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Find the current in the 5 ohm resistance using Norton’s theorem.(a) 1A(b) 1.5A(c) 0.25A(d) 0.5A |
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Answer» The correct option is (d) 0.5A To explain I would say: Shorting all voltage sources and opening all current sources we have: RN=(3||6)+10 = 12 ohm. Since the 5 ohm is the load resistance, we short it and find the resistance through the short. If we apply source transformation between the 6 ohm resistor and the 1A source, we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. Let us consider I1 flowing in the first mesh and I2 flowing in the second mesh. The mesh equations are: 9I1-6I2=4 -6I1+16I2=6 On solving these equations simultaneously, we get I2=0.72A, which is the short circuit current. Connecting the current source in parallel to RN which is in turn connected in parallel to the load resistance=5ohm, we get Norton’s equivalent circuit. Using current divider: I = 0.72*12/(12+5) = 0.5 A. |
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