Electric Circuits 8th Edition 138

Electric Circuits 8th Edition 138 - tion once we know the...

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100_O I ir'r250_O 114 Techniques ofCircuii Analysis 300 {) we compare these two possible rcference nodes by mea]ls of the lollovring sets of equations. The first set pertains to the sircuit shown in Fig. 4.31, and the second set is based on the circuit shown in Fig.4.32. Note that, in addition to selecting the reterence node, we defined the three node voltages "1, ?,r, and ?.,3 and indicated that nodes 1 and 3 form a super- node, because they are cormected by a dependent voltage source.It is understood that a node voltage is a rise from the reference node;tlerefore, in Fig.4.31, we have not placed tlle node voltage polarity refer- erces on the circuit diagram. The second Dode that merits considemtion as the reference node is the lower node in the circuit, as shoM h Fig. 4.32. It is attractive because rt has the most branches connected to it, and the node- voltage equations are thus easier to $'rite. However, to find eitlei the current in the 300 f,, resistor or
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Unformatted text preview: tion once we know the node voltages oa and oc. For example, the curent in the 300 O iesistor is (.,. - r,,)/300. $hereac tbe voltage acfo.< the re<is Llr_ Figur 4.31 a Th circujt shown jn Fjg. 4.29, wjth a 300 0 -ij 1500 rol] o "b 250 o ,.:q! o \ " l2s6v moo ( )''' '*o n( \ Figure 4.32 A The circujtshown in Fis.4.29 with an attemative rference node. set 1 (Fig 4.31) At the supernode, r) ot D o -(D - 128) r00 250 200 400 500 ,r\ + 256 ^ 150 b X ) - D D ) - D l i j : : : + - - + - - + 300 250 .100 1r2 + 128 u3 From the supemode, the constraint equatron $ t)t : Ut )UrI : Ltl -. Set 2 (Fig4.32) 1)" 0" - 256 ,a - 0b oa 0c 2c0- 150 - 1oo - 3oo ='' ,c u. + 128 1).- rb 4oo ' 5oo - 250 - 3oo -' 500 From the suDemode. the constraint equation is 50(r" ,") ub : 50ia : -ff You should veify that the solution of either set leads to a power calculation of 16.57 W dissipated in the 300 O rcsistor. 6 128V...
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