Simultaneous Equations for Circuit Analysis
Mathematics for Electronics
What does it actually mean to obtain a solution for a “simultaneous” system of equations? For example, if 2x + y = 7 and x − y = −1, what do the solution values (x = 2 ; y = 3) represent?
If we were to graph both these linear equations on a Cartesian (x, y) coordinate system, where would the solution (2,3) be located on the graph?
Load lines are useful tools for analyzing transistor amplifier circuits, but they may be hard to understand at first. To help you understand what “load lines” are useful for and how they are determined, I will apply one to this simple tworesistor circuit:

We will have to plot a load line for this simple tworesistor circuit along with the “characteristic curve” for resistor R_{1} in order to see the benefit of a load line. Load lines really only have meaning when superimposed with other plots. First, the characteristic curve for R_{1}, defined as the voltage/current relationship between terminals A and B:

Next, I will plot the load line as defined by the 1.5 kΩ load resistor. This “load line” expresses the voltage available between the same two terminals (V_{AB}) as a function of the load current, to account for voltage dropped across the load:

At what value of current (I_{R1}) do the two lines intersect? Explain what is significant about this value of current.
Load lines are useful tools for analyzing transistor amplifier circuits, but they may be applied to other types of circuits as well. Take for instance this dioderesistor circuit:

The diode’s characteristic curve is already plotted on the following graph. Your task is to plot the load line for the circuit on the same graph, and note where the two lines intersect:

What is the practical significance of these two plots’ intersection?
Suppose you were given the following two equations and asked to find solutions for x and y that will satisfy both at the same time:


If we manipulate the second equation so as to solve for y, we will have a definition of y in terms of x that we may use for substitution in the first equation:

Show the process of substitution into the first equation, and how this leads to a single solution for x. Then, use that value of x to solve for y, resulting in a solution set valid for both equations.
An interesting and useful property in mathematics is the transitive property:

Simply stated, two variables must be equal to one another if they are both equal to a common (third) variable. While not particularly profound or breathtaking in scope, this property is nevertheless useful in solving certain mathematical problems.
Suppose you were given the following two equations and asked to find solutions for x and y that will satisfy both at the same time:


Manipulate both of these equations to solve for y, and then explain how you could apply the transitive principle to solve for x.
Suppose you were given the following two equations and asked to find solutions for x and y that will satisfy both at the same time:


Now, you know that we may do anything we want to either equation as long as we do the same thing to both sides (on either side of the “equal” sign). This is the basic rule we follow when manipulating an equation to solve for a particular variable. For example, we may take the equation y + x = 8 and subtract x from both sides to yield an equation expressed in terms of y:

Following the same principle, we may take two equations and combine them either by adding or subtracting both sides. For example, we may take the equation y − x = 3 and add both sides of it to the respective sides of the first equation y + x = 8:

What beneficial result comes of this action? In other words, how can I use this new equation 2y = 11 to solve for values of x and y that satisfy both of the original equations?
Many circuit analysis techniques require the solution of “systems of linear equations,” sometimes called “simultaneous equations.” This question is really a series of practice problems for solving simultaneous linear equations, the purpose being to give you lots of practice using various solution techniques (including the solution facilities of your calculator).
Systems of two variables:
x + y = 5  x − y = −6  2x + y = 7 
x − y = 1  2x − y = 4  x − y = 2 
3x − 2y = −1  −10x + 2y = 0  3x − 5y = −13 
5x + y = −6  −3x − 5y = −28  −x + 2y = 5 
1000x − 500y = 0  −15000x + 2200y = −66200  9100x − 5000y = 24 
550x + 2500y = 5550  7900x − 2800y = 28300  −5200x − 2700y = −6.5 
Systems of three variables:
x − y + z = 1  3x + 2y − 5z = −21  x + y + z = 0 
−x − y + z = −1  x − 3y + z = 8  2x − y − 4z = −9 
x + y + z = 3  −x − y − z = −12  −2x + 2y − z = 12 
x + y − 2z = −12  −4x − 3y + 2z = −32  19x − 6y + 20z = −33 
3x − 2y + z = 19  x − 2y + 3z = −1  4x + 5y − 3z = −17 
−4x + 3y − 5z = −45  −2x + 7y − z = 3  −7x + 2y − 8z = 9 
890x − 1000y + 2500z = −1500  2750x − 6200y + 4500z = 17500 
3300x + 7200y − 5100z = 21500  −10000x + 5300y − 1000z = 8100 
−x + y − z = 0  6x − 2y − 3z = 5 
Suppose you needed to choose a fixed resistor value (R) to make a voltage divider circuit, given a known potentiometer resistance value, the source voltage value, and the desired range of adjustment:

Solve for R, and show the equation you set up in order to do it.
Hint: remember the series resistor voltage divider formula . . .

Suppose you needed to choose a potentiometer value (R) to make a voltage divider circuit, given a known fixed resistor value, the source voltage value, and the desired range of adjustment:

Solve for R, and show the equation you set up in order to do it.
Hint: remember the series resistor voltage divider formula . . .

An engineer needs to calculate the values of two resistors to set the minimum and maximum resistance ratios for the following potentiometer circuit:

First, write an equation for each circuit, showing how resistances R_{1}, R_{2}, and the 10 kΩ of the potentiometer combine to form the ratio [a/b]. Then, use techniques for solving simultaneous equations to calculate actual resistance values for R_{1} and R_{2}.
The voltage gain of a commonemitter transistor amplifier is approximately equal to the collector resistance divided by the emitter resistance:

Knowing this, calculate the necessary resistance values for the following fixedvalue resistor (R_{2}) and potentiometer (R_{1}) to give this commonemitter amplifier an adjustable voltage gain range of 2 to 8:

The voltage gain of a commonemitter transistor amplifier is approximately equal to the collector resistance divided by the emitter resistance:

Knowing this, calculate the necessary resistance values for the following fixedvalue resistors (R_{1} and R_{2}) to give this commonemitter amplifier an adjustable voltage gain range of 4 to 7:

Suppose you needed to choose two resistance values to make a voltage divider with a limited adjustment range. One of these resistors will be fixed in value (R_{1}), while the other will be variable (a potentiometer connected as a rheostat R_{2}):

Set up a system of simultaneous equations to solve for both R_{1} and R_{2}, and show how you arrived at the solutions for each.
Hint: remember the series resistor voltage divider formula . . .

Suppose you needed to choose two resistance values to make a voltage divider with a limited adjustment range:

Set up a system of simultaneous equations to solve for both R_{1} and R_{2}, and show how you arrived at the solutions for each.
Hint: remember the series resistor voltage divider formula . . .

The voltage gain of an inverting operational amplifier circuit is defined by the ratio of feedback to input resistance:


Calculate the necessary values of R_{1} and R_{2} to limit the minimum and maximum voltage gain of this opamp circuit to 5 and 30, respectively, given a potentiometer in the middle with a fullspan resistance of 5 kΩ:

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