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Figure 521 Equivalentcircuit representation of rst-order circuits
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Transient Analysis
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Figure 522 A more involved RC circuit
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that the switch has been closed for a suf cient time (we shall assume so) Thus: vC (t) = V2 VC (0) = V2 t 0 (533)
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After the switch closes, the circuit on the left-hand side of Figure 522 must be accounted for Figure 523 depicts the new arrangement, in which we have moved the capacitor to the far right-hand side, in preparation for the evaluation of the equivalent circuit Using the Th venin-to-Norton source transformation technique e (introduced in 3), we next obtain the circuit at the top of Figure 524, which can be easily reduced by adding the two current sources and computing the equivalent parallel resistance of R1 , R2 , and R3 The last step illustrated in the gure is the conversion to Th venin form Figure 525 depicts the nal appearance e of the equivalent circuit for t 0
V1 V2 + R1 R2
+ RT = R1 R2 R3 R1 R3 RT V1 V2 R2 vC _ C
Figure 523 The circuit of Figure 645 for t 0
Now we are ready to write the differential equation for the equivalent circuit:
Figure 524 Reduction of the circuit of Figure 523 to Th venin equivalent form e
RT + VT
dvC 1 1 + vC = VT dt RT C RT C = RT C vC (0) = V2
t 0 (534)
The complete solution is then computed following the usual procedure, as shown below
VC _
vC (t) = Ke t/ + f (t) vC (0) = Ke0 + VT K = vC (0) VT = V2 VT vC (t) = (V2 VT )e t/RT C + VT The method illlustrated above is now applied to two examples (535)
Figure 525 The circuit of Figure 522 in equivalent form for t 0
Part I
Circuits
EXAMPLE 58 Turn-Off Transient of DC Motor
Problem
Determine the motor voltage for all time in the simpli ed electric motor circuit model shown in Figure 526 The motor is represented by the series RL circuit in the shaded box
RB + VB
t=0 ibt
+ Rm Lm
Solution
Known Quantities: Battery voltage, resistor, and inductor values Find: The voltage across the motor as a function of time
L = 3 H; VB = 100 V
Schematics, Diagrams, Circuits, and Given Data: RB = 2 Assumptions: The switch has been closed for a long time
; RS = 20
; Rm = 08
Analysis: With the switch closed for a long time, the inductor in the circuit of Figure
526 behaves like a short circuit The current through the motor can then be calculated by the current divider rule in the modi ed circuit of Figure 527, where the inductor has been replaced with a short circuit and the Th venin circuit on the left has been replaced by its e Norton equivalent: 1 1 100 VB Rm 08 im = = = 3472 1 2 1 1 RB 1 1 1 + + + + RB Rs Rm 2 20 08
VB RB
RB RS
+ Rm vm
This current is the initial condition for the inductor current: iL (0) = 3472 A Since the motor inductance is effectively a short circuit, the motor voltage for t < 0 is equal to vm (t) = im Rm = 278 V t <0
When the switch opens and the motor voltage supply is turned off, the motor sees only the shunt (parallel) resistance Rs , as depicted in Figure 528 Remember now that the inductor current cannot change instantaneously; thus, the motor (inductor) current, im , must continue to ow in the same direction Since all that is left is a series RL circuit, with resistance R = Rs + Rm = 208 , the inductor current will decay exponentially with time constant = L/R = 01442 s: iL (t) = im (t) = iL (o)e t/ = 347e t/01442 t >0
+ Rm vm Lm
The motor voltage is then computed by adding the voltage drop across the motor resistance and inductance: diL (t) vm (t) = Rm iL (t) + L dt = 08 347e t/01442 + 3 = 6941e t/01442 t >0 347 e t/01442 01442 t >0
The motor voltage is plotted in Figure 529
Comments: Notice how the motor voltage rapidly changes from the steady-state value of
278 V for t < 0 to a large negative value due to the turn-off transient This inductive kick is typical of RL circuits, and results from the fact that, although the inductor current cannot change instantaneously, the inductor voltage can and does, as it is proportional to
5
Transient Analysis
100 0 100 200 300 400 500 600 700 1
Motor voltage (V)
05
05 1 Time (s)
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