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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:
    Capacitance and Inductance

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

    Draw the waveform of the current in a 12 micro farad capacitor when the capacitor voltage is described by the piecewise curve shown to the right.
    Irwin. page 284 prob 6.12

    This curve for the voltage was of course graphed using the Piecewise program.

    • Press [Diamond] [WINDOW] to enter the graph range variable editing screen.
    • Enter 0 for xmin
    • Enter 16 for xmax
    • Enter 1 for xscl
    • Enter -12 for ymin
    • Enter 12 for ymax
    • Enter 2 for yscl
    • press [2nd] [QUIT] to return to the home screen.
    Assuming you have the Cust() and PceWise programs from circuits.92g, follow these steps to define the input voltage.
    • Press [2nd] [Custom] to activate the custom menu.
    • Press [F2] [ENTER] [ENTER] to start the PceWise program.
    • Since our function does not start until time equals zero, graph the line 0 on the interval from negative infinity to 0.
    • Press [Enter] twice to go to the next range.
    To reach the maximum amplitude of the period the slope equals +2 with a y intercept of zero.
    • Graph this part of the function from 0 to: 6.
    • Graph the line: 2x
    • Press [Enter] twice to go to the next range.
    To reach the minimum amplitude of the period the slope equals -5 with a y intercept of 42.
    • Graph this part of the function from 6 to: 10.
    • Graph the line: -5+42
    • Press [Enter] twice to go to the next range.
    To reach the axis in the period, the slope equals 4x/3 with a y intercept of -64/3.
    • Graph this part of the function from 10 to: 16.
    • Graph the line: 4x/3-64/3
    • Press [Enter] twice to go to the final range.
    To finish out this piecewise function we need to define the line from 2 to positive infinity.
    • Graph this part of the function from 2 to: infinity [2nd] [j].
    • Graph the line: 0
    • Press [Enter] twice to finish the program.
    The program has written the piecewise function and stored it in y1(x).
    • To view the graph press [Diamond] [GRAPH].
    • Press [F1] [9] to bring up the graph formats screen.
    • Select SIMUL for the Graph Order.
    • Select ON for the Grid.
    • Press [ENTER] to redraw the graph.

    The PceWise program, by default, always writes its functions into y1(x) with the variable name fun.

    Before we can graph the waveform of the current we have to know what to do with the voltage. The current through a capacitor in relation to the capacitance and the voltage is:

    Let's add this equation to the formula sheet for safe keeping.
    Since the voltage has been so well defined by the piecewise curve, all we need to do is to take the derivative of the voltage and multiply by the capacitance.
    • Press [DIAMOND] [Y=] to enter the y(x) editor.
    • At y2 enter the equation 12E-6*d(y1(x),x) and press [ENTER].
    • Push the up arrow to highlight the equation in y1 (the voltage), and deselect it by pressing [F4].
    • Press [F2] [A] (ZoomFit) to graph the current.

    As expected a constant slope from the graph of the voltage provides a constant value for the current.
    Given the current through a previously uncharged 4 micro-farad capacitor, graph the voltage, power, and energy of the capacitor.
    Irwin. page 258 example 6.4

    The graph of the current was created using the pcewise program.

    • From -infinity to: 0
      Graph the line:0.
    • From 0 to:2
      Graph the line:16E-6*x/2.
    • From 2 to:4
      Graph the line:-8E-6.
    • From 4 to:infinity
      Graph the line:0.
    • Set the graph window variables as shown.
    To graph the voltage, power, and energy, from the current we need the relationships.Given the current, the voltage can be found by:

    Then the power in watts can be found by:

    Finally the energy stored in joules is given by:

    These equations are added to the text session for easy retrieval. Since the current is defined in the y(x) editor as y1(x) we can use this as reference when graphing the voltage.
    • Press [DIAMOND] [Y=] to enter the y(x) editor.
    • Turn off y1(x) (the function for the current) by pressing [F4].
    • For y2(x) enter the formula for the voltage referencing y1(x) as the current.
    • Press [F2] [A] (ZoomFit) to see the graph of the voltage.
    After the graph is drawn you can use the Maximum function to find the height of this function by pressing [F5] [4].
    To graph the power we can use either the relationship involving voltage alone or we can use the relationship involving voltage and current. To test that these are in fact equal, they have both been placed in the y(x) editor as y3(x) and y4(x) respectively. y1(x) is still referenced as the current and y2(x)is referenced as the voltage. 4E-6 is the value of the capacitor.
    • After the functions have been entered, deselect the other functions by highlighting them and pressing [F4].
    • Press [F2] [A] (ZoomFit) to graph the power.
    • After the graph is drawn you can press [Diamond] [WINDOW] to find the maximum and minimum of the function.
    To graph the energy we use the relationship involving voltage. With y2(x) referenced as the voltage. 4E-6 is the value of the capacitor.
    • After the function has been entered, deselect the other functions by highlighting them and pressing [F4].
    • Press [F2] [A] (ZoomFit) to graph the energy.
    • After the graph is drawn you can press [Diamond] [WINDOW] to find the maximum and minimum of the function.


    How much charge is accumulated if 12 volts is applied to a 50 pf capacitor.
    Irwin. page 256 example 6.1

    This problem uses the relationship between voltage, charge, and capacitance.

    Q = C*V

    The formula is added to the text for easy retrieval. Since this is a simple multiplication problem the answer is:

    Q = 50E-12*12
    = 6E-10
    = 600 pico-coulombs


    Draw the waveform of the voltage in a 10 milli henry inductor when the inductor current is described by the piecewise curve shown to the right.
    Irwin. page 262 example 6.5

    This curve for the current was of course graphed using the Piecewise program.

    • Press [Diamond] [WINDOW] to enter the graph range variable editing screen.
    • Enter 0 for xmin
    • Enter 4 for xmax
    • Enter 1 for xscl
    • Enter 0 for ymin
    • Enter 20 for ymax
    • Enter 2 for yscl
    • press [2nd] [QUIT] to return to the home screen.
    Assuming you have the Cust() and PceWise program from circuits.92g, follow these steps to define the input voltage.
    • Press [2nd] [Custom] to activate the custom menu.
    • Press [F2] [ENTER] [ENTER] to start the PceWise program.
    • Since our function does not start until time equals zero, graph the line 0 on the interval from negative infinity to 0.
    • Press [Enter] twice to go to the next range.
    To reach the maximum amplitude of the period the slope equals +10 with a y intercept of zero.
    • Graph this part of the function from 0 to: 2.
    • Graph the line: 10x
    • Press [Enter] twice to go to the next range.
    To reach the axis in the period, the slope equals -10 with a y intercept of 40.
    • Graph this part of the function from 2 to: 4.
    • Graph the line: -10x+40
    • Press [Enter] twice to go to the final range.
    To finish out this piecewise function we need to define the line from 4 to positive infinity.
    • Graph this part of the function from 4 to: infinity [2nd] [j].
    • Graph the line: 0
    • Press [Enter] twice to finish the program.
    The program has written the piecewise function and stored it in y1(x).
    • To view the graph press [Diamond] [GRAPH].
    • Press [F1] [9] to bring up the graph formats screen.
    • Select SIMUL for the Graph Order.
    • Select ON for the Grid.
    • Press [ENTER] to redraw the graph.

    The PceWise program, by default, always writes its functions into y1(x) with the variable name fun.

    Before we can graph the waveform of the voltage we have to know what to do with the current. The voltage through an inductor in relation to the inductance and the current is:

    Let's add this equation to the formula sheet for safe keeping.
    Since the current has been so well defined by the piecewise curve, all we need to do is to take the derivative of the current and multiply by the inductance.
    • Press [Diamond] [Y=] to enter the y(x) editor.
    • At y2 enter the equation 10E-3*d(y1(x),x) and press [ENTER].
    • Push the up arrow to highlight the equation in y1 (the current), and deselect it by pressing [F4].
    • Press [F2] [A] (ZoomFit) to graph the current.

    As expected a constant slope from the graph of the current provides a constant value for the voltage. By pressing [F3] and tracing the function we see that the voltage is a square wave with maximum and minimum values of .1 volts


    The voltage across a 200 milli-henry inductor is given as:

    v(t) = (1-3t)e-3t mV t > 0
    = 0 t < 0

    Find and plot the waveforms for current, energy, and power.
    Irwin. page 264 example 6.7

    The graph of the current was created using the pcewise program.

    • From -infinity to: 0
      Graph the line:0.
    • From 0 to:infinity
      Graph the line:(1-3t)e-3t.
    • Set the graph window variables as shown.
    To graph the current, power, and energy, from the voltage we need the relationships.Given the voltage, the current can be found by:

    Then the power in watts can be found by:

    Finally the energy stored in joules is given by:

    These equations are added to the text session for easy retrieval. Since the current is defined in the y(x) editor as y1(x) we can use this as reference when graphing the voltage.
    • Press [Diamond] [Y=] to enter the y(x) editor.
    • Turn off y1(x) (the function for the voltage) by highlighting the function and pressing [F4].
    • For y2(x) enter the formula for the current referencing y1(x) as the voltage.
    • Press [F2] [A] (ZoomFit) to see the graph of the current.
    After the graph is drawn you can use the Maximum function to find the height of this function by pressing [F5] [4].
    To find the equation of the current press
    [Diamond] [Q] to go to the home screen.
    • Store the expression for the voltage in the variable named v(x).
    • Apply the formula for finding the current through an inductor given the voltage. Store this expression in the variable named i(x) for later use.
    • Evaluate i(x)
    .
    To graph the power we can use either the relationship involving current alone or we can use the relationship involving current and voltage. To test that these are in fact equal, they have both been placed in the y(x) editor as y3(x) and y4(x) respectively. y1(x) is still referenced as the voltage and y2(x) is referenced as the current. .2 = 200E-3 is the value of the inductor.
    • After the functions have been entered, deselect the other functions by highlighting them and pressing [F4].
    • Press [F2] [A] (ZoomFit) to graph the power.
    • After the graph is drawn you can press [F5] [4] to find the maximum of the function.
    To find the equation of the power press
    [Diamond] [Q] to go to the home screen.
    • Apply the formula for finding the power used by an inductor given the current.
    • Apply the formula for finding the power used by an inductor given the voltage and current.
    The value of the inductor = .2 = 200E-3
    Notice that these two expressions are exactly the same.
    To graph the energy we use the relationship involving current. With y2(x) referenced as the current. .2 = 200E-3 is the value of the inductor.
    • After the function has been entered, deselect the other functions by highlighting them and pressing [F4].
    • Press [F2] [A] (ZoomFit) to graph the energy.
    • After the graph is drawn you can press [F5] [4] to find the maximum of the function.
    To find the equation of the energy press
    [Diamond] [Q] to go to the home screen.
    • Apply the formula for finding the energy stored by an inductor given the current.
    The value of the inductor = .2 = 200E-3


    Find the expression for Vo of the RC Op-Amp circuit.
    Irwin. page 275 fig 6.19a

    Using the same technique that we used for the earlier op-amp circuits, we can find Vo. The KCL equation at the inverting terminal, written as the sum of the currents leaving the node set equal to zero is:


    Remember that for Ideal op-amps we say that
    V+ = V- = V

    Since V+ is grounded: V = 0
    i+ = i- = 0.
    • The KCL equation for the inverting terminal evaluated at V- = 0 and i- = 0. Now that the calculus features are involved in the equations, the variables v1 and vo are written as functions of t.
    • Using the TI-92's solve command we solve for vo.

    What is important to notice is that the output voltage is now the derivative of the input voltage multiplied by the constant R*C.


    Find the expression for Vo of the RC Op-Amp circuit.
    Irwin. page 275 fig 6.19b

    Using the same technique that we used for the earlier op-amp circuits, we can find Vo. The KCL equation at the inverting terminal, written as the sum of the currents leaving the node set equal to zero is:


    Remember that for Ideal op-amps we say that

    V+ = V- = V
    Since V+ is grounded: V = 0
    i+ = i- = 0.
    • The KCL equation for the inverting terminal evaluated at V- = 0 and i- = 0. Now that the calculus features are involved in the equations, the variables v1 and vo are written as functions of t.
    • Using the TI-92's solve command we solve for vo.
    • To get vo(t) by itself we must integrate both sides of the equation.

    What is important to notice is that the output voltage is now the integral of the input voltage divided by the constant R*C.


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