ssrs barcodelib For the circuit of Fig. 7-33, obtain the current iL , for all values of t. in Software

Make QR Code JIS X 0510 in Software For the circuit of Fig. 7-33, obtain the current iL , for all values of t.

For the circuit of Fig. 7-33, obtain the current iL , for all values of t.
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Fig. 7-33 For t < 0, the 50-V source results in inductor current 50=20 2:5 A. The 5-A current source is applied for t > 0. As t ! 1, this current divides equally between the two 10-
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resistors, whence iL 1 2:5 A. The time constant of the circuit is  0:2 10 3 H 1 ms 20
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and so, with t in ms and using iL 0 iL 0 2:5 A, iL iL 0 iL 1 e t= iL 1 5:0e 100t 2:5 A
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Finally, using unit step functions to combine the expressions for t < 0 and t > 0, iL 2:5u t 5:0e 100t 2:5 u t A
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The switch in Fig. 7-34 has been in position 1 for a long time; it is moved to 2 at t 0. the expression for i, for t > 0.
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With the switch on 1, i 0 50=40 1:25 A. With an inductance in the circuit, i 0 i 0 . after the switch has been moved to 2, i 1 10=40 0:25 A. In the above notation, B i 1 0:25 A and the time constant is  L=R 1=2000 s. A i 0 B 1:00 A Then, for t > 0, A
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i 1:00e 2000t 0:25
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CHAP. 7]
FIRST-ORDER CIRCUITS
Fig. 7-34
Fig. 7-35
The switch in the circuit shown in Fig. 7-35 is moved from 1 to 2 at t 0. t > 0.
Find vC and vR , for
With the switch on 1, the 100-V source results in vC 0 100 V; and, by continuity of charge, vC 0 vC 0 . In position 2, with the 50-V source of opposite polarity, vC 1 50 V. Thus, B vC 1 50 V A vC 0 B 150 V 1 s 200
 RC and Finally, KVL gives vR vC 50 0, or
vC 150e 200t 50 V
vR 150e 200t
Obtain the energy functions for the circuit of Problem 7.22.
wC 1 Cv2 1:25 3e 200t 1 2 mJ C 2 t 2 vR dt 11:25 1 e 400t mJ wR 0 R
A series RC circuit, with R 5 k
and C 20 mF, has two voltage sources in series, v1 25u t V v2 25u t t 0 V Obtain the complete expression for the voltage across the capacitor and make a sketch, if t 0 is positive.
The capacitor voltage is continuous. For t 0, v1 results in a capacitor voltage of 25 V. For 0 t t 0 , both sources are zero, so that vC decays exponentially from 25 V towards zero: vC 25e t=RC 25e 10t
t 0
In particular, vC t 0 25e 10t (V). For t ! t 0 , vC builds from vC t 0 towards the nal value 25 V established by v2 : vC vC t 0 vC 1 e t t =RC vC 1 25 1 e10t 1 e 10t Thus, for all t, vC 25u t 25e 10t u t u t t 0 25 1 e10t 1 e 10t u t t 0 V See Fig. 7-36.
0 0 0
t ! t 0
FIRST-ORDER CIRCUITS
[CHAP. 7
Fig. 7-36
Supplementary Problems
7.25 The capacitor in the circuit shown in Fig. 7-37 has initial charge Q0 800 mC, with polarity as indicated. the switch is closed at t 0, obtain the current and charge, for t > 0. Ans: i 10e 25 000t A ; q 4 10 4 1 e 25 000t C If
Fig. 7-38 Fig. 7-37
A 2-mF capacitor, with initial charge Q0 100 mC, is connected across a 100-
resistor at t 0. Calculate the time in which the transient voltage across the resistor drops from 40 to 10 volts. Ans: 0:277 ms In the RC circuit shown in Fig. 7-38, the switch is closed on position 1 at t 0 and then moved to 2 after the passage of one time constant. Obtain the current transient for (a) 0 < t < ; b t > . Ans: a 0:5e 200t A ; b 0:516e 200 t  (A) A 10-mF capacitor, with initial charge Q0 , is connected across a resistor at t 0. Given that the power transient for the capacitor is 800e 4000t (W), nd R, Q0 , and the initial stored energy in the capacitor. Ans: 50
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