Example 1: Figure 8-N1a shows a plot of the voltage across the inductor in Figure 8-N1b. a) Determine the equation that represents the inductor voltage as a function of time. b) Determine the value of the resistance R. c) Determine the equation that represents the inductor current as a function of time. Figure 8-N1 Solution: Part a) The inductor voltage is represented by an equation of the form ()for 0for 0atDtvtEFe t−<⎧=⎨+≥⎩ where D, E, F and a are unknown constants. The constants D, E and F are described by () when 0Dvt t=< , ()limtEvt→∞= , ()0limtEFv→++= t From the plot, we see that D = 0, E = 0, and E + F = 4 V Consequently, ()0 for 04 for 0attvtet−<⎧=⎨≥⎩ To determine the value of a , we pick a time when the circuit is not at steady state. One such point is labeled on the plot in Figure 8-N1. We see ()0.14 2 Vv = , that is, the value of the voltage is 2 volts at time 0.14 seconds. Substituting these into the equation for gives ()vt()()0.14ln 0.524 50.14aea−=⇒=−= Consequently ()50 for 04 for 0ttvtet−<⎧=⎨≥⎩ Part b) Figure 8-N2a shows the circuit immediately after the switch opens. In Figure 8-N2b, the part of the circuit connected to the inductor has been replaced by its Thevenin equivalent circuit. The time constant of the circuit is given by 45tLRRτ==+ Also, the time constant is related to the exponent in ()vt by 5ttτ−=− . Consequently 1551 4RRτ5+== ⇒ = Ω Figure 8-N2 Part c) The inductor current is related to the inductor voltage by () () ()010tit v d iLττ=+∫Figure 8-N3 Figure 8-N3 show the circuit before the switch opens. The closed switch is represented by a short circuit. The circuit is at steady state and the voltage sources have constant voltages so the inductor acts like a short circuit. The inductor current is given by ()60.4 A15it== In particular, . The current in an inductor is continuous, so . Consequently, ()0 0.4 Ai −=()(00ii+= −)()0 0.4 Ai = Returning to the equation for the inductor current, we have ()()055114 0.4 1 0.4 0.6 0.245tttit e d e eττ−−=+=−+=−−∫5− In summary, ()50.4 for 00.6 0.2 for 0ttitet−<⎧=⎨−≥⎩Example 2: Figure 8-N4a shows a plot of ()vt, the voltage across the 24 kΩ resistor in Figure 8-N4b. a) Determine the equation that represents ()vt as a function of time. b) Determine the value of the capacitance C. Figure 8-N4 Part a) The voltage is represented by an equation of the form ()for 0for 0atDtvtEFe t−<⎧=⎨+≥⎩ where D, E, F and a are unknown constants. The constants D, E and F are described by () when 0Dvt t=< , ()limtEv→∞= t, ()0limtEF vt→++= From the plot, we see that D = 6, E = 3.6, and E + F = 6 V Consequently, ()6for3.6 24 for 0attvtet− 0<⎧=⎨+≥⎩ To determine the value of a , we pick a time when the circuit is not at steady state. One such point is labeled on the plot in Figure 8-N4. We see ()0.00247 4 Vv =, that is, the value of thevoltage is 2 volts at time 0.00247 seconds or 2.47 ms. Substituting these into the equation for gives ()vt ()()0.00247ln 0.16674 3.6 2.4 7250.00247aea−=+ ⇒= =− Consequently ()7256for3.6 2.4 for 0ttvtet− 0<⎧=⎨+≥⎩ Part b) Figure 8-N5a shows the circuit immediately after the switch closes. In Figure 8-N5b, the part of the circuit connected to the capacitor has been replaced by its Thevenin equivalent circuit. The time constant of the circuit is given by ()311 10tRCCτ==× Also, the time constant is related to the exponent in ()vt by 725ttτ−=− . Consequently ()()3931111 10 125 10 125 nF725725 11 10CC−=× ⇒ = = × =× Figure 8-N5Example 3: Figure 8-N6a shows a plot of ()vt, the voltage across one of the 5 Ω resistors in Figure 8-N6b. c) Determine the equation that represents ()vt as a function of time. d) Determine the value of the capacitance C. Figure 8-N6 Part a) The voltage is represented by an equation of the form ()for 0for 0atDtvtEFe t−<⎧=⎨+≥⎩ where D, E, F and a are unknown constants. The constants D, E and F are described by () when 0Dvt t=< , ()limtEv→∞= t, ()0limtEF vt→++= From the plot, we see that D = 0, E = 1.2, and E + F = 6 V Consequently, ()0for1.2 4.8 for 0attvtet− 0<⎧=⎨+≥⎩ To determine the value of a , we pick a time when the circuit is not at steady state. One such point is labeled on the plot in Figure 8-N6. We see ()0.72 2 Vv = , that is, the value of the voltage is 2 volts at time 0.7.2 seconds. Substituting these into the equation for gives ()vt()()0.72ln 0.16672 1.2 4.8 2.50.72aea−=+ ⇒ = =− Consequently ()2.50for1.2 4.8 for 0ttvtet− 0<⎧=⎨+≥⎩ Part b) Figure 8-N7a shows the circuit immediately after time t = 0. In Figure 8-N7b, the part of the circuit connected to the capacitor has been replaced by its Thevenin equivalent circuit. The time constant of the circuit is given by 8tRCCτ== Also, the time constant is related to the exponent in ()vt by 725ttτ−=−. Consequently ()1118 0.05 F2.5 2.5 8 20CC=⇒= == Figure