CHAPTER 1. DIRECT CURRENT CIRCUITS15V1− V2 =2RIA− RIB(1.35)V2 =−RIA +3RIB,(1.36)V1− V2V2=2R−R−R 3RIAIB≡RIAIB,(1.37)R−1 =1R3/51/51/52/5,(1.38)IA =1R35(V1− V2)+15V2 =1R35V1−25V2(1.39)IB =1R15(V1− V2)+25V2 =1R15V1 +15V2 .(1.40)The voltage across AB is given simply byVAB =IBR(1.41)=15(V1 + V2).(1.42)1.4Equivalent CircuitsEquivalent circuits is often the hardest concept and most numerically intensive in the course.Learning them well could make a diﬀerence on your midterm exam. Look in several booksuntil you ﬁnd the explanation you understand best.Since Ohm’s law and Kirchoﬀ’s equations are linear, we can replace any DC circuit by asimpliﬁed circuit. Just like a combination of resistors and Ohm’s law could give an equivalentresistor, a combination of circuit elements and Kirchoﬀ’s laws can give an equivalent circuit.Two possibilities are shown in ﬁgure 188.8.131.52.1Thevenin’s and Norton’s TheoremsA Thevenin equivalent circuit contains an equivalent voltage source VTh in series with anequivalent resistor RTh. A Norton equivalent circuit contains an equivalent current sourceIN in parallel with an equivalent resistor RN.1.4.2Determination of Thevenin and Norton Circuit ElementsOne approach to determine the equivalent circuits is:1. Thevenin – calculate the open-circuit voltage VAB = VTh.