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As a step in an exercise, I'm attempting to compute the current going through R3 caused by the current source in following circuit:

enter image description here


My attempt: I simplify the circuit as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

  • I3 splits into two currents, one going through R4 and another through R5. Using a current divider I find that (3/5)I3 = 5mA, so that I3 = 8.33mA.

  • KVL in the top quadrant gives $$-4I_1-2(I_1-I_2)+2(I_3-I_1) = 0 \implies -4I_1+I_2+I_3 = 0.$$

  • KVL in the lower-right quadrant gives $$-2(I_2-I_3)+2(I_1-I_2)=0\implies I_1-2I_2+I_3=0.$$

  • Combining the above two equations gives \$-5I_1+3I_2 = 0\implies I_1 = (3/5)I_2\$.

  • Substituting into the second equation gives $$\frac{3}{5}I_2-2I_2 + I_3 = 0 \implies \frac{7}{5}I_2 = I_3 \implies I_2 = 5.95mA.$$

  • Finally, the current that flows through R3 is given by \$I_2-I_3 = 2.38mA\$.


The correct answer, however, is \$0.91mA\$. Where did I go wrong?

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  • \$\begingroup\$ How can any current in this circuit be larger than 5mA? \$\endgroup\$
    – G36
    Jan 26 at 7:33
  • \$\begingroup\$ @G36 that did cross my mind. Using the current divider I do end up with \$I_3=8.33\$. \$\endgroup\$
    – Sam
    Jan 26 at 8:04
  • \$\begingroup\$ This is wrong it cannot be the case. Notice that R4 is in series with the current source. Thus, I_R4 = 5mA. So, the I_s1 current can "only" split into three currents. 5mA = I_R5 + I_R2 + I_R1. And we can find (I_R2 + I_R5) easily. (I_R2 + I_R5) = 5mA * 4kΩ/(1.5kΩ + 4kΩ) = 5mA*(4/5.5) = 3.6363mA. And because R5 = R2+(R3+R6)||R7). The current will split in half, Therefore, I_R2 = (I_R2 + I_R5)/2 = 1.8181mA. And again because R3+R6 = R7 this current (I_R2) will also split in half. Thus, I_(R3+R6) = 1.8181mA/2 = 0.909 = 0.91mA. And we are done. \$\endgroup\$
    – G36
    Jan 26 at 8:33

2 Answers 2

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> My attempt: I simplify the circuit

Don't stop where you did; carry on simplifying. After all we are EEs and we should use our EE skills first before resorting to math: -

enter image description here

In the 2nd image above, I converted the 5 mA current source and parallel R5 to a voltage source of 15 volts in series with R5 (Thevenin's theorem).

In the 3rd image I compressed R1, R2, R3, R6 and R7 to one value. I am left with a very simple voltage divider equation. I can now calculate the voltage at the marked spot to be 5.456 volts.

And, that will be the same voltage at the junction of R5 and R2 in your original diagram. You can then continue to find the current through R3. Do you need any further help? I get 0.909096 mA for the current through R3 BTW.

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I apply the principle of superposition of effects. A partial solution is as follows:

enter image description here

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