Question

Question 3 Solve for the open circuit voltage drop from a to b in Circuit 9 ΚΩ 4 ΚΩ 6 ΚΩ 3 2 mA 8 V. 8ΚΩ 13.6V 132 13 ν Π

Answer 1

Answer:13.2V

Redrawing the given network:

gkor 4 uur Ja Uz Gkor ia Ал N, a 2MA ④ 8v 6 Г от л 845

Clearly V₂= 8

Now, by nodal analysis

V1 V3 V1 - 8 + = 20.3611V1 -0.111113 9

$\frac{V_{3}-V_{1}}{9}+\frac{V_{3}-8}{6}+\frac{V_{3}}{8}=0\Rightarrow -0.1111V_{1}+0.4027V_{3}=1.3333$

by solving the above two equations

$V_{1}=13.21V$

$V_{3}=6.95V$

Hence open-circuit voltage drop from a to b is 13.21