Linear Circuit Analysis:

Charge and Current

Work, Power, Voltage, and
Resistance

Kirchoff's Laws

Y Delta, Node, and Loop

Circuits with Operational
Amplifiers

Network Theorems

Analysis of Diode Circuits

Capacitance and Inductance

First-Order Transient Circuits

AC Steady State Analysis

Steady State Power

The Power Factor

Introduction

adv.92g
circuits.92g
delta2yt.92g

Linear Circuit Analysis:
Network Theorems

Using Superposition to find V_o in the circuit.
Irwin. page 176 example 4.4

To find V_o we can consider the contributions from the sources one at a time and then add them together to find the final voltage. When working with a single source the other independent sources are turned off. A voltage source that has been turned off is represented by a short. No voltage is dropped. A current source that has been turned off is represented by an open circuit. No current flows.

With the 2A current source turned off we have the resulting circuit. Looking carefully we notice that now the 6 ohm resistor is in series with the 2 ohm resistor. These two resistors are in parallel with the 4 ohm resistor. We can use the parallel function and voltage division to find the voltage across the (6+2)//4 part of the circuit.

Using the parallel function, the circuit is transformed into a simple voltage source with two resistors in series.
Using voltage division we find the voltage across the (6+2)//4 resistors of the circuit. Separately this is the voltage across the the 4 ohm resistor which is equal to the voltage across the 6+2 ohm combination.
Using voltage division again we find the voltage across the 6 ohm resistor alone from the single voltage source.

With the 6V voltage source turned off we have the resulting circuit. Looking carefully we notice that now the 2 ohm and 4 ohm resistors are in parallel. This combination in turn is in series with the other 2 ohm resistor. We can use the parallel function and current division to find the voltage across the 6 ohm resistor.

Using addition and the parallel function, the circuit is transformed into a simple current source with two resistors in parallel.
Using current division, we find the current across the 6 ohm resistors of the circuit.
Using Ohm's law, we find the voltage across the 6 ohm resistor alone from the single current source.
By the theory of superposition, The total voltage across the 6 ohm resistor is the sum of the voltage provided by the voltage source and the voltage provided by the current source.

Use superposition to find V_o in the circuit.
Irwin. page 206 prob 4.8

With the 2A current source turned off we have the resulting circuit. Looking carefully we notice that now the 6 ohm resistor is in series with the 12 ohm resistor. These two resistors are in parallel with the 4 ohm resistor. We can use the parallel function and voltage division to find the voltage across the (6+12)//4 part of the circuit.

Using the parallel function, the circuit is transformed into a simple voltage source with two resistors in series.
Using voltage division we find the voltage across the (6+12)//4 resistors of the circuit. Separately this is the voltage across the the 4 ohm resistor which is equal to the voltage across the 6+12 ohm combination.

With the 6V voltage source turned off we have the resulting circuit. This circuit is a good candidate for solving by using nodal analysis. At the node V_A The KCL equation written as the sum of the currents leaving the node = 0 is:

At the node V_O2 The KCL equation written as the sum of the currents leaving the node = 0 is:

Store the KCL equation for node V_A as the variable v1.
Store the KCL equation for node V_O2 as the variable v2.

With the custom menu, press [F1] [5] to call the simul function. Enter v1 and v2 as the equations. Enter va and vo2 as the variables. Press [ENTER].
To find the final voltage across the 4 ohm resistor simply add the contributing parts from the voltage source and the current source.

Use superposition to find V_o in the circuit.
DeCarlo. page 159 prob 5.11a

With the 26 amp source turned off we get the resulting circuit. This portion can easily be solved with loop analysis. For the top loop, the KVL equation is written as the sum of all the voltage drops equals zero.

For the bottom loop, the KVL equation is written as the sum of all the voltage drops equals zero.

Store the KVL equation for the top loop in the variable named v1.
Store the KVL equation for the bottom loop in the variable named v2.
Using the simul function, enter v1 and v2 as the equations, and solve for ix and iy.
vo1 from the independent voltage source is equal to the current (iy) times the resistance (1 ohm).

With the 52V voltage source turned off we have the resulting circuit. This circuit is an excellent candidate for solving by using nodal analysis. The only node equation that needs to be written is at V_o2. Pay special attention to the direction of the dependent voltage source. At the node V_o2, the KCL equation written as the sum of the currents leaving the node equals zero is:

With this circuit configuration i_x is easily defined by:

Evaluate the KCL equation for the node using the with operator to define ix.
Solve this equation for vo2.
The final voltage is the sum of the voltages provided by the independent voltage source and the independent current source.

Using source transformation find V_o.
Irwin. page 183 example 4.6

If we have embedded within a network a current source i in parallel with a resistor R (referred to as a Norton equivalent segment), we can replace this combination with a voltage source of value v = iR in series with the resistor R (referred to as a Thévenin equivalent segment). The reverse is also true; that is, a voltage source v in series with a resistor R (Thévenin equivalent segment) can be replaced with a current source value i = v/R
in parallel with the resistor R (Norton equivalent segment). To solve this problem we simply apply Ohm's law, addition, addition in parallel and finally current division. Starting from the left side of the circuit.

Transform the 12V source in series with the 3 ohm resistor to a 4A source in parallel with the 3 ohm resistor.
Combine the 3 ohm resistor in parallel with the 6 ohm resistor.
Transform the 4A source in parallel with the 2 ohm resistor to an 8V source in series with a 2 ohm resistor.
Add the two 2 ohm resistors in series together to make a 4 ohm resistor.

Transform the 8V source in series with the 4 ohm resistor to a 2A source in parallel with the 4 ohm resistor.
The two 2A sources can be added together to make one 4A source.
Transform the 4A source in parallel with the 4 ohm resistor to an 16V source in series with a 4 ohm resistor.

Add the two 4 ohm resistors in series together to make an 8 ohm resistor.
Finally by voltage division we can find the voltage across the 8 ohm resistor to be equal to eight volts.

As you can see by the TI-92 screen shots, the math involved in source transformation is somewhat trivial.

Also,you can't tell a whole lot about what is happening with the circuit. To help combat this and make the idea of source transformation easier to understand, I have written another program entitled ThevnNor.92g that is available in circuits.92g. The program name has been added to the custom menu for easy access.

This program is designed to handle more than just source transformation. However, it works great for it.After being prompted for the initial source and element, the program will prompt you for source and element values. Then the program will draw the circuit segments. When the circuit segments are drawn you can press enter to either add elements or sources, transform sources, start over, or quit. Take the time to sketch the circuit from the problem and follow along with the example.

Use Thévenin's theorem to find V_o in the network.
Irwin. page 207 prob 4.25

Thévenin's theorem tells us that we can replace the entire network, exclusive of the load, by an equivalent circuit that contains only an independent voltage source in series with a resistor in such a way that the current-voltage relationship at the load is unchanged. For this problem we say that the 8 ohm resistor as the load. Step one is to treat the load as an open circuit and find the voltage across the open circuit, V_oc.

Because of KVL around the loop on the left side, the voltage at v1 can be found by voltage division.
KCL at node v2 lets us find the voltage at v2.
The open circuit voltage is now simply v1-v2.

Step two is to treat the load as a short circuit and find the current across the short circuit. I_sc can be solved for easily with loop analysis. The equation for the first loop written as the sum of all the voltage drops equals zero is:

The equation for the loop around the current source written as the sum of all the voltage drops equals zero is:

The KCL equation for the node V₂ written as the sum of all the currents entering the node equals zero is:

Store the equation for the first loop in the variable named e1.
Store the equation for the loop around the current source in the variable named e2.
Store the KCL equation for the node V₂ in the variable named e3.

Use the simul function to solve for the currents. Enter {e1,e2,e3} for the list of equations. Enter {i1,i2,i3} as the list of variables.

For our problem, I_sc = i2

The technique for finding the Thevenin's resistance is to "turn off" all of the independent sources (Voltage sources are represented by a short; Current sources are represented by an open) and then find the equivalent resistance across the load. For this problem the 3 ohm and 6 ohm resistors are in parallel. They are in series with the 2 ohm resistor.

Using the parallel function and addition we find the Thevenin's resistance for the network.
In finding the Thevenin's resistance, the short circuit current, and the open circuit voltage you really only need to find two of the three as these three items obey Ohm's Law.

Knowing Voc and Rth we can now redraw the circuit as shown.

The voltage across the load resistor is now just a simple case of voltage division.

Use Thevenin's theorem to find V_o in the network.
Irwin. page 209 prob 4.47

Thevenin's theorem tells us that we can replace the entire network, exclusive of the load, by an equivalent circuit that contains only an independent voltage source in series with a resistor in such a way that the current-voltage relationship at the load is unchanged. For this problem we say that the 4 ohm resistor as the load. Step one is to treat the load as an open circuit and find the voltage across the open circuit, V_oc.

Using node analysis we find the equation for the sum of the currents leaving the node V₁ is:

The dependent current source is defined by the equation as:

The voltage at V_oc is easily defined as:

Input the equation for the currents at node V₁ and store in the variable named e1.
Set the equation for the definition of the dependent source equal to zero and store in the variable named e2.
Set the equation for the definition of V_oc equal to zero and store in the variable named e3.

Use the simul function with {e1,e2,e3} as the list of equations and {v1,ix,voc} as the list of variables.

The short circuit current can determined by using nodal analysis. Then after finding the voltage for V₂ divide the answer by 4 ohms to find I_sc. The KCL equation for the node V₁ as the sum of all the currents leaving the node is written as:

The controlled voltage source is still defined as:

The voltage at V₂ can be defined as:

Input the equation for the currents at node V₁ and store in the variable named e1.
Set the equation for the definition of the dependent source equal to zero and store in the variable named e2.
Set the equation for the definition of V₂ equal to zero and store in the variable named e3.

Use the simul function with {e1,e2,e3} as the list of equations and {v1,ix,v2} as the list of variables.
Divide V₂ by 4 ohms to get I_sc

The Thevenin's resistance for the circuit is now just the open circuit voltage divided by the short circuit current.

Now that we know the Thevenin's voltage and the Thevenin's resistance we can remodel the original circuit as shown.

Finding the voltage across the load resistor is now just a simple matter of voltage division.

Linear Circuit Analysis: Network Theorems

Linear Circuit Analysis:
Network Theorems