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Using Norton's equivalent

The same problem is then worked out with Norton's equivalent, as shown in figures 5.29, 5.30, 5.31, 5.32. Finally, $ i=\frac{10}{9}\;Amp$

Figure 5.29:
Figure 5.30:

Figure 5.31:

Figure 5.32:

ynsingh 2007-07-25