Critically Damped Case (α = ω0)

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$$v(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}$$

(8.33)

Critically Damped Case (α = ω0)

For α = ω0, L = 4R² C. The roots are real and equal so that the response is

$$v(t) = (A_1 + A_2 t)e^{-\alpha t}$$

(8.34)

Underdamped Case (α < ω0)

When α < ω0, L < 4R² C. In this case the roots are complex and may be expressed as

$$s_{1,2} = -\alpha \pm j\omega_d \tag{8.35}$$

where

$$\omega_d = \sqrt{\omega_0^2 - \alpha^2} \tag{8.36}$$

The response is

$$v(t) = e^{-at}(A_1 \cos \omega_d t + A_2 \sin \omega_d t)$$

(8.37)

The constants A1 and A2 in each case can be determined from the initial conditions. We need v(0) and dv(0)∕dt. The first term is known from Eq. (8.27b). We find the second term by combining Eqs. (8.27) and (8.28), as

dv(0) _____ dt ⁼⁰

• I0 + C

___ V0 R

or

$$\frac{dv(0)}{dt} = -\frac{(V_0 + RI_0)}{RC}$$

(8.38)

The voltage waveforms are similar to those sho wn in Fig. 8.9 and will depend on whether the circuit is overdamped, underdamped, or critically damped.

Having found the capacitor voltage v(t) for the parallel RLC circuit as shown above, we can readily obtain other circuit quantities such as indi vidual element currents. F or e xample, the resistor current is iR = v∕R and the capacitor current is iC = C dv∕dt. We have selected the capacitor voltage v(t) as the key variable to be determined first in order to take advantage of Eq. (8.1a). Notice that we first found the inductor current i(t) for the RLC series circuit, whereas we first found the capacitor voltage v(t) for the parallel RLC circuit.

Example 8.5

In the parallel circuit of Fig. 8.13, find v(t) for t > 0, assuming v(0) = 5 V, i(0) = 0, L = 1 H, and C = 10 mF. Consider these cases: R = 1.923 Ω, R = 5 Ω, and R = 6.25 Ω.

Solution:

Solution:
\nCASE 1 If

$$R = 1.923 \Omega$$

,
\n

$$\alpha = \frac{1}{2RC} = \frac{1}{2 \times 1.923 \times 10 \times 10^{-3}} = 26$$

\n

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 10 \times 10^{-3}}} = 10$$

Since α > ω0 in this case, the response is overdamped. The roots of the characteristic equation are

$$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -2, -50$$

and the corresponding response is

$$v(t) = A_1 e^{-2t} + A_2 e^{-50t}$$

(8.5.1)

We now apply the initial conditions to get A1 and A2.

$$v(0) = 5 = A_1 + A_2 \tag{8.5.2}$$ $$v(0) = 5 = A_1 + A_2$$ $$\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{1.923 \times 10 \times 10^{-3}} = -260$$

But differentiating Eq. (8.5.1),

$$\frac{dv}{dt} = -2A_1e^{-2t} - 50A_2e^{-50t}$$ $$At t = 0,$$ $$-260 = -2A_1 - 50A_2 \tag{8.5.3}$$

From Eqs. (8.5.2) and (8.5.3), we obtain A1 = −0.2083 and A2 = 5.208. Substituting A1 and A2 in Eq. (8.5.1) yields

$$v(t) = -0.2083e^{-2t} + 5.208e^{-50t}$$

(8.5.4)

■ CASE 2 When R = 5 Ω,

en

$$R = 5 \Omega$$

,
\n

$$\alpha = \frac{1}{2RC} = \frac{1}{2 \times 5 \times 10 \times 10^{-3}} = 10$$

while ω0 = 10 remains the same. Since α = ω0 = 10, the response is critically damped. Hence, s1 = s2 = −10, and

$$v(t) = (A_1 + A_2 t)e^{-10t}$$

\n(8.5.5)

To get A1 and A2, we apply the initial conditions

$$v(0) = 5 = A_1 \tag{8.5.6}$$ $$v(0) = 5 = A_1$$ $$\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{5 \times 10 \times 10^{-3}} = -100$$

But differentiating Eq. (8.5.5),

$$\frac{dv}{dt} = (-10A_1 - 10A_2t + A_2)e^{-10t}$$

At t = 0,

$$-100 = -10A_1 + A_2 \tag{8.5.7}$$

From Eqs. (8.5.6) and (8.5.7), A1 = 5 and A2 = −50. Thus,

$$v(t) = (5 - 50t)e^{-10t} \text{ V}$$

(8.5.8)

■ CASE 3 When R = 6.25 Ω,

$$\text{hen } R = 6.25 \, \Omega,$$

\n

$$\alpha = \frac{1}{2RC} = \frac{1}{2 \times 6.25 \times 10 \times 10^{-3}} = 8$$

while ω0 = 10 remains the same. As α < ω0 in this case, the response is underdamped. The roots of the characteristic equation are

$$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -8 \pm j6$$

Hence,

$$v(t) = (A_1 \cos 6t + A_2 \sin 6t)e^{-8t}$$

(8.5.9)

We now obtain A1 and A2, as

$$v(0) = 5 = A_1 \tag{8.5.10}$$ $$\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{6.25 \times 10 \times 10^{-3}} = -80$$

But differentiating Eq. (8.5.9),

$$\frac{dv}{dt} = (-8A_1 \cos 6t - 8A_2 \sin 6t - 6A_1 \sin 6t + 6A_2 \cos 6t)e^{-8t}$$

At $t = 0$ ,

$$-80 = -8A_1 + 6A_2$$

(8.5.11)

From Eqs. (8.5.10) and (8.5.11), A1 = 5 and A2 = −6.667. Thus,

$$v(t) = (5 \cos 6t - 6.667 \sin 6t)e^{-8t}$$

(8.5.12)

Notice that by increasing the value of R, the degree of damping decreases and the responses differ. Figure 8.14 plots the three cases.

For Example 8.5: responses for three degrees of damping.

Practice Problem 8.5

In Fig. 8.13, let R = 2 Ω, L = 0.4 H, C = 25 mF, v(0) = 0, i(0) = 50 mA. Find v(t) for t > 0.

Answer: −2te⁻10*^t* V.

Example 8.6 Find v(t) for t > 0 in the RLC circuit of Fig. 8.15.

Solution:

When t < 0, the switch is open; the inductor acts like a short circuit while the capacitor behaves like an open circuit. The initial voltage across the capacitor is the same as the voltage across the 50-Ω resistor; that is,

$$v(0) = \frac{50}{30 + 50}(40) = \frac{5}{8} \times 40 = 25 \text{ V}$$

(8.6.1)

The initial current through the inductor is

$$i(0) = -\frac{40}{30 + 50} = -0.5 \text{ A}$$

The direction of i is as indicated in Fig. 8.15 to conform with the direc tion of I0 in Fig. 8.13, which is in agreement with the convention t hat current flows into the positive terminal of an inductor (see Fig. 6.23). We need to express this in terms of dv∕dt, since we are looking for v.

need to express this in terms of

$$dv/dt$$

, since we are looking for v.
\n

$$\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{25 - 50 \times 0.5}{50 \times 20 \times 10^{-6}} = 0$$

\n(8.6.2)

When t > 0, the switch is closed. The voltage source along with the 30-Ω resistor is separated from the rest of the circuit. The parallel RLC circuit acts independently of the voltage source, as illustrated in Fig. 8.16. Next, we determine that the roots of the characteristic equation are

circuit acts independently of the voltage source, as illustrated in
\nNext, we determine that the roots of the characteristic equation
\n

$$\alpha = \frac{1}{2RC} = \frac{1}{2 \times 50 \times 20 \times 10^{-6}} = 500$$

\n

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{0.4 \times 20 \times 10^{-6}} = 354$$

\n

$$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$$

\n

$$= -500 \pm \sqrt{250,000 - 124,997.6} = -500 \pm 354$$

or

$$s_1 = -854, \qquad s_2 = -146$$

The circuit in Fig. 8.15 when t > 0. The parallel RLC circuit on the right-hand side acts independently of the circuit on the left-hand side of the junction.

Since α > ω0, we have the overdamped response

$$v(t) = A_1 e^{-854t} + A_2 e^{-146t}$$

(8.6.3)

At t = 0, we impose the condition in Eq. (8.6.1),

$$v(0) = 25 = A_1 + A_2$$

$\Rightarrow$ $A_2 = 25 - A_1$ (8.6.4)

Taking the derivative of v(t) in Eq. (8.6.3),

$$\frac{dv}{dt} = -854A_1e^{-854t} - 146A_2e^{-146t}$$

Imposing the condition in Eq. (8.6.2),

$$\frac{dv(0)}{dt} = 0 = -854A_1 - 146A_2$$

or

$$0 = 854A_1 + 146A_2 \tag{8.6.5}$$

Solving Eqs. (8.6.4) and (8.6.5) gives

$$A_1 = -5.156, \qquad A_2 = 30.16$$

Thus, the complete solution in Eq. (8.6.3) becomes

$$v(t) = -5.156e^{-854t} + 30.16e^{-146t}$$

V

Refer to the circuit in Fig. 8.17. Find v(t) for t > 0.

Answer: 50(e⁻10*^t* − e⁻2.5*^t* ) V.