Finding current using Kirchoff Law, but getting unsolvable system

Question

Hello, the original circuit above is where I am trying to find I1, and I simplify it the figure below and assuming the following current directions.

Using Kirchoff current and voltage laws I end up with

However, plugging in the equations into my calculator it says there is no solution. Is it with my kirchoff current equations (first 4 equations) because if I add up all current law equations I end up with 0 = 0. Can anyone tell me what I am doing wrong? Thanks for any help.

EDIT:

As other mentioned I need more KVL equations. But I also need my KVL to be independent loops, so I need at least 3 independent KVL equations.

Answer 1

If you use KVL in the other loop (36v i5 and i2) then use that, the other kvl equations and two of the kcl equations it should work.

Adding all the kcl equations should give 0=0. They don't include the voltage so they can't actually solve for anything here.

Using star-delta transforms and mesh analysis can make it a bit easier to solve (transform the delta that doesn't have I1 in it to a star/wye).

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

First, convince yourself that the above re-drawn schematic is the same as your problem original. I may have the numbering off (actually, I certainly do have the numbering off), but it is the approach which is important.

So:

We can make a quick substitution by combining R4 to R6 as R9=12 ohms because they're in series. I probably could also reduce R9 and R3 in parallel, but I'll leave them as-is for now.

Next, write out KCL and Ohm's law (assume currents are flowing "down" through resistors, up through V0):

\begin{equation} I_0 - I_1 - I_2 = 0\\ I_1 - I_3 - I_8 - I_9 = 0\\ I_2 - I_3 - I_7 - I_9 = 0\\ \end{equation} \begin{equation} I_1 = \frac{V_a - V_b}{R_1}\\ I_2 = \frac{V_a - V_c}{R_2}\\ I_3 = \frac{V_b - V_c}{R_3}\\ I_7 = \frac{V_c}{R_7}\\ I_8 = \frac{V_b}{R_8}\\ I_9 = \frac{V_b - V_c}{R_9}\\ V_a = V_0 \end{equation}

Substituting back in:

\begin{equation} I_0 - \frac{V_a - V_b}{R_1} - \frac{V_a - V_c}{R_2} = 0\\ \frac{V_a - V_b}{R_1} - \frac{V_b - V_c}{R_3} - \frac{V_b}{R_8} - \frac{V_b - V_c}{R_9} = 0\\ \frac{V_a - V_c}{R_2} - \frac{V_b - V_c}{R_3} - \frac{V_c}{R_7} - \frac{V_b - V_c}{R_9} = 0\\ V_a = V_0 \end{equation}

A little bit of re-writing (Gn = 1/Rn):

\begin{equation} I_0 + G_1 V_b + G_2 V_c = (G_1 + G_2) V_0\\ (G_1 + G_3 + G_8 + G_9) V_b - (G_3 + G_9) V_c = G_1 V_0\\ (G_3 + G_9) V_b - (G_3 - G_2 - G_7 + G_9) V_c = G_2 V_0 \end{equation}

We have three equations with three unknowns: I0, Vb, and Vc. Once you've solved for these you can calculate I1 easily using R1, Va, and Vb. And yes, this is a solvable system. I'll stop short of just posting the number solution.

Incidentally, this approach is known as Modified Nodal Analysis and is used in SPICE circuit simulation software. It basically adds an extra unknown current for each voltage source, and then adds an extra equation for the difference between the nodal voltages. I simply did some extra "inline plugging in" of the source voltage equation to reduce the set of equations/unknowns to 3. Yes, this approach might look like extra work because you are solving for voltages first, but it is a much more systematic approach, quite robust, and in the long run I find it faster to actually do.

Answer 3

Here is an easy way

^{simulate this circuit – Schematic created using CircuitLab}

This is an unbalanced Wheatstone bridge

Here is a clever way to find Current in each Branch

^{simulate this circuit}

Now I am going to Write A KCL equation for node C and D

I will use the potentials marked in the schematic

So for C

Sum of current converging on a point = 0

$$\frac { X-36 }{ 2 } \quad +\quad \frac { X-0 }{ 12 } \quad +\quad \frac { X-Y }{ 6\quad } \quad =\quad 0$$

Likewise for D

$$\frac { Y-36 }{ 9 } \quad +\quad \frac { Y-0 }{ 18 } \quad +\quad \frac { Y-X }{ 6\quad } \quad =\quad 0$$

Solve these 2 equations

You get

X = 30

Y = 27

Now we can get our answers

$${ I }_{ 1 }\quad =\quad \frac { { V }_{ A }\quad -{ V }_{ C } }{ R } \quad =\quad \frac { 36-X }{ 2 } =\frac { 36-X }{ 2 } =\quad 6(0.5)\quad =\quad 3\quad A$$

So you now can find every current

Answer 4

Here is an alternative for solving the circuit.

Made with Maple sheet. Used 2nd picture of OP question. Use eq6 OR eq7.

Made with microcap v12