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2t 25. The voltage across an inductor is given by: v(t)=0 for t ≤0; v(t) = 20 te Obtain the expression for current in the inductor. [Hint. Solve this problem in the same way as shown in Example 1.86.] 26. Find equivalent capacitance of the combination shown in Fig. P1.16 and Fig. P1.17. 6 μF HH 2 μF: 4 μF 4 µF 6 µF Fig. P1.16 (4+4) x6 8+6 + 2 x6 oa 8=2= 3C for t≥ 0. 2C 4H C 10 C 2 12 C HH ↑ Fig. P1.17 = 2.85 µF. In Fig. P1.17, Cab = 14C] IniH bo

2t 25. The voltage across an inductor is given by: v(t)=0 for t ≤0; v(t) = 20 te Obtain the expression for current in the inductor. [Hint. Solve this problem in the same way as shown in Example 1.86.] 26. Find equivalent capacitance of the combination shown in Fig. P1.16 and Fig. P1.17. 6 μF HH 2 μF: 4 μF 4 µF 6 µF Fig. P1.16 (4+4) x6 8+6 + 2 x6 oa 8=2= 3C for t≥ 0. 2C 4H C 10 C 2 12 C HH ↑ Fig. P1.17 = 2.85 µF. In Fig. P1.17, Cab = 14C] IniH bo

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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25. The voltage across an inductor is given by: v(t)=0 for t ≤0; v(t) = 20 te 2¹ for f ≥ 0. Obtain the expression for current in the inductor. [Hint. Solve this problem in the same way as shown in Example 1.86.] 26. Find equivalent capacitance of the combination shown in Fig. P1.16 and Fig. P1.17. 6 µF HH 2 μF = 4 μF 4 μF 6 µF Fig. P1.16 (4+4) ×6 8+6 + 2×6 oa 3C HH 2C CH 4H CI 10 C 12 C Fig. P1.17 - 2.85 µF. In Fig. P1.17, Cab = 14C] bo
Transcribed Image Text:25. The voltage across an inductor is given by: v(t)=0 for t ≤0; v(t) = 20 te 2¹ for f ≥ 0. Obtain the expression for current in the inductor. [Hint. Solve this problem in the same way as shown in Example 1.86.] 26. Find equivalent capacitance of the combination shown in Fig. P1.16 and Fig. P1.17. 6 µF HH 2 μF = 4 μF 4 μF 6 µF Fig. P1.16 (4+4) ×6 8+6 + 2×6 oa 3C HH 2C CH 4H CI 10 C 12 C Fig. P1.17 - 2.85 µF. In Fig. P1.17, Cab = 14C] bo
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