Lecture21 - Chapter 4

# Lecture21 - Chapter 4 - EE216 C ircuit Analysis I Le cture...

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1 EE216 Circuit Analysis I Lecture # 21 – Chapter 4

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2 Norton’s Equivalent Norton’s theorem simplifies a network in terms of currents instead of voltages For current analysis, Norton’s theorem can be used to reducea network to a simple parallel circuit , with a current source and resistance
3 Norton’s Equivalent I N is the short-circuit current through terminals A and B. To find R N wemust find V OC ; then Norton’s Resistance: R N = V OC /I N Alternatively , R N is the resistanceacross terminals A and B when all independent sources have been disabled.

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4 Thevenin-Norton Conversion Norton from Thevenin R N = R TH I N = V TH /R TH Thevenin from Norton R TH = R N V TH = I N x R N
5 Thevenin’s Equivalent with Dependant Sources Determination of V TH remains same. Determination of R TH : Method 1: remains same, i.e. find I sc . Method 2: cannot be applied since dependant sources cannot be removed. Method 2 Alternate: disableindependent current/voltagesources and apply a test voltage or current source . Usually, it is easiest to apply 1V or 1A test source.

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6 Thevenin’s Equivalent with Dependant Sources (V TH ) V TH is voltagebetween EF is same as between AB 0 I 2 1 V 8 5 40 V X A A = - + - - 5 V 40 I A X - = V V V TH A 20 = =
7 Thevenin’s Equivalent with Dependant Sources (R TH thru I sc Method) I SC = 2I X + I X + 8 = 32 A (I X = 8 A) R Th = V Th /I SC = 20/32 = 0.625

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