Linear Circuit Analysis:
First-Order Transient Circuits

For t < 0 the capacitor is considered to be fully charged.The capacitor appears to the rest of the citcuit as an open. Therefore, no current flows through the capacitor. The voltage across the capacitor for t < 0 can be found by using voltage division.

• Use the VoltDiv function to find the voltage at V_c(t) for t < 0.

• Enter the KCL equation for the node above the capacitor.
• Multiply by 10 and subtract by 5v(t)/3 to isolate the differential term.

• Press [2nd] [VAR-LINK] to enter the Var-Link screen.
• Press the down arrow until the program named setcustm is highlighted.
• Press [F1] [7] to unlock this program.

• Press [APPS] [7] [2] and open the setcustm program located in the adv folder.
• On line 7 of the program editor, in front of the word "None" type in "Circuits\Cust",
• Press [2nd] [QUIT] to return to the home screen.

• Press [APPS] [7] [2] and open the cust program located in the circuits folder.
• On line 3 from the bottom of the program editor, add the line
  Item "adv\SetCustm()"
• Press [2nd] [QUIT] to return to the home screen.

• Enter t for the independent variable.
• The switch is open at t = 0, so the initial value for the independent variable is 0.
• The dependent variable for this case is the voltage above the capacitor, v.
• The initial value that we found for the voltage before the switch is thrown is 8.
• The expression for dV(t)/dt was found to be -5v/3.
• Press [ENTER] twice.

• Press [ESC] to return to the home screen.
• Press [F1] [3] to bring the answer to the screen history.
• Use the Graph function to graph the answer and then ZoomBox to make the graph fill the screen.

• Using the Trace feature we see that the capacitor discharges fairly rapidly when the switch is opened. In less than three seconds the voltage stored by the capacitor has fallen from 8 volts (the initial voltage) down to .1 volts.

For t < 0 the inductor is considered to be fully charged.The inductor appears to the rest of the citcuit as a short. Therefore, no voltage is dropped across the inductor. The initial current through the inductor can be found by using superposition from the two seperate sources. Voltage division from each source. Then Ohm's law to transform to a current.

With the individual voltage sources turned off, one at a time, the resulting circuits are shown.

• The voltage V_o from the 12 volt source is found using voltage division and the parallel function.
• The voltage V_o from the 4 volt source is found using voltage division and the parallel function.
• Since the 4 volt source is reverse biased, the algebraic sum of the two sources involves subtraction.
• The initial current through the inductor is the same current that goes through the 2 ohm resistor. The current is found via Ohm's law.

• Enter the KVL equation set equal to 0
• Divide by 2, subtract 2i(t), and add 6 to solve for di/dt.
• Press [F2] [4] [ENTER] to change to the adv custom menu.
• Press [2nd] [CUSTOM] to bring up the custom menu.
• Press [F6] [2] [ENTER] to start the differential equation solver.

• Enter t (time) as the independent variable.
• Enter 0 as the initial value for time.
• Enter i (current) as the dependent variable.
• Enter 4/3 as the initial value for the current.
• Enter -2i + 6 as the expression for the derivative.

• Press [ESC] to return to the home screen.
• Press [F1] [3] to bring the answer to the screen history.
• With the expand feature we multiply the answer by 2 (ohms) to find the voltage. Use the Graph function to graph the answer and then ZoomBox to make the graph fill the screen.

Before the switch is closed we know that the initial current is zero. Also we know that the initial time is zero. Therefore we already have both of or initial conditions for the differential equation.

• With the circuits custom menu active press [F2] [3] [Enter] to start the ThevNor program.
• Select Voltage and Resistor as the initial source and elements and press [ENTER].
• Enter 15 for the source value and 150 for the resistor value.
• After the first screen is drawn, press [ENTER] and select 1 to transform to a Norton equivalent.

• Press [ENTER] and select 3 to add an element.
• Enter 300 for the resistor value.
• Press [ENTER] twice to draw the new circuit.

• Press [ENTER] and select 1 to transform back to a Thévenin circuit.

The sin(10⁴t) part of the voltage source was ignored for this part of the problem as it has no effect on source transformation when only resistors are involved. From the TI-92 screen we can see that we can replace everything to the left of the switch with the circuit segment shown.

• Enter the KVL equation set equal to zero.
• Isolate the derivative on the left side of the equation.

• With the adv custom menu in place, press[F6] [2] [ENTER] to start the HODE1IV program.
• Enter t as the independent variable.
• The switch is open at t = 0, so the initial value for the independent variable is 0.
• The dependent variable for this case is the current through the inductor, i.
• The initial value that we are told for the current before the switch is thrown is 0.
• The expression for di(t)/dt was found to be 1000*sin(10^4*t)-20000*i.
• Press [ENTER] twice.

• After the program returns the answer press[ESC] to return to the home screen.
• Recall the answer by pressing [F1] [3] [ENTER].
• Use the TI-92's Expand function to get the equation into a more friendly form.

• Use the graph function to graph the current.

The voltage from the source is also shown in the graph. The amplitude was multiplied by .005 to make it fit on the graph.( y1(x)=.005*10sin(10^4*x) )The graph shows how the inductor forces the current to lead the voltage of the source.