This preview shows page 1. Sign up to view the full content.
Unformatted text preview: i2 V0 R3i2 0 41 EE1002 Introduction to Circuits and Systems Case: with a current source in one of the branches. Suppose there was a current source between in D and D’ as given in the Figure. C R1
B i1 2A R2 D D’ Vs R3 R4 i2
A i2 R3i2 (i1 i2 ) (i1 i2 ) R4 (i1 i2 ) Vs R2i1 2A i1 R1i1 To apply the KVL to the mesh ABCDD’A , we need the voltage across the current source (between DD’). However, unlike a resistor, we cannot express the voltage across the current source in terms of the mesh currents. This problem is solved by applying KVL for the loop ABCDA. Sum of voltage rises around this loop: VS R1i1 R2i1 V0 R3i2 0 (Equation 1) We need to get another independent equation for solving the system. As can be seen from the figure, i1 i2 2 42 (Equation 2) EE1002 Introduction to Circuits and Systems We can solve for both the mesh currents and from there determine the voltage and current in all the parts of the circuit. Case: mesh analysis with controlled sources Such situations arise in study of transistor amplifiers. Method for handling such situations: 1) Tr...
View Full Document
This document was uploaded on 01/20/2014.
- Winter '14