Solution:

← Back to Fundamentals of Electric Circuits Overview \alpha = R/(2L) = 30/6 = 5

and $\omega_0 = \frac{1}{\sqrt{3\frac{1}{27}}} = 3$ Since 5 is greater than 3, we have the overdamped case \_\_\_\_\_\_ *i*(*t*) = *C*

s_{1,2} = -5 \pm \sqrt{5^2 - 9} = -1, -9, \qquad v(0) = 0,

$v(\infty) = 6 \text{ V}, \qquad i(0) = 0$ *dv*(*t*) \_\_\_\_ *dt* , where where \n

v(t) = A_1 e^{-t} + A_2 e^{-9t} + 6

$$\n$$

v(0) = 0 = A_1 + A_2 + 6

$$\n$$

i(0) = 0 = C(-A_1 - 9A_2)

$$which yields *A*1 = −9*A*2. Substituting this into the above, we get 0 = 9*A*2 − *A*2 + 6, or *A*2 = 0.75 an*d A*1 = −6.75.$$

v(t) = (-6.75e^{-t} + 0.75e^{-9t} + 6)u(t) V

for all $0 < t < 2$ s. At *t* = 1 s, *v*(1) = −6.75*e*⁻¹ + 0.75*e*⁻⁹ + 6 = −2.483 + 0.0001 + 6 = 3.552 V. At *t* = 2 s, *v*(2) = −6.75*e* −2 + 0 + 6 = 5.086 V. Note that from 2 < *t* < 4 s, *V*Th = 0, which implies that *v*(∞) = 0. Therefore, *v*(*t*) = (*A*3*e*⁻(*t* − 2) + *A*4*e*⁻9(*t* − 2))*u*(*t* − 2) V. At *t* = 2 s, *A*3 + *A*4 = 5.086.

i(t) = (A_3e^{-(t-2)} - 9A_4e^{-9(t-2)})

$$\n$$

i(t) = \frac{(-A_3e^{-(t-2)} - 9A_4e^{-9(t-2)})}{27}

$$and$$

i(2) = \frac{(6.75e^{-2} - 6.75e^{-18})}{27} = 33.83 \text{ mA}

$$Therefore, −*A*3 − 9*A*4 = 0.9135. Combining the two equations, we get −*A*3 − 9(5.086 − *A*3) = 0.9135, which leads to *A*3 = 5.835 and *A*4 = −0.749.$$

v(t) = (5.835e^{-(t-2)} - 0.749e^{-9(t-2)}) u(t-2)

V At *t* = 3 s, *v*(3) = (2.147 − 0) = 2.147 V. At *t* = 4 s, *v*(4) = 0.7897 V. - 5. **Evaluate.** A check between the values calculated above and the plot shown in Figure 8.37 shows good agreement within the obvious level of accuracy. - 6. **Satisfactory?** Yes, we have agreement and the results can be presented as a solution to the problem. Practice Problem 8.12 Find *i*(*t*) using *PSpice* for 0 < *t* < 4 s if the pulse voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.38. # **Solution:** When the switch is in position *a*, the 6- Ω resistor is redundant. The schematic for this case is shown in Fig. 8.41(a). To ensure that current **Figure 8.41** For Example 8.13: (a) for dc analysis, (b) for transient analysis. *i*(*t*) enters pin 1, the inductor is rotated three times before it is placed in the circuit. The same applies for the capacitor. We insert pseudo -components VIEWPOINT and IPROBE to determine the initial capacitor voltage and initial inductor current. We carry out a dc *PSpice* analysis by select ing **Analysis/Simulate.** As shown in Fig. 8.41(a), we obtain the initial capacitor voltage as 0 V and the initial inductor current *i*(0) as 4 A from the dc analysis. These initial values will be used in the transient analysis. When the switch is moved to position *b*, the circuit becomes a sourcefree parallel *RLC* circuit with the schematic in Fig. 8.41(b). We set the initial condition IC = 0 for the capacitor and IC = 4 A for the inductor. A current marker is inserted at pin 1 of the inductor. We select **Analysis/ Setup/Transient** to open up the *Transient Analysis* dialog box and set *Final Time* to 3 s. After saving the schematic, we select **Analysis/ Transient**. Figure 8.42 sho ws the plot of *i*(*t*). The plot agrees with *i*(*t*) = 4.8*e*⁻*^t* − 0.8*e*⁻6*^t* A, which is the solution by hand calculation. Plot of *i*(*t*) for Example 8.13. Refer to the circuit in Fig. 8.21 (see Practice Prob. 8.7). Use *PSpice* to Practice Problem 8.13 obtain *v*(*t*) for 0 < *t* < 2. **Answer:** See Fig. 8.43.