Monthly Archives: September 2013

Mesh Analysis - Supermesh

Resistive Circuits, , , , , , ,

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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Solve By Source Definitions, KCL and KVL

Resistive Circuits, , , ,

Find the voltage across the current source and the current passing through the voltage source.
Problem 1213
Assume that I_1=3A, R_1=2 \Omega, R_2=3 \Omega, R_3=2 \Omega,I_1=3A, V_1=15 V,

Solution
R_1 is in series with the current source; therefore, the same current passing through it as the current source:
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Find Voltage Using Voltage Division Rule

Resistive Circuits, , , , ,


Determine voltage across R_2 and R_4 using voltage division rule.
Assume that
V_1=20 V, R_1=10 \Omega, R_2=5 \Omega, R_3=30 \Omega and R_4=10 \Omega
solve using voltage division rule

Solution:
Please note that the voltage division rule cannot be directly applied. This is to say that:
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