Network Theory Tutorial
Network Theory Useful Resources
Selected Reading
- Network Theory - Filters
- Two-Port Parameter Conversions
- Two-Port Networks
- Network Theory - Coupled Circuits
- Parallel Resonance
- Network Theory - Series Resonance
- Response of AC Circuits
- Response of DC Circuits
- Maximum Power Transfer Theorem
- Network Theory - Norton’s Theorem
- Thevenin’s Theorem
- Superposition Theorem
- Network Topology Matrices
- Network Theory - Network Topology
- Star to Delta Conversion
- Delta to Star Conversion
- Equivalent Circuits Example Problem
- Network Theory - Equivalent Circuits
- Network Theory - Mesh Analysis
- Network Theory - Nodal Analysis
- Electrical Quantity Division Principles
- Network Theory - Kirchhoff’s Laws
- Network Theory - Passive Elements
- Network Theory - Active Elements
- Example Problems
- Network Theory - Overview
- Network Theory - Home
Network Theory Useful Resources
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- Who is Who
- Computer Glossary
- HR Interview Questions
- Effective Resume Writing
- Questions and Answers
- UPSC IAS Exams Notes
Equivalent Circuits Example Problem
Equivalent Circuits Example Problem
In the previous chapter, we discussed about the equivalent circuits of series combination and parallel combination inspanidually. In this chapter, let us solve an example problem by considering both series and parallel combinations of similar passive elements.
Example
Let us find the equivalent resistance across the terminals A & B of the following electrical network.
We will get the equivalent resistance across terminals A & B by minimizing the above network into a single resistor between those two terminals. For this, we have to identify the combination of resistors that are connected in series form and parallel form and then find the equivalent resistance of the respective form in every step.
The given electrical network is modified into the following form as shown in the following figure.
In the above figure, the letters, C to G, are used for labelpng various terminals.
Step 1 − In the above network, two 6 Ω resistors are connected in parallel. So, the equivalent resistance between D & E will be 3 Ω. This can be obtained by doing the following simppfication.
$$R_{DE} = frac{6 imes 6}{6 + 6} = frac{36}{12} = 3 Omega$$
In the above network, the resistors 4 Ω and 8 Ω are connected in series. So, the equivalent resistance between F & G will be 12 Ω. This can be obtained by doing the following simppfication.
$$R_{FG} = 4 + 8 = 12 Omega$$
Step 2 − The simppfied electrical network after Step 1 is shown in the following figure.
In the above network, two 3 Ω resistors are connected in series. So, the equivalent resistance between C & E will be 6 Ω. This can be obtained by doing the following simppfication.
$$R_{CE} = 3 + 3 = 6 Omega$$
Step 3 − The simppfied electrical network after Step 2 is shown in the following figure.
In the above network, the resistors 6 Ω and 12 Ω are connected in parallel. So, the equivalent resistance between C & B will be 4 Ω. This can be obtained by doing the following simppfication.
$$R_{CB} = frac{6 imes 12}{6 + 12} = frac{72}{18} = 4 Omega$$
Step 4 − The simppfied electrical network after Step 3 is shown in the following figure.
In the above network, the resistors 2 Ω and 4 Ω are connected in series between the terminals A & B. So, the equivalent resistance between A & B will be 6 Ω. This can be obtained by doing the following simppfication.
$$R_{AB} = 2 + 4 = 6 Omega$$
Therefore, the equivalent resistance between terminals A & B of the given electrical network is 6 Ω.
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