Problem from book - Page 1
Forum section: electronics / beginners

author: filipst
author_rank: Newbie
timestamp: May 17, 2023, 10:41:29 am
content: Hi,I can't figure out how to read the circuit from 2.10Can anyone give me some hints?Thank you.I think my main issue is not understanding dependent sources properly.

author: Andy Watson
author_rank: Super Contributor
timestamp: May 17, 2023, 11:08:14 am
content: You are told that \$i_\phi\$ = 2A, the dependent current source produces \$2 i_\phi\$ = 4A. You should now be able to label the currents flowing in all the branches.

author: Ian.M
author_rank: Super Contributor
timestamp: May 17, 2023, 11:34:55 am
content: For convenience take the bottom node as the 0V reference. Ohm's law + known current through 30Ω resistor gives you voltage of top left node. KCL for that node (see Andy's reply) gives you current through the 10Ω resistor, and thus voltage across it. You now have all the node voltages. Apply KCL to top right node to get current through Vs. With all node voltages and branch currents known, grind through calculating the powers.

author: armandine2
author_rank: Frequent Contributor
timestamp: May 17, 2023, 12:18:49 pm
content: Practically, I guess this isn't impossible to demonstrate - with a power supply and an electronic load

author: barshatriplee
author_rank: Regular Contributor
timestamp: May 17, 2023, 03:27:58 pm
content: I just solved number 'a' like this. Sorry. Could not help with the others.

author: Picuino
author_rank: Super Contributor
timestamp: May 17, 2023, 06:38:45 pm
content: V_R1 = 30Ω · 2A = 60VI_Vs = 5A - 2A = 3ABecause the dependent current source and resistor 2, being in series, do not change the current flowing through that branch, which is the one coming out of the independent current source minus the one flowing through the first resistor.3A - 2·2A + I_R2 = 0 -> I_R2 = 4A - 3A = 1AV_R2 = 10Ω · 1A = 10VVs = V_R1 + V_R2 = 60V + 10V = 70VP_Vs = Vs · Is = 70V · 3A = 210WI'm sorry I don't put more component references, but the circuit does not name the components.

author: filipst
author_rank: Newbie
timestamp: May 18, 2023, 09:46:21 pm
content: Thanks for the answers! I think I understand that one now.Here's another one:I can't verify c.I got the solutions for this book, but they appear to be for another edition because the problems are the same but the values are different.But I followed the steps in the solution for these values and still I get more power developed than dissipated.

author: Andy Watson
author_rank: Super Contributor
timestamp: May 19, 2023, 01:42:49 am
content: There is a typo, possibly more then one, in the question. The specification of \$i_0\$ = 2A is inconsistent with the values of resistance given in the problem. It is also unnecessary for solving the problem, however, I'm guessing that the textbook has not yet covered the required methods for solving the problem without knowing \$i_0\$ (you need to know about superposition, thevenin, or star-delta transforms to solve this without knowing \$i_0\$.)Try using \$i_0\$ = 1.171 A - it should give you a consistent answer.Or post the answers given in the text and we can try to guess where the typo is!

author: Picuino
author_rank: Super Contributor
timestamp: May 19, 2023, 06:57:39 am
content: Solving by the mesh method (with 3 meshes):180 = 5(i1-i0) + 70(i1-i2)0 = 60·i0 + 25(i0-i2) + 5(i0-i1)0 = 70(i2-i1) + 25(i2-i0) + 35·i2Solving the system:i0 = 1.171 Ai1 = 5.404 Ai2 = 3.135 ANot consistent with the i0=2A data