Page 156 - Schaum's Outline of Theory and Problems of Electric Circuits

FIRST-ORDER CIRCUITS
CHAP. 7]
R 2 t=ðR 1 CÞ 145
v 2 ðtÞ¼ e uðtÞ
R 1
Alternate Approach
The unit step response may also be found by the Laplace transform method (see Chapter 16).

EXAMPLE 7.17 Passive phase shifter. Find the relationship between v 2 and v 1 in the circuit of Fig. 7-45(a).
Let node D be the reference node. Apply KCL at nodes A and B to find
ðv A v 1 Þ
dv A
KCL at node A: C þ ¼ 0
dt R
dðv B v 1 Þ v B
KCL at node B: C þ ¼ 0
dt R
Subtracting the second equation from the first and noting that v 2 ¼ v A v B we get
dv 2 dv 1
v 2 þ RC ¼ v 1 RC
dt dt
EXAMPLE 7.18 Active phase shifter. Show that the relationship between v 2 and v 1 in the circuit of Fig. 7-45(b)is
the same as in Fig. 7-45(a).
Apply KCL at the inverting (node A) and non-inverting (node B) inputs of the op amp.
ðv A v 1 Þ ðv A v 2 Þ
KCL at node A: þ ¼ 0
R 1 R 1
ðv B v 1 Þ dv B
KCL at node B: þ C ¼ 0
R dt
From the op amp we have v A ¼ v B and from the KCL equation for node A, we have v A ¼ðv 1 þ v 2 Þ=2. Substituting
the preceding values in the KCL at node B, we find

dv 2 dv 1
v 2 þ RC ¼ v 1 RC
dt dt

Solved Problems

7.1 At t ¼ 0 , just before the switch is closed in Fig. 7-20, v ¼ 100 V. Obtain the current and
C
charge transients.

Fig. 7-20

1
With the polarities as indicated on the diagram, v R ¼ v C for t > 0, and 1=RC ¼ 62:5s . Also,

þ
v C ð0 Þ¼ v C ð0 Þ¼ 100 V. Thus,
v R ¼ v C ¼ 100e 62:5t ðVÞ i ¼ v R ¼ 0:25e 62:5t ðAÞ q ¼ Cv C ¼ 4000e 62:5t ðmCÞ
R