Category Archives: Resistive Circuits

Solved problems in resistive circuits such as nodal analysis, mesh analysis, superposition, source transformation, Thevenin and Norton theorems and so on

Mesh (Current) Analysis Problem

Solve the circuit by mesh analysis and find the current I_x and the voltage across R_2.
mesh analysis problem

Solution

Mesh Analysis

There are four meshes in the circuit. So, we need to assign four mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.
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Superposition method - Circuit with two sources

Find I_x using superposition rule:
Main cuircuit to be analyzed using superposition method

Solution

Superposition

The superposition theorem states that the response (voltage or current) in any branch of a linear circuit which has more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are turned off (made zero).
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Find Thevenin's and Norton's Equivalent Circuits

Find Thevenin's and Norton's Equivalent Circuits:
1254-1
Suppose that R_1=5\Omega, R_2=3\Omega and I_S=2 A.

Solution

The circuit has both independent and dependent sources. In these cases, we need to find open circuit voltage and short circuit current to determine Norton's (and also Thevenin's) equivalent circuits.
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Solve Using Current Division Rule

Find current of resistors, use the current division rule.
Problem 1246 (1)
Suppose that R_1=2 \Omega, R_2=4 \Omega, R_3=1 \Omega, I_S=5 A and V_S=4 V

Solution:
R_2 and R_3 are parallel. The current of I_S is passing through them and it is actually divided between them. The branch with lower resistance has higher current because electrons can pass through that easier than the other branch. Using the current division rule, we get
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Mesh Analysis - Supermesh

Solve the circuit and find the power of sources:
Problem 1226 - 1
V_S=10V, I_S=4 A, R_1=2 \Omega, R_2=6 \Omega, R_3=1 \Omega, R_4=2 \Omega.

Solution:
There are three meshes in the circuit. So, we need to assign three mesh currents. It is better to have all the mesh currents loop in the same direction (usually clockwise) to prevent errors when writing out the equations.
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