The problem has given the following circuit:

enter image description here

And asked to find the value(s) of Ix if the elements absorb the power(s) given.

I tired to set up KCL/KVL but I keep getting stuck.

I first drew this: enter image description here

And set KVL1 = (Ra)(i1)-(Rb)(i2) KVL2 = (Rb)(i2)+(Rc)(i3)

After that I am confused on how to continue without knowing the resistances.

  • \$\begingroup\$ Your KVL's are missing Vo and R1 and I would think you should write a KVL equation something like \$ \text{ KVL1: } V_r + V_s + i \cdot R_t = 0 \$. KVL1 self has no value. \$\endgroup\$ – Huisman Sep 15 at 17:27
  • \$\begingroup\$ Try to represent those boxes (which I assume are passive components) as Thevenin equivalents, a resistor in series with a voltage supply. \$\endgroup\$ – jDAQ Sep 15 at 18:03
  • \$\begingroup\$ This block diagram suggests you have2 sources 48V and B, 8W with 2 loads A+C= 88W thus the 48 V supply delivers 80W so Ix is simply 80W/48V = answer \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 15 at 19:02
  • \$\begingroup\$ B might not be an ideal source, and have some internal resistance. The two loads, A,C, could be more than just a resistive load. You should also include the power losses on the resistor R1. \$\endgroup\$ – jDAQ Sep 15 at 19:20

You could use the fact that the power in a closed system is conserved. That is, the sum, $$\sum_i P_i = 0$$ Where, $$ 0=P_{V_0}+P_{R_1}+P_A+P_B+P_C.$$ So, $$ 0=-I_XV_0+R_1I_X^2+48-8+40.$$ Hence, $$ 0=-48I_X+4I_X^2+80.$$ This will give you two possible values for \$I_X\$. Maybe both results are ok, or there is a way to further determine which of the \$I_X\$ is the right one. If someone wants to further deduce it, please go-ahead.

  • \$\begingroup\$ Thanks so much! \$\endgroup\$ – John K Sep 15 at 19:15

Redrawing things can sometimes help a little. Here's how I might attempt to redraw your schematic:


simulate this circuit – Schematic created using CircuitLab

Note that you have 5 equations and 5 unknowns. (Substitution could reduce that, if you like.) A quick check suggests that you will find two equally valid answers for the problem, though. You should also find (in both cases) that \$I_\text{B}\$ is negative (goes opposite to the indicated arrow.)


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