size of the resistor, the higher the power rating it has!

Example:

Suppose the voltage difference across the resistor is going to be 6V, and a current of 20mA is going to pass through it. For the given situation, what should be the suitable power rating of the resistor?

$P = V \times I$

$P = 6 \times (20 \times 10^{-3})$

$P = 0.12\, W$

From the above example, we can infer that a resistor of 1/8 W is ‘just’

3. What is meant by finding equivalent resistance ?

When it comes to circuit solving, we will encounter lot of complex combination of resistors present in the circuit. Finding ‘equivalent’ of such combination of resistors means that, we must be able to replace that whole thing with just a single resistor without changing any of the other parameters (current, potential difference across given points, etc. ) in the circuit.

Note that, in Fig.1 (a) and (b), except for the number of resistors, there is no change in other parameters (I remains I, E remains E, $\Delta$V remains $\Delta$V)

Now, how to actually calculate this value of R_eq is what we need to study in this article

4. Resistors in Series

Resistors are said to be in Series when the current flowing through them is the same. Done!

Now, with reference to the above figure,

The voltage drop across Req should be equal to the sum of the voltage drops across R₁ and R₂, since our main aim is not to disturb the potential difference between points A and B ($\Delta V_{AB}$)

\Delta V_{AB} = \Delta V_{Req} = \Delta V_1 + \Delta V_2

i R_{eq} = i R_1 + i R_2 \quad \text{(current remains same)}

{R_{eq} = R_1 + R_2}

Important Note:

From the above relation, we can infer that we can use series combination if we need a resistance value greater than the individual resistances (i.e., R_eq > R₁ & also R_eq > R₂)

So, Resistors in Series just add up directly!

The figure below shows the series combination of R1 and R2 resistors having resistances of 1.5 kΩ and 1.75 kΩ. The equivalent resistance R_eq in this case is measured by a multimeter in Tinkercad. The value for R_eq comes out to be 3.25 kΩ

5. Resistors in Parallel

Resistors are said to be in parallel, when they have same potential difference across them. Done!

The potential difference across Req should also be equal to the potential differences across R₁ and R₂ (i.e., $\Delta$V), since our main aim is not to disturb the potential difference between points A and B

Here, the quantity that is varying is current, and by applying KCL starting from A, we have the relation,

$i = i_1 + i_2$

$\frac{\Delta V}{R_{eq}} = \frac{\Delta V}{R_1} + \frac{\Delta V}{R_2} \quad \text{(PD remains same)}$

$\boxed{\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}}$

Important Note :

From the above relation, we can infer that we can use parallel combination if we need a resistance value even lower than the individual resistances. (i.e. R_eq < R₁ & also R_eq < R₂)

Breadboard Connections for parallel combination (in Tinkercad):

6. Example on Series & Parallel Combinations

Question-1

Find the equivalent resistance of the given setup across points A and C

Solution:

Step – 1: Both the 4 Ω resistors are connected across the same points B and C. Hence, both are in parallel combination. R_eq for just this combination will be 2 Ω

Step-2: Now, 5Ω and 2 Ω are in series combination. Req of this will be 7Ω

Step-3 : Finally, we have 7 Ω resistor between A and C. This is our final Req between points A and C

Question – 2: Find the effective resistance between points A and B for the figure below

Solution:

Step 1 : Same current passes through AF and FE, which makes both the 3 ohm resistors in series. Req for this will be 3Ω+3Ω = 6 Ω

Step 2 : Two 6 Ω resistors are connected across same points A and E which makes them in parallel. R_eq for this will be 3ohm

Step 3 : Again, both 3 Ω are in series. R_eq will be 6 Ω

Step 4: Both 6 Ω are in parallel. R_eq will be 3 Ω

Step-5 : 3 Ω and 3 Ω are in series. R_eq of this will be 6 Ω

Step-6: two 6 Ω resistors are in parallel. R_eq of this will be 3 Ω. Keep on simplifying further.

Step-7 :  3 Ω and 3Ω  are in series

Step-8: 6 Ω and 3Ω are in parallel. Solving this combination, we get R_eq of this as 2ohms.

Final Answer : So, the overall equivalent resistance across points A and B is 2 Ω

Conclusion:

Through this article, we tried to learn about resistors and how to perform the basic analysis of circuits involving resistors. According to me, understanding electronics right from the root will be really beneficial.

Learning to deal with Resistors, Capacitors, Inductors, and integrated circuits helps us to build the

The Complete Guide on Resistors:

• Part-1: What Is a Resistor & How to Use It in Circuits
• Part 2 (You are here): Color Coding & Combining Resistors: Series-Parallel Combinations
• Part-3: How to Solve Resistor Networks – Methods & Examples

About the Author

I am a Mechanical Engineering student at IIITDM Kancheepuram, interested in teaching, as simple as that! I love to simplify the so-called complicated topics and present them to you in very lucid language. I’m also the author of ‘ Life Lessons from Physics ‘, a book aimed at showing what Physics would have to tell if it were made to give a lecture on Life.