Grouping of resistance?

A more complex resistor network can be solved by systematic grouping of resistors. Resistors R2 and R3 are connected in series. They are in parallel with resistor R1. To solve the network, the resistors are separated in two groups.

How to calculate the equivalent resistance value for resistors in parallel?

Resistors are often connected in series or parallel to create more complex networks. An example of 3 resistors in parallel is shown in the picture above. The voltage across resistors in parallel is the same for each resistor. The current however, is in proportion to the resistance of each individual resistor. The equivalent resistance of several resistors in parallel is given by:

$\frac{1}{R_{eq}}=\sum_{i=1}^{n}\frac{1}{R_{i}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+..+\frac{1}{R_{n}}$

The current through each resistor is given by:

$I_{i}=\frac{V}{R_{i}}$

To quickly calculate the equivalent resistance value of two resistors in parallel

How to solve a network with resistors in parallel and series?

A more complex resistor network can be solved by systematic grouping of resistors. In the picture below three resistors are connected. Resistors R2 and R3 are connected in series. They are in parallel with resistor R1. To solve the network, the resistors are separated in two groups. Group 1 consists of only R1. Group 2 consists of R2 and R3.

how to solve resistor networks

The equivalent resistance of group 2 is easily calculated by the sum of R2 and R3:

$R_{group2}=R_{2}+R_{3}$

This leads to the simplified circuit with two resistors in parallel. The equivalent resistance value of this circuit is easily calculated:

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{group2}}$

${R_{eq}}=\frac{R_{1}\cdot R_{group2}}{R_{1}+R_{group2}}$

${R_{eq}}=\frac{R_{1}\cdot (R_{2}+R_{3})}{R_{1}+R_{2}+R_{3}}$

example

A circuit designer needs to install a resistor with 9 ohms and can choose from the value(.., 10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82, ..). The value of 9 ohms is unfortunately not available in this series. He decides to connect to standard values in parallel with an equivalent resistance of 9 ohms. The equivalent resistance value for 2 resistors in parallel is calculated with these steps:

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

$\frac{1}{R_{eq}}=\frac{R_{2}}{R_{1}R_{2}}+\frac{R_{1}}{R_{1}R_{2}}$

$\frac{1}{R_{eq}}=\frac{R_{1}+R_{2}}{R_{1}R_{2}}$

$R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$

The above equation shows that if R1 is equal to R2 Req is half of the value of one of the two resistors. For a Req of 9 ohms, R1 and R2 should therefore have a value of 2x9=18 ohms. This happens to be a standard value from the E-series.

$R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{{18}\cdot{18}}{18+18}=9 \Omega$

As a solution finally, the designer connects two resistors of 18 ohms in parallel as shown in the figure right.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals

NCERT Solutions for Class 12 Maths Chapter 7 Integrals: Engineering aspirants and students appearing for CBSE Class 12 board exams must take the NCERT Mathematics textbooks seriously and finish them from top to bottom. They must also go through the NCERT solutions for Class 12 Maths and Class 11 Maths to have a better understanding of the various concepts. In this article, we will provide you with NCERT Solutions for Class 12 Maths Chapter 7 – Integrals which have been designed by the best teachers in India.