Use Nodal only. Write the 2 equations and solve them for y, Vi and V2. Show your calculations in solving (including Cramer\'s rule if you use it...otherwise no bonus points) 6A 1/9 10A 1/2 1 50V(+ 1/4 10v 51. V2-1) 2V 2) 33V 3) 41V 4) 57V 5) 6V Solution writing KCL for the node where voltage is v2: (v2-10*v)/(1/9)+6+(v2-v1)/((1/2))=0 ==>9*(v2-10*v)+6+2*(v2-v1)=0 ==>9*v2-90*v+6+2*v2-2*v1=0 ==>90*v+2*v1-11*v2=6….(1) writing KCL for the node where voltage is v1: 10=(v1-v2)/(1/2) +(v1/(1/4)) ==>10=2*(v1-v2)+4*v1 ==>6*v1-2*v2=10….(2) voltage across the current source of 10 A is v ==>10*v-v=v1 ==>v1=9*v…(3) substituting equation 3 in equation 1 and 2 we get 90*v+18*v-11*v2=6 ==>108*v-11*v2=6….(4) substituting equation 3 in equation 2: 54*v-2*v2=10….(5) equation 4-equation 5*2: -11*v2+4*v2=6-2*10 ==>-7*v2=-14 ==>v2=2 volts then from equation 5, v=(10+2*v2)/54 =0.26 volts then v1=9*v =2.33 volts hence answers are: Q50. V1=7/3 volts option 5 is correct. Q51. V2=2 volts option 1 is correct. Q52. v=0.26 volts .

1. Use Nodal only. Write the 2 equations and solve them for y, Vi and V2. Show your calculations
in solving (including Cramer's rule if you use it...otherwise no bonus points) 6A 1/9 10A 1/2 1
50V(+ 1/4 10v 51. V2-1) 2V 2) 33V 3) 41V 4) 57V 5) 6V
Solution
writing KCL for the node where voltage is v2:
(v2-10*v)/(1/9)+6+(v2-v1)/((1/2))=0
==>9*(v2-10*v)+6+2*(v2-v1)=0
==>9*v2-90*v+6+2*v2-2*v1=0
==>90*v+2*v1-11*v2=6….(1)
writing KCL for the node where voltage is v1:
10=(v1-v2)/(1/2) +(v1/(1/4))
==>10=2*(v1-v2)+4*v1
==>6*v1-2*v2=10….(2)
voltage across the current source of 10 A is v
==>10*v-v=v1
==>v1=9*v…(3)
substituting equation 3 in equation 1 and 2 we get
90*v+18*v-11*v2=6
==>108*v-11*v2=6….(4)

2. substituting equation 3 in equation 2:
54*v-2*v2=10….(5)
equation 4-equation 5*2:
-11*v2+4*v2=6-2*10
==>-7*v2=-14
==>v2=2 volts
then from equation 5,
v=(10+2*v2)/54
=0.26 volts
then v1=9*v
=2.33 volts
hence answers are:
Q50. V1=7/3 volts
option 5 is correct.
Q51.
V2=2 volts
option 1 is correct.
Q52. v=0.26 volts