We were given a circuit with 7 resistors and 3 voltage sources

enter image description here

For emf sources, \$E_1=120V\$, \$E_2=60V\$, and \$E_3=30V\$ while for resistors in ohms,\$R_1=10\$, \$R_2=5\$, \$R_3=20\$, \$R_4=8\$, \$R_5=12\$, \$R_6=6\$, \$R_7=8\$. Nodes are indicated in small letters (a-j). Loops \$abefa\$, \$abcdefa\$ and \$aghcdijfa\$ are assumed clockwise while the loop \$cbedc\$ is counterclockwise.

I applied Kirchoff's Current Rule in nodes \$b\$ and \$e\$:

\begin{align} I_1+I_3 &= I_2\\ I_4+I_5 &= I_2 \end{align}

And then I apply the Voltage Rule: Loop \$abefa\$: $$120-I_1R_1-I_2R_2-30-I_4R_4=0$$

Loop \$cbedc\$: $$60-I_3R_3-I_2R_2-30-I_5R_5=0$$

Loop \$abcdefa\$:


Since I have 5 equations already with 5 unknows, hopefully I should solve these values.

From 1st and 2nd equations, \$I_4 = I_2-I_5 = I_1+I_3-I_5\$. Setting \$I_4\$, and \$I_2\$ in terms of \$I_1\$, \$I_3\$ and \$I_5\$ only and substituting them to equations 3, 4 and 5, I get:

\begin{align} 23I_1+13I_3-8I_5&=90\\ 18I_1-12I_3-20I_5&=60\\ 5I_1+25I_3+12I_5&=30 \end{align}

Luckily my calculator supports systems of equations for 3 uknowns, but then I get a math error, which could indicate infinitely many solutions? Maybe I incorrectly applied Kirchoff's rule but I don't know where. Also, I verified if the circuit works properly using an online circuit simulator and yes it worked well with indicated current and voltage drop values within each resistor. https://www.falstad.com/circuit/ (The circuit setup installed in the simulator is identical to the image). We are asked to solve this using 4 methods (Nodal, Maxwell, Superposition and Kirchoff) and I started with Kirchoff.

  • \$\begingroup\$ hey, fixed your line breaks for you. \$\endgroup\$ Aug 11, 2019 at 14:07
  • \$\begingroup\$ So what is your question? \$\endgroup\$
    – JRN
    Aug 11, 2019 at 14:31
  • \$\begingroup\$ I get math error in solving the values, that means I can't solve them \$\endgroup\$ Aug 11, 2019 at 14:41
  • 1
    \$\begingroup\$ Try solving the equations without a calculator. \$\endgroup\$
    – Finbarr
    Aug 11, 2019 at 15:15
  • \$\begingroup\$ @Finbarr I've tried solving this manually, reduced these 3 systems of equations to 2 systems of equations. The result is I got 2 identical equations. If this is really the case (infinitely many solutions), how come the simulator shows values for voltage drops and current within each resistor? \$\endgroup\$ Aug 11, 2019 at 15:27

2 Answers 2


You have four relatively obvious current loops that could use KVL. That's all you need for mesh analysis. (You don't need to include any KCL statements for the nodes.)


simulate this circuit – Schematic created using CircuitLab

I've labeled each clockwise-going current loop inside each independent loop, which hopefully will be obvious to you. The following equations will make it explicit and if the above loops aren't obvious from the schematic then they will be from the equations below. I will start each loop from their lower-left corner position and I will follow around that loop in clockwise (with the current) direction:

$$\begin{align*} 0\:\text{V}-I_1\cdot R_6-\left(I_1-I_2\right)\cdot R_3-\left(I_1-I_4\right)\cdot R_1&= 0\:\text{V}\\\\ 0\:\text{V}+V_3-\left(I_2-I_4\right)\cdot R_2-\left(I_2-I_1\right)\cdot R_3-V_2-\left(I_2-I_3\right)\cdot R_5 &= 0\:\text{V}\\\\ 0\:\text{V}-\left(I_3-I_4\right)\cdot R_4-\left(I_3-I_2\right)\cdot R_5-I_3\cdot R_7&= 0\:\text{V}\\\\ 0\:\text{V}+V_1-\left(I_4-I_1\right)\cdot R_1-\left(I_4-I_2\right)\cdot R_2-V_3-\left(I_4-I_3\right)\cdot R_4 &= 0\:\text{V} \end{align*}$$

From the above, you should be able to solve for all of \$I_1\$, \$I_2\$, \$I_3\$, and \$I_4\$. With those in hand, node voltages (if you set a ground point on one of the nodes, anyway) and the net currents in devices can all be readily computed.


A glance at your 3 equations shows that the first is just the sum of the other two. So you don't have a set of 3 independent simultaneous equations with a unique solution.

The reason the equations are not independent is that the loop abcdefa is the superimposition of the other two.

You will need to involve loops aghcba and defjid and I6 and I7. But beware, aghcdijfa is again not independent of the other loops.

Obviously I6 and I7 are also involved. You would be adding 2 new variables, I6 and I7, but also 3 new, independent equations. So removing one of your original equations would give you enough independent ones for a solution.


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