I'm stuck with this simple problem (please be aware that this is my first non trivial circuit!):


simulate this circuit – Schematic created using CircuitLab

Due to Kirchoff law (I chose clockwise as positive) I've written: $$-i_1+i_0+i_2=0$$ $$i_4-i_5-i_0=0 $$

$$i_1-i_3-i_4=0$$ $$i_3-i_2+i_5=0$$ So I go on with $$i_1-i_2=i_4-i_5$$ $$i_1=i_3+i_4$$ Then, for tensions: $$V_1+V_3+V_2=0$$ $$V_4+V_5+V_3=0$$ $$V_0+V_1+V_4=0$$

I obtain the system of equations: $$i_3(R_1+R_2+R_3)+R_1 i_4+ R_2 i_5=0$$ $$i_3 R_3+ i_4R_4 + i_5 R_5=0$$ $$V_0+R_1 (i_3+i_4)+ R_4 i_4 =0$$

I've solved the problem, but trying some simulators I realize that the result is correct if all the resistors are of the same value, and completely wrong otherwise. What's the problem? Thanks a lot for your precious help!

  • \$\begingroup\$ @try-catch-finally what I am asking here is: what's wrong in my assumption? where did i uncorrectly applied laws? \$\endgroup\$ – Surfer on the fall Dec 26 '16 at 10:22
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    \$\begingroup\$ Poor translation. "Tension" would be "voltage" in english. \$\endgroup\$ – JRE Dec 26 '16 at 10:42
  • \$\begingroup\$ I just saw the same question 2 days ago. \$\endgroup\$ – Bradman175 Dec 26 '16 at 11:03
  • \$\begingroup\$ Hint: Its a balanced Wheatstone bridge so R3 could be any value - with no voltage difference between its ends the current will always be zero. It works as long as R1/R4 = R2/R5 (not necessarily all the same value). \$\endgroup\$ – JIm Dearden Dec 26 '16 at 14:11

I got the error. If I suppose current flows down through R3, then the second equation for tensions should follow the sign convention. It should be $$V_4+V_5-V_3=0$$, instead of $$V_4+V_5+V_3=0$$


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