Linear Circuit Analysis:
Series and Parallel Circuits - Y, Delta, Node, and Loop

Find the equivalent resistance of the network.
Irwin. page 57 example 2.22

To help in reducing the combination of resistors to a single equivalent resistor, We are going to use delta-to-wye conversion.

These conversions require the use of 6 formulas and some careful algebra.

For consistency and completeness, let's add these equations to the formula text. These formulas never change and you are always using either one set of formulas or the other. In an effort to make this type of problem easier to understand, I have written a program group entitled delta2yt.92g.

• The program is started by typing delta2y on the entry line.
• Press [2] to convert from a delta configuration to a Y configuration.

• Enter 18 in the field for R1 and press the down arrow.
• Enter 12 in the field for R2 and press the down arrow.
• Enter 6 in the field for R3 and press [ENTER] twice.

• The program has completed the math for us and displays the values of the resistors in their correct orientation.
• Press [3] to quit and return to the Home Screen.

• Press [APPS] [7] [3] to open a new program session.
• Press the Down Arrow twice and enter the name for the Program. (cust)
• Press [ENTER] twice

• Start entering the code.
• Enclose everything within the Custom    EndCustm commands.
• Use Title commands for headers to be called by the Function keys.
• Use Item commands for the specific functions.
• Parameters are listed to help you remember. Press [2nd] [QUIT] to exit the Program Editor and return to the Home Screen.

• On the Entry Line type cust() .
• Press [ENTER] [2nd] [CUSTOM].
• Use the Function Keys to find and call the function or program you want.

This example will incorporate several of the ideas that we have covered so far.

• Find the resistance of the middle branch of the circuit with addition and the parallel function.
• Use the value found to calculate the total resistance of the circuit. Store this value in r.

• By KCL at the top of the circuit we can define the current going to the resistors. This value is stored in i.
• By ohms law we can now define V_o in terms of V_x.
• Solve this expression for V_x.

• By using Voltage division we can find another expression for V_x in terms of V_o.
• Setting the two expressions that we have for .V_x equal to each other, we can solve for V_o.
• Applying the formula for power in terms of voltage and resistance, we find the power to the 12 ohm resistor is 3 watts.

Nodal analysis is simply a use of KCL at all nodes in order to set up a system of equations. Then we can solve the system of equations by using the matrix identity:

• Sum all of the currents leaving the node V₁. If the current is not known directly, it can be written in terms of Voltage divided by resistance.
• Multiply to clear fractions.
• Subtract the equation by 36 to get the equation in standard form.

Now we have the equation in a form that we can recognize. Let's store the coefficients in a row of a matrix for use in solving later.

• Press [APPS] [6] [3] to open a new Data/Matrix editing session.

• Select Matrix from the Data/Matrix choice and press the down arrow twice.
• Enter the variable name of the matrix and press the down arrow.
• Enter the number of Rows as 2 and press the down arrow.
• Enter the number of columns as 2 and press [ENTER] twice.

• Enter the coefficients for V1 and V2 from the first equation on the first row.
• Press [2nd] [QUIT] to return to the Home Screen.

• Sum all of the currents leaving the node V₁. If the current is not known directly, it can be written in terms of Voltage divided by resistance.
• Multiply this equation by 12 to clear fractions and add by 24 to get the equation in standard form.

Now we have the equation in a form that we can recognize. Let's store the coefficients in row 2 of matrix a for use in solving.

• Press [APPS] [6] [2] to reopen the Matrix Editor.
• Enter the coefficients for the second equation on row 2 of the matrix.
• Press [2nd] [QUIT] to return to the Home screen.

• Press [a] [ENTER] to display the matrix on the home screen.
• Multiply the inverse of a times the column vector of scalars obtained from the right side of the equations.
• By voltage division we can find the voltage across the 2 ohm resistor on the far right of the network.

When doing nodal analysis where a voltage source is present, we simply define the voltage at that node as the voltage that is supplied by that source. All other techniques are the same.

• Write the KCL equation for node 1. All currents are defined to be leaving the node.
• Multiply by 4 and subtract 8 to get the equation in standard form.

• Write the KCL equation for node 2. All currents are defined to be leaving the node.
• Multiply by 4 and add 16 to get the equation in standard form.

• The equation for node 3 is solved by inspection since there is an independent voltage source supplying voltage to the node.
• Write the KCL equation for node 4. All currents are defined to be leaving the node.
• Multiply by -2 to get the equation in standard form.

• Open the Matrix Editor and put in the values for the coefficients that we found from the equations. For this example I have stored this matrix as the variable named a.

• Write the final equation as a column vector of unknowns equal to A^-1 times the column vector of scalars from the right hand side of the equations.

When dependant sources are involved, their currents must be defined in terms of known values.

• By KCL at node V₂, we find an equation in terms of I_s and V₂. All currents are expressed as leaving the node.
• Multiply by 4 and add 16 to get the equation in standard form.
• By KCL at node V₁, all currents are expressed as leaving the node.

• The voltage for the dependant voltage source is defined in terms of V₁ and V₂.
• Subtracting 4I_x from both sides puts the equation in standard form.
• By Ohm's law we can define I_x in terms of V₁.
• Multiplying by 2 and subtracting 2I_x puts this equation in standard form.

• c1 = the coefficients of I_s.
• c2 = the coefficients of I_x.
• c3 = the coefficients of V₁.
• c4 = the coefficients of V₂.

For this example I have stored this matrix as the variable named a.

• Write the final equation as a column vector of unknowns equal to A^-1 times the column vector of scalars from the right hand side of the equations.

Loop analysis is simply a use of KVL in all meshes of a network in order to set up a system of equations. Then we can solve the system of equations by using the matrix identity:

All equations are written as the sum of the voltage drops equals the sum of the voltage gains. All mesh currents are defined to be traveling in a clockwise direction. Voltages across resistors is defined as resistance times current.

• Sum of the voltages around mesh one.
• Sum of the voltages around mesh two.
• Sum of the voltages around mesh three.

• The coefficients of I₁ are stored in column 1.
• The coefficients of I₂ are stored in column 2.
• The coefficients of I₃ are stored in column 3.

• To verify that we have entered the coefficients correctly press [a] [ENTER].

• Write the final equation as a column vector of the unknown currents equal to a^-1 times a column vector of scalars from the right hand side of the equations.

This network contains an independent current source. Therefore we will apply the technique of a super mesh to solve this circuit. All equations are written as the sum of the voltage drops equals the sum of the voltage gains. All mesh currents are defined to be traveling in a clockwise direction. Voltages across resistors is defined as resistance times current.

• The super mesh equation is written to include meshes 1 and 3. This mesh equation ignores the branch containing the independent current source.
• The equation for mesh 2 is written as the sum of voltage drops equals zero around the loop.
• The third equation is found by inspection. The sum of the currents in this branch must equal the current provided by the independent current source.

• The coefficients of I₁ are stored in column 1.
• The coefficients of I₂ are stored in column 2.
• The coefficients of I₃ are stored in column 3.

• To verify that we have entered the coefficients correctly press [a] [ENTER].
• Write the final equation as a column vector of the unknown currents equal to a^-1 times a column vector of scalars from the right hand side of the equations.

This Network contains a dependant current source. To solve for the unknowns we will need to use a supermesh around the dependant current source and write an equation to define the dependant current source. All equations are written as the sum of the voltage drops equals the sum of the voltage gains. All mesh currents are defined to be traveling in a clockwise direction. Voltages across resistors is defined as resistance times current.

• Equation 1 is the supermesh around the dependant current source. It is a KVL equation, written as the sum of the voltage drops equal to the voltage gains.
• To define the current of the dependant current source in terms of I₁ and I₂, equation 2 is a KCL equation at the top center node of the circuit, written as the sum of the currents leaving the node equals the sum of the currents entering the node.
• Subtract this equation by (I₁ + 2*V_o) to put the equation in standard form.
• To define the amount of current provided by the dependant current source, we write an Ohm's law equation relating the resistance times the current that goes through it.
• Subtract this equation by 2*I₂ to put the equation in standard form.

• The coefficients of I₁ are stored in column 1.
• The coefficients of I₂ are stored in column 2.
• The coefficients of V_o are stored in column 3.

• To verify that we have entered the coefficients correctly press [a] [ENTER].
• Write the final equation as a column vector of the unknown currents and voltage equal to a^-1 times a column vector of scalars from the right hand side of the equations.