# Getting the wrong result of a current through a resistor

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

My attempt: I simplify the circuit as follows:

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?

• How can any current in this circuit be larger than 5mA?
– G36
Commented Jan 26 at 7:33
• @G36 that did cross my mind. Using the current divider I do end up with $I_3=8.33$.
– Sam
Commented Jan 26 at 8:04
• 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.
– G36
Commented Jan 26 at 8:33