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1 kn 1 kn 31 kn ' 31 kn 2V 10v E 2 kn 3 2 kn 2 kn3 Vx 31 kn V. 4 mA Using I1, 12, 13, 14, 15, and 16 for the mesh currents in mA, write the equation for the application of KVL around the requested meshes. If it is not possible to write such an equation please enter "NP" without the quotes:

1 kn 1 kn 31 kn ' 31 kn 2V 10v E 2 kn 3 2 kn 2 kn3 Vx 31 kn V. 4 mA Using I1, 12, 13, 14, 15, and 16 for the mesh currents in mA, write the equation for the application of KVL around the requested meshes. If it is not possible to write such an equation please enter "NP" without the quotes:

Power System Analysis and Design (MindTap Course List)
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.39P
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KVL around i1

KVL around i2

KVL around i3

KVL around i4

KVL around i5

KVL around i6

please help me with this question, the solution attach with the question is know incorrect, thanks.

1 ko 31 kM 31 kn 3) 3 1 kN 2V 10V E 2 kN 4 mA 32 kn 2 kn 3Vỵ 31 kN V. Using I1, 12, 13, 14, 15, and 16 for the mesh currents in mA, write the equation for the application of KVL around the requested meshes. If it is not possible to write such an equation, please enter “NP" without the quotes:
Transcribed Image Text:1 ko 31 kM 31 kn 3) 3 1 kN 2V 10V E 2 kN 4 mA 32 kn 2 kn 3Vỵ 31 kN V. Using I1, 12, 13, 14, 15, and 16 for the mesh currents in mA, write the equation for the application of KVL around the requested meshes. If it is not possible to write such an equation, please enter “NP" without the quotes:
Solution: For meh l-s Call aurents in mA os all resistors in Kn) -i -(i-iz) - 2ra=01 but Vx= 2 (i s-i6) - hence -i- 2iti2- 4is+426=o funally - 2i+i2- 4is+4i6=0 4 i) For mesh 24 a62- i) - i2-ig)-1u=o -2i,+i+i3 -1° =0 | or muh3a -i3-i)-i, -2liz -ic) = 0 - 4i3+i2+2i = o for mesh 4- iy= 4mA (no kVL)eqcir) %3D fgor muth 5 -2(is-4) +10-2(is-i6)= 0 -4is +2+2i6 = ° eq lv) for mush 6 -2(16-is) – 2(i6-i2)- is=• -si6+223+2is =0 (i) eg cri).
Transcribed Image Text:Solution: For meh l-s Call aurents in mA os all resistors in Kn) -i -(i-iz) - 2ra=01 but Vx= 2 (i s-i6) - hence -i- 2iti2- 4is+426=o funally - 2i+i2- 4is+4i6=0 4 i) For mesh 24 a62- i) - i2-ig)-1u=o -2i,+i+i3 -1° =0 | or muh3a -i3-i)-i, -2liz -ic) = 0 - 4i3+i2+2i = o for mesh 4- iy= 4mA (no kVL)eqcir) %3D fgor muth 5 -2(is-4) +10-2(is-i6)= 0 -4is +2+2i6 = ° eq lv) for mush 6 -2(16-is) – 2(i6-i2)- is=• -si6+223+2is =0 (i) eg cri).
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