ECE 201 Spring 2010Homework 9 SolutionsProblem 37(a)The following loop equation can be written,−(I1− 0.75)200 − 300I1− (I1+ 0.1)500 = 0⇒ I1= 0.1 AP = V IPs1= 200(0.75 − 0.1)0.75= 97.5 WPs2= 500(0.1 + 0.1)0.1= 10 W(b)Again, writing the loop equation and comparing the equation obtained withthat given in the problem,−(I1− 0.4)R1− 600I1− (I1+ 0.1)R2= 0(R1+ R2+ 600)I1= 0.4R1− 0.1R2⇒ R1+ R2= 14004R1− R2= 600⇒ R1= 400 ΩR2= 1000 Ω1Problem 42(a)The following loop equations can be written,21 − 20I1− 80(I1− I2) = 024 + 80I2− 80(I1− I2) = 0⇒ I1= 0.15 AI2= −0.075 AVR3= (I1− I2)R3= 0.225 × 80 = 18 VPR3= 18(0.225) = 4.05 W(b)Again, writing the loop equations,21 − (I1+ I2)20 − 80I2− 24 = 021 − (I1+ I2)20 − 80I1= 0⇒ I1= 0.225 AI2= −0.075 AVR3= I1R3= 0.225 × 80 = 18 VPR3= 18(0.225) = 4.05 WProblem 52(a)Assume clockwise loop currents I1and I2in the two rightmost loops respec-tively. The following equations can then be written for the two loops,−(I1− Is1)R1− rmIx− R2(I1− I2) = 0Ix= Is1− I1−R2(I2− I1) − R3I2− Vs2= 02Putting in matrix form,rm− R1− R2R2R2−(R2+ R3)I1I2=(rm− R1)Is1Vs2(b)Putting the respective literal values,−80 4040 −120I1I2=−4040⇒ I1= 0.4 AI2= −0.2 A(c)VA= (1 − 0.4)100 = 60 VVB= 0.6(40 ) = 2 4 V(d)Ps1= 60 × 1 = 60 WPs2= 40 × 0.2 = 8 W(e)Pdep= 60 × 0.6 × −0.4 = −14.4
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