Circuit Theory/Simultaneous Equations/Example 5

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example 5 circuit

Find currents and voltages. Assume the sources and resistance values are knows. Solve symbolically.

From a practical point of view, putting more than one power supply in a circuit is bad practice. Almost everyone tries to design circuits with only one power supply.

From a modeling point of view, transistors at the first level of approximation are a type of power supply. So putting multiple power supplies in a circuit is practice or preparation for more complicated circuits.


symbols have been added

There is one voltage source, in series with a resistor to get it's current from: i_{R1}. The two current sources had no clear resistor in parallel with them from which to get their voltage, so voltages for each had to be labeled: i_{I1} and i_{I2}.


loops and the + - added

There are three loops. Remember that the + and - right now are not guessing the polarity of the answer, but are capturing the layout of the circuit.

There are no trivial loops (components in parallel).


constant EMF sections highlighted
Junctions chosen and current directions put on drawing

There are two trivial junctions where series components share the same current. One is between the voltage source and R_1. The other is between current source I_2 and R_5.

There are four non-trivial junctions, see the areas constant EMF that are shaded in different colors.

This means that three junctions that can be used. It doesn't matter which three were chosen. But three have to be chosen and then currents placed relative to the voltage polarity on the resistors. The two branches where there are current sources don't need additional current symbols made up for them, just like the voltage source does need a voltage symbol made up for it.

Count Eq[edit]

We are being told that the resistors and voltage sources have values, and are thus to be treated as knowns. We just don't know what the specific values are so we have to work symbolically.

R_1 = \surd
R_2 = \surd
R_3 = \surd
R_4 = \surd
R_5 = \surd
I_1 = \surd
I_2 = \surd
V_1 = \surd
i_{R1} = ?
i_{R2} = ?
i_{R3} = ?
i_{R4} = ?
v_{R1} = ?
v_{R2} = ?
v_{R3} = ?
v_{R4} = ?
v_{R5} = ?
v_{I1} = ?
v_{I2} = ?

There are 11 unknowns. There are 5 equations from the resistors, 3 from the loops and 3 from the junctions. So the problem can be solved explicitly.

Terminal Eq[edit]

v_{R1} = i_{R1} * R_1
v_{R2} = i_{R2} * R_2
v_{R3} = i_{R3} * R_3
v_{R4} = i_{R4} * R_4
v_{R5} = -I_2 * R_5
The current I_2 is negative, because it is going into the negative terminal of R_5.

When there are lots of power sources, the direction of their currents or the polarity of the sources has to be captured by the equations.

If this direction is mistaken, often the answer magnitudes turn out correct but the sign is wrong. The point is that the answers are very similar. The problem is usually that the sign convention is messed up some how.

Loop Eq[edit]

L_1: v_{R2} + v_{R1} - V_1 = 0
L_2: v_{R3} + v_{R4} + v_{I1} - v_{R2} = 0
L_3: v_{R5} + v_{I2} - V_{R4} = 0

The voltage polarities of the current sources were chosen arbitrarily, but a choice has to be made. The choice has to be documented on the drawing, and then reflected in the equations. Otherwise the equations can not be checked.

Junction Eq[edit]

J_1:  i_{R1} - i_{R2} - i_{R3} = 0
J_2:  i_{R3} + I_2 - i_{R4} = 0
J_3:  i_{R4} - I_2 - I_1 = 0

Solve the Equations[edit]


The algebra solution is massive, messy, hard to check and doesn't inspire anyone. Not going to attempt it.

Differential Equations[edit]

There are none in this problem, but three quarters of this course is going over circuits this complex with capacitors and inductors instead of resistors and looking at the differential equations.

Symbolic Computations[edit]

Mathematica solution. Click here for text to enter into Mathematica.

Wolfram Alpha doesn't work with more than 6 or 7 equations no matter what their form. Going to do in Mathematica. Then do in MuPAD. Then compare the answers.

MathWorks MuPAD solution .. click here for the text to cut and paste .. MuPad Solution contains variable solution

MuPAD's answers match with Mathematica, but the format of the more complicated answers is very different. The goal would then be to use some kind of simplify command, but alternative symbols can be used to express the same answer. Look at V_{I2} answers. They are the same, but they use different symbols.

In fact, here is another answer:

V_{I1} = \frac{(R_1 + R_2)}{R_1*R_2}*(\frac{V_1}{R_1}-I_1)-R_3*I_1-R_4*(I_1+I_2)

Mathematica solution simplied with the simplify command .. looks closer to MuPAD solution

Playing around with simplify and expand commands helps a little. There are all sorts of options associated with the simplify command that lead into a domain of symbolic computation that looses focus on this course. There is no immediate agreement among the packages to the solution of complicated expressions. A visual check of everything but V_{I1} is a match.

The only other quick way to check is to plug in a random set of numbers and see if the two solutions come up with the same numeric solution. But enough time has been spent on this.

Numeric Solution[edit]

There is no numeric solution because no numbers were given.


Simulation is impossible with out numbers.

Build Intuition[edit]

  • Complicated expressions have no universal form that symbolic computation agrees upon. So checking them is an algebric process, or can trust that inserting numbers and showing they match means the symbolic expressions are the same.
  • The direction choices when the variables were defined implied that the V_1 was dumping energy into the circuit. Looking at the solution, it appears that I_{R1}, the current through V_1 is positive. Thus it remains that V_1 is pumping energy into the circuit.
  • The direction choices when the variables were defined implied that the I_2 was dumping energy into the circuit. Looking at the solution, it appears that V_{I2}, the voltage across I_2 is positive. Thus it remains that I_2 is pumping energy into the circuit.
  • The direction choices when the variables were defined implied that the I_1 was taking energy out of the circuit and charging it's batteries. However, looking at the solution, it appears that V_{I2}, the voltage across I_1 is negative. Thus it seems that in reality I_1 is also pumping energy into the circuit.