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De nition The Norton equivalent current is equal to the short-circuit current that would ow were the load replaced by a short circuit

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An explanation for the de nition of the Norton current is easily found by considering, again, an arbitrary one-port network, as shown in Figure 357, where the one-port network is shown together with its Norton equivalent circuit It should be clear that the current, iSC , owing through the short circuit replacing the load is exactly the Norton current, iN , since all of the source current in the circuit of Figure 357 must ow through the short circuit Consider the circuit of Figure 358, shown with a short circuit in place of the load resistance Any of the techniques presented in this chapter could be employed to determine the current iSC In this particular case, mesh analysis is a convenient tool, once it is recognized that the short-circuit current is a mesh current Let i1 and i2 = iSC be the mesh currents in the circuit of Figure 358 Then, the following mesh equations can be derived and solved for the short-circuit current: (R1 + R2 )i1 R2 iSC = vS R2 i1 + (R2 + R3 )iSC = 0 An alternative formulation would employ nodal analysis to derive the equation vS v v v = + R1 R2 R3 leading to v = vS R2 R3 R1 R3 + R 2 R3 + R 1 R2

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Recognizing that iSC = v/R3 , we can determine the Norton current to be: iN = v v S R2 = R3 R1 R3 + R 2 R3 + R 1 R2

Thus, conceptually, the computation of the Norton current simply requires identifying the appropriate short-circuit current The following example further illustrates this idea

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Computing the Norton Current 1 2 3 4 Replace the load with a short circuit De ne the short circuit current, iSC , to be the Norton equivalent current Apply any preferred method (eg, nodal analysis) to solve for iSC The Norton current is iN = iSC

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R2 = 3

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Assumptions: Assume reference node is at the bottom of the circuit Analysis: We rst compute the Th venin equivalent resistance We zero the two sources e by shorting the voltage source and opening the current source The resulting circuit is shown in Figure 360 We can clearly see that RT = R1 R2 + R3 = 6 3 + 2 = 4 Next we compute the Norton current Following the Norton current methodology box, we rst replace the load with a short circuit, and label the short-circuit current, iSC The circuit is shown in Figure 361 ready for node voltage analysis Note that we have identi ed two node voltages, v1 and v2 , and that the voltage source requires that v2 v1 = V The unknown current owing through the voltage source is labeled i Applying KCL at nodes 1 and 2, we obtain the following set of equations:

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To eliminate one of the three unknowns, we substitute v2 V = v1 in the rst equation: I v2 V i =0 R1 node 1

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