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Message: <blockquote data-quote="Little DJ" data-source="post: 10836688" data-attributes="member: 320663">Network Analysing Methods : Part 01The networks worked with so far had a single voltage source and could be easily analysed using techniques such as Kirchhoff’s voltage law and Kirchhoff’s current law. The methods used in determining the operation of complex networks will include branch-current analysis, mesh (or loop) analysis, and nodal analysis. Although any of the above methods may be used, there are certain circuits which are more easily analysed using one particular approach. In using the techniques outlined above, it is assumed that the networks are linear bilateral networks. The term linear indicates that the components used in the circuit have voltage-current characteristics which follow a straight line.<img src="http://3.bp.blogspot.com/-DPsLUB1HA8U/TkvzJGKcnvI/AAAAAAAAAFM/Cb-aM7s_0fk/s400/qqq.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />(b) Non-linear V-I Characteristics The term bilateral indicates that the components in the network will have characteristics which are independent of the direction of the current through the element or the voltage across the element. A resistor is an example of a linear bilateral component since the voltage across a resistor is directly proportional to the current through it and the operation of the resistor is the same regardless of the direction of the current.                             Constant-Current Sources                                                <img src="http://2.bp.blogspot.com/-UqCEz7YA-CI/Tkvz_RQvBjI/AAAAAAAAAFQ/sVjcVpdw9yY/s1600/pp.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />A constant current source maintains the same current in its branch of the circuit regardless of how components are connected external to the source.The direction of the current source arrow indicates the direction of conventional current in the branch. The magnitude and the direction of current through a voltage source varies according to the size of the circuit resistances and how other voltage sources are connected in the circuit. For current sources, the voltage across the current source depends on how the other components are connected.<img src="http://2.bp.blogspot.com/-rPl0g8Ngyu0/Tkv0xjivY7I/AAAAAAAAAFU/hRbwI-EvNcQ/s1600/jj.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />If R = 50 OhmsVR= Vs = 2A x 50 Ohm = 100 VIf there is more than one source: <img src="http://3.bp.blogspot.com/-hGUyOAqO5U4/Tkv1g-5nEWI/AAAAAAAAAFY/ZwYy8FNq54M/s1600/tt.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Is = 2mAV1= 1k x 2mA = 2VV2= 2k x 2mA = 4VApplying Kirchoff's Voltage Law:-2V+10V-4V+Vs = 0Vs = -4VFrom the above result, you see that the actual polarity of Vs is opposite to that assumed.<img src="http://2.bp.blogspot.com/-rBWwG6PPJLI/Tkv2v86HWEI/AAAAAAAAAFc/d3aUDmGoWH4/s320/o.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Because the 5V supply is effectively across the load resistor, I1 = (5V)/10 ohm = 0.5A (In the direction assumed)Applying Kirchhoff’s current law at point a,I2  = 0.5 A + 2.0 A = 2.5 AFrom Kirchhoff’s voltage law,-10 V + VS + 5 V = 0 VVs= 5VThe constant-current source determines the current in its branch of the circuit.The magnitude and polarity of voltage appearing across a constant-current source are dependent upon the network in which the source is connected.Source ConversionsIdeal constant current  source has no internal resistance included as part of the circuit. Voltage sources always have some series resistance, although in some cases this resistance is so small in comparison with other circuit resistance that it may effectively be ignored when determining the operation of the circuit. Similarly, a constant-current source will always have some shunt (or parallel) resistance. If this resistance is very large in comparison with the other circuit resistance, the internal resistance of the source may once again be ignored. An ideal current source has an infinite shunt resistance.<img src="http://4.bp.blogspot.com/-tx02RMKImF4/Tkv561qpx2I/AAAAAAAAAFg/8W1kvmB0ds0/s400/rr.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />If the internal resistance of a source is considered, the source, whether it is a voltage source or a current source, is easily converted to the other type. The current source of figure is equivalent to the voltage source ifI= E/Rsand the resistance in both sources is Rs.Similarly, a current source may be converted to an equivalent voltage source by lettingE = I x RsThese results may be easily verified by connecting an external resistance, RL, across each source. The sources can be equivalent only if the voltage across RL is the same for both sources. Similarly, the sources are equivalent only if the current through RL is the same when connected to either source.<img src="http://2.bp.blogspot.com/-84pdwVuouXY/Tkv7YMDgeCI/AAAAAAAAAFo/opxkBlkVOS8/s200/uu.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />The current across the load resistor:                       IL= (Rs/(Rs+RL))x I                                                                                                             But when converting the source                                       I = E/Rs                                                                                                                                                                             IL= (Rs/(Rs+RL))x (E/Rs)   IL= E/(Rs+RL)  The voltage across the load resistor:    VL = IL x RL   VL= E/(Rs+RL) x RL<img src="http://2.bp.blogspot.com/-ikSFYZq5T2A/Tkv7N4ZX1XI/AAAAAAAAAFk/CiuYm7n1yM0/s200/rrr.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />The voltage across the  load resistor:   VL = RL/(Rs+RL)x E The current across the  load resistor:   IL= E/(Rs+RL) The voltages and currents across the load resistor is same in both occasions. Therefore this could conclude that the load current and voltage drop are the same whether the source is a voltage source or an equivalent current source.For More Information: <a href="http://www.electronicworkspace.com/2011/08/network-analysing-methods-part-01.html" target="_blank">Click Here</a></blockquote>

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