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First-Order Circuits
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7.1 INTRODUCTION Whenever a circuit is switched from one condition to another, either by a change in the applied source or a change in the circuit elements, there is a transitional period during which the branch currents and element voltages change from their former values to new ones. This period is called the transient. After the transient has passed, the circuit is said to be in the steady state. Now, the linear di erential equation that describes the circuit will have two parts to its solution, the complementary function (or the homogeneous solution) and the particular solution. The complementary function corresponds to the transient, and the particular solution to the steady state. In this chapter we will nd the response of rst-order circuits, given various initial conditions and sources. We will then develop an intuitive approach which can lead us to the same response without going through the formal solution of di erential equations. We will also present and solve important issues relating to natural, force, step, and impulse responses, along with the dc steady state and the switching behavior of inductors and capacitors.
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Assume a capacitor has a voltage di erence V0 between its plates. When a conducting path R is provided, the stored charge travels through the capacitor from one plate to the other, establishing a current i. Thus, the capacitor voltage v is gradually reduced to zero, at which time the current also becomes zero. In the RC circuit of Fig. 7-1(a), Ri v and i C dv=dt. Eliminating i in both equations gives dv 1 v 0 dt RC 1
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The only function whose linear combination with its derivative can be zero is an exponential function of the form Aest . Replacing v by Aest and dv=dt by sAest in (1), we get   1 1 st st Ae A s sAe est 0 RC RC from which s 1 0 RC or s 1 RC (2)
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Given v 0 A V0 , v t and i t are found to be 127
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FIRST-ORDER CIRCUITS
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[CHAP. 7
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v t V0 e t=RC ; t>0 dv V0 t=RC e ; i t C dt R
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3 t>0 4
The voltage and current of the capacitor are exponentials with initial values of V0 and V0 =R, respectively. As time increases, voltage and current decrease to zero with a time constant of  RC. See Figs. 7-1(b) and (c).
EXAMPLE 7.1 The voltage across a 1-mF capacitor is 10 V for t < 0. At t 0, a 1-M
resistor is connected across the capacitor terminals. Find the time constant , the voltage v t , and its value at t 5 s.  RC 106 10 6 s 1 s v t 10e t V ; t > 0 v 5 10e 5 0:067 V
Fig. 7-1 EXAMPLE 7.2 A 5-mF capacitor with an initial voltage of 4 V is connected to a parallel combination of a 3-k
and a 6-k
resistor (Fig. 7-2). Find the current i in the 6-k
resistor.
Fig. 7-2
CHAP. 7]
FIRST-ORDER CIRCUITS
The equivalent resistance of the two parallel resistors is R 2 k
. The time constant of the circuit is RC 10 2 s. The voltage and current in the 6-k
resistor are, respectively, v 4e 100t V and i v=6000 0:67e 100t mA
ESTABLISHING A DC VOLTAGE ACROSS A CAPACITOR
Connect an initially uncharged capacitor to a battery with voltage V0 through a resistor at t 0. The circuit is shown in Fig. 7-3(a).
Fig. 7-3
For t > 0, KVL around the loop gives Ri v V0 which, after substituting i C dv=dt , becomes dv 1 1 v V dt RC RC 0 with the initial condition v 0 v 0 0 5b t>0 5a
The solution should satisfy both (5a) and (5b). The particular solution (or forced response) vp t V0 satis es (5a) but not (5b). The homogeneous solution (or natural response) vh t Ae t=RC can be added and its magnitude A can be adjusted so that the total solution (6a) satis es both (5a) and (5b). v t vp t vh t V0 Ae t=RC From the initial condition, v 0 V0 A 0 or A V0 . Thus the total solution is 6a
FIRST-ORDER CIRCUITS
[CHAP. 7
v t V0 1 e t=RC u t V i t 0 e t=RC u t R
[see Fig. 7-3 b [see Fig. 7-3 c
6b 6c
EXAMPLE 7.3 A 4-mF capacitor with an initial voltage of v 0 2 V is connected to a 12-V battery through a resistor R 5 k
at t 0. Find the voltage across and current through the capacitor for t > 0. The time constant of the circuit is  RC 0:02 s. Following the analysis of Example 7.2, we get v t 12 Ae 50t From the initial conditions, v 0 v 0 12 A 2 or A 10. v t 12 10e 50t V i t 12 v =5000 2 10 3 e 50t A 2e 50t mA The current may also be computed from i C dv=dt . And so the voltage increases exponentially from an initial value of 2 V to a nal value of 12 V, with a time constant of 20 ms, as shown in Fig. 7-4(a), while the current decreases from 2 mA to zero as shown in Fig. 7-4(b). Thus, for t > 0,
Fig. 7-4