Skip to main content
10 events
when toggle format what by license comment
Nov 21, 2019 at 1:17 comment added Tony Stewart EE since 1975 It works but the steady state is to the right of his curve But the Thevenin appears not to work.. The Norton equivalent with 5A//1.6 Ohms works
Nov 21, 2019 at 0:20 comment added Vinicius ACP @G36 So, in a "real number world", this circuit simply does not work? I mean, if I have the second circuit situation, will the entire circuit be "off" or something like that?
Nov 20, 2019 at 22:59 comment added Vinicius ACP @TonyStewartSunnyskyguyEE75 Ok. $$ v=3.744 \mathrm V \implies i \approx 2.66 \mathrm A \\ \begin{alignat}{1} \\ v &= 8-v_R \\ v &= 8-1.6 \cdot i \\ v &= 8-1.6 \cdot 2.66\\ 3.744 &= 3.744 \end{alignat} $$ Without using the Thévenin's equivalent: $$ \begin{alignat}{1} v_R &= v-3.2\\ 1.6 \cdot i_R &= 3.744-3.2 \\ 1.6 \cdot (3-2.66) &= 0.544 \\ 1.6 \cdot 0.34 &= 0.544 \\ 0.544 &= 0.544 \\ \end{alignat} \\3.2+0.544=3.744=v $$ I can't apply these values in \$ v=3.2-v_R \$ because they are not solutions of $$ 4.8v^2-27.8v+41.6=0 $$ So, where is the mistake?
Nov 20, 2019 at 22:04 comment added Tony Stewart EE since 1975 plug in my answer and you will see that your KCL statement is not right
Nov 20, 2019 at 22:02 comment added Vinicius ACP @TonyStewartSunnyskyguyEE75 What mistake? I simply applied it: $$ -3.2+v_R+v = 0 \implies v = 3.2 - v_R $$
Nov 20, 2019 at 21:57 comment added Tony Stewart EE since 1975 try v=3.744 then i {=2.66} + 0.34 = 3A
Nov 20, 2019 at 21:49 history edited G36 CC BY-SA 4.0
edited body
Nov 20, 2019 at 21:42 comment added Tony Stewart EE since 1975 The complex numbers mean you made a mistake early in KVL and this answer is wrong/incomplete for v > 3V
Nov 20, 2019 at 21:40 comment added Vinicius ACP But what is the interpretation of the complex voltages and currents I got? Do they have any physical meaning?
Nov 20, 2019 at 21:24 history answered G36 CC BY-SA 4.0