This is the circuit

Using Kirchhoff's 2nd law, taking closed loop $\underline{A}BD\underline{A}$,

$$2i_1-3i_2+i_3=0 .....(i)$$

Taking closed loop $\underline{D}BC\underline{D}$,

$$-6i_1+10i_2+17i_3=0 .....(ii)$$

Taking closed loop $\underline{A}BCD \underline{A}$,

$$-8i_1+13i_2+16i_3=0 .....(iii)$$

But this system of equations is not solvable; what mistake did I make? Is $\underline{A}BCD \underline{A}$ not a valid closed loop to take?

  • $\begingroup$ Does this answer your question? physics.stackexchange.com/q/91463 $\endgroup$ – Amirhosein Rezaee Apr 28 at 6:47
  • 1
    $\begingroup$ Saying that the system you have found is "not solvable" is an overstatement, and doesn't apply here. That term implies that the system does not have any compatible solutions. Your system does have (an infinity of) solutions, and it just packs information to decide between them. The technical term is that it is indeterminate. $\endgroup$ – Emilio Pisanty Apr 28 at 11:33

It is not solvable because you allow the current to enter the points $A$ and $C$, but don't have any equation connecting the current $I_{A/C}$ and voltage $U_{AC}$. In other words, you haven't chosen a loop that comes through $E$ (the battery).

In Kirchhoff equations, it's important to choose independent loops. But in your case $ABCDA = -ABDA+DBCD$.

To solve the problem, you can keep any 2 of the 3 loops you considered and add any loop involving point $E$.


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