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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:
    Series and Parallel Circuits - Kirchoff's Laws

    ------------

    Find the equivalent resistance.
    Irwin. page 89 prob 2.28

    There are several ways to find the equivalent resistance. However the most general form follows the equation:


    Where Ri is a list of resistive devices.
    Since we will need to find the equivalent resistance on several occasions, this will be an appropriate function to have on the TI-92.
    From the home screen:

    • Press [APPS] [7] [3].
    • From the Program, Function choice, select Function.
    • Press the down arrow twice and enter the name of the function (parallel).
    • Press [ENTER] twice.
    • Input the function parameter (a), and the code for the function ( (sum(a^(-1)))^-1 ).
    • Press [2nd] [QUIT] to exit the program editor.
    The parameter passed here is a list. This is done so that we can find the equivalent resistance of any number of resistors.
    • Press [Diamond] [ENTER] to return the decimal representation.

    Resistors in series add in series. Resistors in parallel add in parallel. Starting from the right of the circuit and working to the left, we find the equivalent resistance.
    Also shown is the equivalent resistance can be found in one single line.


    Given the network, find the current through each resistor.
    Irwin. page 47 example 2.17

    The concept of current division is expressed here. The current through each resistive element can be expressed as:



    Where ITotal is the total current, R1 is the resistor for which we wish to find the current through, and R2 is the equivalent resistance of every other element affecting the current flow.
    Current division is another formula to add to the Text file and good place for a user defined function. From the home screen:
    • Press [APPS] [7] [3].
    • From the Program, Function choice, select Function.
    • Press the down arrow twice and enter the name of the function (currdiv).
    • Press [ENTER] twice.
    • Input the function parameters (i,s1,s2), and the code for the function ( i*s2/(s1+s2) ).
    • Press [2nd] [QUIT] to exit the program editor.
    Finding the currents is now just a simple issue of entering the parameters in the function.
    • The current through the 60 Ohm resistor is .6 Amps.
    • The current through the 120 Ohm resistor is .9 Amps.

    Find Ix in the network.
    Irwin. page 87 prob 2.13

    This problem is used to show the idea behind Kirchoff's Current Law. (KCL) This law simply states that all of the current coming into a node must equal the current leaving that node.


    Looking at the top node of the circuit, the currents come into the node from R2 and the independent current source. The sum of the currents coming in is:
    .
    The currents leave the node from R1 and the dependant current source. The sum of the currents leaving the node is:
    .
    • From the Home Screen press [F2] [ENTER] (solve).
    • Enter the equation by pressing [6] [+] [i] [x] [=] [1] [+] [4] [i] [x].
    • Solve for Ix by pressing [,] [i] [x] [)] [ENTER].
    Also shown is how the equation can be entered and solved for manually.


    What is the voltage dropped by each resistor?
    Irwin. page 39 example 2.13

    The concept of voltage division is expressed here. The voltage across each resistive element can be expressed as:


    Where VTotal is the total voltage, R1 is the resistor for which we wish to find the voltage across, and R2 is the equivalent resistance of every other element affecting the voltage.
    Voltage division is another formula to add to the Text file and good place for a user defined function. From the home screen:
    • Press [APPS] [7] [3].
    • From the Program, Function choice, select Function.
    • Press the down arrow twice and enter the name of the function (voltdiv).
    • Press [ENTER] twice.
    • Input the function parameters (v,s1,s2), and the code for the function ( v*s1/(s1+s2) ).
    • Press [2nd] [QUIT] to exit the program editor.
    Finding the voltages is now just a simple issue of entering the parameters in the function.
    • The voltage through the 90 Ohm resistor is 6.75 Volts.
    • The voltage through the 30 Ohm resistor is 2.25 Volts.


    Find Vx in the circuit.
    Irwin. page 87 prob 2.17

    This exercise is used to demonstrate Kirchoff's Voltage law.(KVL) Kirchoff's second law states that the algebraic sum of the voltages around any loop is zero.

    Starting at node a, we simply add the voltages algebraically for each element around the loop.
    Vab = -1 volt, Vbc = -Vx volts, Vcd = 12 volts,
    Vde = -1 volt, and Vea = -4Vx volts.
    We can use the zeros function on the TI-92 to solve for Vx. Also shown is hoe the equation can be solved for manually.


    Determine the expression for the gain, (The ratio of the output voltage to the input voltage).
    Irwin. page 66 exampel 2.28

    Though this circuit may look complicated, we can solve this problem using the ideas covered previously.

    To simplify the circuit, first lets replace R3,R4,and R5 with a single resistor RL such that:
    • By applying KVL to loop one we can write the first equation for the voltage for Vi.
    • By observing Ohm's law we can write the equation for the voltage across R2.
    • Solve the equation for Vi, for I1. now we have I1 in terms of known voltages and resistors.
    • Rewrite Vg with the new value for I1.
    • By applying KCL at node 4 we find an expression for the output voltage Vo.
    • We have an expression for Vg that we can input to the equation for Vo.
    • Now the gain, Which is the ratio of Vo to Vi can be found by dividing both sides of the equation by Vi.
    Reasonable values for this type of circuit are:
    • R1 = 100 ohms.
    • R2 = 1 kohms.
    • gm = 0.04 S.
    • R3 = 50 kohms.
    • R4 = R5 = 10 kohms.
    • Find the value for RL by using the parallel function and the known values for R3, R4, and R5.
    • Solve for the numerical value for the gain by in putting the value for RL, the other resistors and the gain from the dependant current source.
    From this we find the magnitude of the gain is 165.29 .

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