Circuit Analysis and Network Theorems 1, Slides of Electronic Circuits Analysis

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A method of electrical circuit analysis in which
the mesh currents are taken as independent
variables and the potential differences around a
mesh are equated to zero.
STEPS:
1. Mark(assign variable) all the parts of the circuit.
2. Draw circulating currents in each loop or mesh.
3. Use KVL to form equations in each loop or mesh.
4. Solve for the unknowns.
MESH ANALYSIS
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Download Circuit Analysis and Network Theorems 1 and more Slides Electronic Circuits Analysis in PDF only on Docsity!

A method of electrical circuit analysis in which

the mesh currents are taken as independent

variables and the potential differences around a

mesh are equated to zero.

STEPS:

1. Mark(assign variable) all the parts of the circuit.

2. Draw circulating currents in each loop or mesh.

3. Use KVL to form equations in each loop or mesh.

4. Solve for the unknowns.

MESH ANALYSIS

In the circuit shown below, solve the current in

each branch using mesh analysis.

Example:

Solution:

Using Cramerโ€™s Rule,

Let ๐ด =

A method of determining the voltage

between nodes in an electrical circuit in terms of

branch currents. KCL is used to form equation in

each node. The number of equations should be

n- 1 , where n is the number of nodes.

NODAL ANALYSIS

In the circuit shown below, solve the current in

each branch using nodal analysis.

Example:

Solution:

ILLUSTRATION

Example 1: In the figure shown, calculate the current through and the voltage across the load impedance ๐‘๐ฟ by using Theveninโ€™s theorem.

Solving for ๐‘‰๐‘กโ„Ž ๐‘ = 40โˆ 0ยฐ + 30โˆ  โˆ’ 90ยฐ = 50โˆ  โˆ’ 36 .87ยฐ๐‘œโ„Ž๐‘š๐‘  ๐ผ =

Thevenin equivalent circuit: ๐‘๐‘‡ = ๐‘๐‘กโ„Ž + ๐‘๐ฟ = 30 โˆ  0 ยฐ + 4 0โˆ  โˆ’ 90ยฐ = 50 โˆ  โˆ’ 53 .13ยฐ ๐‘œโ„Ž๐‘š๐‘  ๐ผ๐ฟ = ๐‘‰๐‘กโ„Ž ๐‘๐‘‡

72โˆ โˆ’ 53 .13ยฐ 50โˆ โˆ’ 53 .13ยฐ = 1. 44 โˆ 0ยฐamps

Example 2: In the circuit shown, find the current through the capacitor using Theveninโ€™s theorem.

  • 5 โˆ 0ยฐ ิก^ ( 5 + ๐‘—5) ๐‘๐‘กโ„Ž = 8. 062 โˆ  29. 745 ยฐ ิก^ ( 7. 071 โˆ  45 ยฐ) ๐‘๐‘กโ„Ž = 3. 801 โˆ  37. 875 ยฐ๐‘œโ„Ž๐‘š๐‘  Solving for ๐‘๐‘กโ„Ž