EE1002 Notes Part1

# Understandthisbyapplyingthesuperpositionprinciple

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Unformatted text preview: g the voltage source, the circuit will be 0.5 i1 0.25 0.75 The current i1 can be obtained using any of the earlier known methods like: a) Current divider rule 46 EE1002 Introduction to Circuits and Systems b) Node voltage analysis c) Mesh current analysis Let us use the current divide rule. The current of 2A, is divided between 0.25 and 0.75 . So the current in the 0.25 is: i1 2 0.75 1.5 A 0.25 0.75 Step2: Keeping the voltage source only and killing the current source, the circuit will be 0.5 3V i2 0.25 0.75 ix As the resultant circuit is one with a single voltage source with two resistances in series, we can find the current easily. 3 3A 0.25 0.75 i2 ix 3 A ix Step3: Find the total current, i i1 i2 1.5 3 1.5 A It can be observed that, applying superposition method resulted in simpler circuits and ease of analysis. However, this may not be the case always and hence, superposition method is seldom used for circuit analysis. 47 EE1002 Introduction to Circuits and Systems One‐ port networks and equivalent circuits While building complex circuits, sometimes, part of the circuit is interfaced with anoth...
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