Page 147 - Schaum's Outline of Theory and Problems of Electric Circuits

FIRST-ORDER CIRCUITS
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7.8 DC STEADY STATE IN INDUCTORS AND CAPACITORS [CHAP. 7

As noted in Section 7.1, the natural exponential component of the response of RL and RC circuits to
step inputs diminishes as time passes. At t ¼1, the circuit reaches steady state and the response is
made of the forced dc component only.
Theoretically, it should take an infinite amount of time for RL or RC circuits to reach dc steady
state. However, at t ¼ 5 , the transient component is reduced to 0.67 percent of its initial value. After
passage of 10 time constants the transient component equals to 0.0045 percent of its initial value, which
is less than 5 in 100,000, at which time for all practical purposes we may assume the steady state has been
reached.
At the dc steady state of RLC circuits, assuming no sustained oscillations exist in the circuit, all
currents and voltages in the circuit are constants. When the voltage across a capacitor is constant, the
current through it is zero. All capacitors, therefore, appear as open circuits in the dc steady state.
Similarly, when the current through an inductor is constant, the voltage across it is zero. All inductors
therefore appear as short circuits in the dc steady state. The circuit will be reduced to a dc-resistive case
from which voltages across capacitors and currents through inductors can be easily found, as all the
currents and voltages are constants and the analysis involves no differential equations.
The dc steady-state behavior presented in the preceding paragraph is valid for circuits containing
any number of inductors, capacitors, and dc sources.

EXAMPLE 7.7 Find the steady-state values of i L , v C1 , and v C2 in the circuit of Fig. 7-13(a).
When the steady state is reached, the circuit will be as shown in Fig. 7-13(b). The inductor current and
capacitor voltages are obtained by applying KCL at nodes A and B in Fig. 7-13(b). Thus,
v A v A v B v A þ 18 v B
Node A: þ þ ¼ 3 or 2v A v B ¼ 0
3 6 6
v B v B v A v B 18 v A
Node B: þ þ ¼ 0 or 4v A þ 5v B ¼ 36
12 6 6
Solving for v A and v B we find v A ¼ 6 V and v B ¼ 12 V. By inspection of Fig. 7-13(b), we have i L ¼ 2 mA, v C1 ¼ 8V,
and v C2 ¼ 6V.

EXAMPLE 7.8 Find i and v in the circuit of Fig. 7-14.
At t ¼ 0, the voltage across the capacitor is zero. Its final value is obtained from dc analysis to be 2 V. The
time constant of the circuit of Fig. 7-14, as derived in Example 7.6, is 0.034 s. Therefore,
v ¼ 2ð1 e 1000t=34 ÞuðtÞ ðVÞ
3
6
dv ð9 10 Þð2 10 Þ 1000t=34 1000t=34
i ¼ C ¼ e uðtÞ ðAÞ¼ 0:53e uðtÞ ðmAÞ
dt 34

7.9 TRANSITIONS AT SWITCHING TIME
A sudden switching of a source or a jump in its magnitude can translate into sudden jumps in
voltages or currents in a circuit. A jump in the capacitor voltage requires an impulse current. Simi-
larly, a jump in the inductor current requires an impulse voltage. If no such impulses can be present,
the capacitor voltages and the inductor currents remain continuous. Therefore, the post-switching
conditions of L and C can be derived from their pre-switching conditions.

EXAMPLE 7.9 In Fig. 7-15(a) the switch S is closed at t ¼ 0. Find i and v for all times.

At t ¼ 0 , the circuit is at steady state and the inductor functions as a short with vð0 Þ¼ 0 [see Fig. 7-15(b)].

The inductor current is then easily found to be ið0 Þ¼ 2 A. After S is closed at t ¼ 0, the circuit will be as shown in
Fig. 7-15(c). For t > 0, the current is exponential with a time constant of ¼ L=R ¼ 1=30 s, an initial value of

þ
ið0 Þ¼ ið0 Þ¼ 2 A, and a final value of 12=3 ¼ 4 A. The inductor’s voltage and current are