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Solution:

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dv(0+)dt=iC(0+)C=20.1=20 V/s\frac{dv(0^{+})}{dt} = \frac{i_C(0^{+})}{C} = \frac{2}{0.1} = 20 \text{ V/s}

Similarly, since L di∕dt = vL, di∕dt = vL∕L. We now obtain vL by applying KVL to the loop in Fig. 8.3(b). The result is

−12+4i(0+)+vL(0+)+v(0+)=0-12 + 4i(0^+) + v_L(0^+) + v(0^+) = 0

or

vL(0+)=12−8−4=0v_L(0^+) = 12 - 8 - 4 = 0

Thus,

di(0+)dt=vL(0+)L=00.25=0 A/s\frac{di(0^{+})}{dt} = \frac{v_L(0^{+})}{L} = \frac{0}{0.25} = 0 \text{ A/s}

(c) For t > 0, the circuit undergoes transience. But as t → ∞, the circuit reaches steady state again. The inductor acts like a short circuit and the capacitor like an open circuit, so that the circuit in Fig. 8.3(b) becomes that shown in Fig. 8.3(c), from which we have

i(∞)=0 A,v(∞)=12 Vi(\infty) = 0 \text{ A}, \qquad v(\infty) = 12 \text{ V}

Practice Problem 8.1 The switch in Fig. 8.4 was open for a long time but closed at t = 0. Determine: (a) i(0+), v(0+), (b) di(0+)∕dt, dv(0+)∕dt, (c) i(∞), v(∞).

Answer: (a) 3.5 A, 7 V, (b) 87.5 A/s, 0 V/s, (c) 21 A, 42 V.

In the circuit of Fig. 8.5, calculate: (a) iL(0+), vC (0+), vR(0+), (b) diL(0+)∕dt, dvC(0+)∕dt, dvR(0+)∕dt, (c) iL(∞), vC(∞), vR(∞). Example 8.2

Solution:

(a) For t < 0, 3u(t) = 0. At t = 0−, since the circuit has reached steady state, the inductor can be replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig. 8.6(a). From this figure we obtain

iL(0−)=0i_L(0^-) = 0

, vR(0−)=0v_R(0^-) = 0 , vC(0−)=−20v_C(0^-) = -20 V (8.2.1)

Although the derivatives of these quantities at t = 0− are not required, it is evident that they are all zero, since the circuit has reached steady state and nothing changes.

Figure 8.6

The circuit in Fig. 8.5 for: (a) t = 0−, (b) t = 0+.

For t > 0, 3u(t) = 3, so that the circuit is no w equivalent to that in Fig. 8.6(b). Since the inductor current and capacitor v oltage cannot change abruptly,

iL(0+)=iL(0−)=0,i_L(0^+) = i_L(0^-) = 0,

vC(0+)=vC(0−)=−20v_C(0^+) = v_C(0^-) = -20 V (8.2.2)

Although the voltage across the 4-Ω resistor is not required, we will use it to apply KVL and KCL; let it be called vo. Applying KCL at node a in Fig. 8.6(b) gives

3=vR(0+)2+vo(0+)43 = \frac{v_R(0^+)}{2} + \frac{v_o(0^+)}{4}

(8.2.3)

Applying KVL to the middle mesh in Fig. 8.6(b) yields

−vR(0+)+vo(0+)+vC(0+)+20=0(8.2.4)-v_R(0^+) + v_o(0^+) + v_C(0^+) + 20 = 0 \tag{8.2.4}

Since vC (0+) = −20 V from Eq. (8.2.2), Eq. (8.2.4) implies that

vR(0+)=vo(0+)(8.2.5)v_R(0^+) = v_o(0^+) \tag{8.2.5}

From Eqs. (8.2.3) and (8.2.5), we obtain

vR(0+)=vo(0+)=4 Vv_R(0^+) = v_o(0^+) = 4 \text{ V}

(8.2.6)

(b) Since L diL∕dt = vL,

diL(0+)dt=vL(0+)L\frac{di_L(0^+)}{dt} = \frac{v_L(0^+)}{L}

But applying KVL to the right mesh in Fig. 8.6(b) gives

vL(0+)=vC(0+)+20=0v_L(0^+) = v_C(0^+) + 20 = 0

Hence,

diL(0+)dt=0(8.2.7)\frac{di_L(0^+)}{dt} = 0\tag{8.2.7}

Similarly, since C dvC∕dt = iC, then dvC∕dt = iC∕C. We apply KCL at node b in Fig. 8.6(b) to get iC:

vo(0+)4=iC(0+)+iL(0+)(8.2.8)\frac{v_o(0^+)}{4} = i_C(0^+) + i_L(0^+) \tag{8.2.8}

Since vo(0+) = 4 and iL(0+) = 0, iC(0+) = 4∕4 = 1 A. Then

dvC(0+)dt=iC(0+)C=10.5=2 V/s\frac{dv_C(0^+)}{dt} = \frac{i_C(0^+)}{C} = \frac{1}{0.5} = 2 \text{ V/s}

(8.2.9)

To get dvR(0+)∕dt, we apply KCL to node a and obtain

3=vR2+vo43 = \frac{v_R}{2} + \frac{v_o}{4}

Taking the derivative of each term and setting t = 0+ gives

0=2dvR(0+)dt+dvo(0+)dt0 = 2\frac{dv_R(0^+)}{dt} + \frac{dv_o(0^+)}{dt}

(8.2.10)

We also apply KVL to the middle mesh in Fig. 8.6(b) and obtain

−vR+vC+20+vo=0-v_R + v_C + 20 + v_o = 0

Again, taking the derivative of each term and setting t = 0+ yields

−dvR(0+)dt+dvC(0+)dt+dvo(0+)dt=0-\frac{dv_R(0^+)}{dt} + \frac{dv_C(0^+)}{dt} + \frac{dv_o(0^+)}{dt} = 0

Substituting for dvC (0+)∕dt = 2 gives

dvR(0+)dt=2+dvo(0+)dt\frac{dv_R(0^+)}{dt} = 2 + \frac{dv_o(0^+)}{dt}

(8.2.11)

From Eqs. (8.2.10) and (8.2.11), we get

dvR(0+)dt=23 V/s\frac{dv_R(0^+)}{dt} = \frac{2}{3} \text{ V/s}

We can find diR(0+)∕dt although it is not required. Since vR = 2iR,

diR(0+)dt=12dvR(0+)dt=1223=13 A/s\frac{di_R(0^+)}{dt} = \frac{1}{2}\frac{dv_R(0^+)}{dt} = \frac{1}{2}\frac{2}{3} = \frac{1}{3} \text{ A/s}

(c) As t → ∞, the circuit reaches steady state. We have the equivalent circuit in Fig. 8.6(a) except that the 3-A current source is now operative. By current division principle,

iL(∞)=22+4⋅3 A=1 Ai_L(\infty) = \frac{2}{2+4} \cdot 3 \text{ A} = 1 \text{ A}

\n

vR(∞)=42+4⋅3 A×2=4 V,vC(∞)=−20 Vv_R(\infty) = \frac{4}{2+4} \cdot 3 \text{ A} \times 2 = 4 \text{ V}, \qquad v_C(\infty) = -20 \text{ V}

\n(8.2.12)

For the circuit in Fig. 8.7, find: (a) iL(0+), vC(0+), vR(0+), Practice Problem 8.2 (b) diL(0+)∕dt, dvC(0+)∕dt, dvR(0+)∕dt, (c) iL(∞), vC(∞), vR(∞).

Answer: (a) −6 A, 0, 0, (b) 0, 20 V/s, 0, (c) −2 A, 20 V, 20 V.