Schaum's Outline of Electric Circuits, 6th edition (Schaum's Outlines)

Schaum's Outline of Electric Circuits, 6th edition (Schaum's Outlines)

Mahmood Nahvi, Joseph Edminister

Language: English

Pages: 504

ISBN: 0071830456

Format: PDF / Kindle (mobi) / ePub


Tough Test Questions? Missed Lectures? Not Enough Time?

Fortunately, there's Schaum's. This all-in-one-package includes more than 500 fully solved problems, examples, and practice exercises to sharpen your problem-solving skills. Plus, you will have access to 25 detailed videos featuring instructors who explain the most commonly tested problems--it's just like having your own virtual tutor! You'll find everything you need to build confidence, skills, and knowledge for the highest score possible.

More than 40 million students have trusted Schaum's to help them succeed in the classroom and on exams. Schaum's is the key to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easy-to-follow, topic-by-topic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills.

This Schaum's Outline gives you

  • 500 fully solved problems
  • Extra practice on topics such as amplifiers and operational amplifier circuits, waveforms and signals, AC power, and more
  • Support for all the major textbooks for electric circuits courses

Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum’s to shorten your study time--and get your best test scores!

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t<0 t>0 t<0 1V 0 t<0 . t>0  ÀeÀ1000t ðVÞ À1 V t>0 t<0 5.46 Repeat Problem 5.45 for v1 ¼ 5.47 In the differential equation 10À2 dv2 =dt þ v2 ¼ vs , vs is the forcing function and v2 is the response. Design an op amp circuit to obtain v2 from vs . Ans: See Fig. 5-24, with R1 ¼ Rf ; RC ¼ 10À2 , and v1 ¼ Àvs . 5.48 Design a circuit containing op amps to solve the following set of equations: Ans: v2 ðtÞ ¼ y 0 þ x ¼ vs1 2y þ x 0 þ 3x ¼ Àvs2 Ans. See Fig. 5-55, with R1 C ¼ R4 C ¼ 1 s, R2 C

6.1 Graph each of the following functions and specify period and frequency. ðaÞ v1 ðtÞ ¼ cos t ðbÞ v2 ðtÞ ¼ sin t ðcÞ v3 ðtÞ ¼ 2 cos 2t ðdÞ v4 ðtÞ ¼ 2 cos ðt=4 À 458Þ ¼ 2 cos ðt=4 À =4Þ ¼ 2 cos½ðt À 1Þ=4Š ðeÞ v5 ðtÞ ¼ 5 cos ð10t þ 608Þ ¼ 5 cos ð10t þ =3Þ ¼ 5 cos 10ðt þ =30Þ (a) See Fig. 6-2(a). T ¼ 2 ¼ 6:2832 s and f ¼ 0:159 Hz. (b) See Fig. 6-2(b). T ¼ 2 ¼ 6:2832 s and f ¼ 0:159 Hz. (c) T ¼ 1 s and f ¼ 1 Hz. See Fig. 6-2(c). (d) See Fig. 6-2(d). T ¼ 8 s and f ¼ 0:125 Hz. (e)

case, find t1 for the maximum current. Underdamped or Oscillatory Case ð < !0 Þ When < !0 , s1 and s2 in the solution to the differential equation suggested qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiin the preceding are complex conjugates s1 ¼ þ j and s2 ¼ À j , where is now given by !20 À 2 . The solution can be written in the exponential form i ¼ eÀ t ðA1 e j t þ A2 eÀj t Þ or, in a readily derived sinusoidal form, i ¼ eÀ t ðA3 cos t þ A4 sin tÞ EXAMPLE 8.3 Repeat Example 8.1 for C ¼ 1 mF. As before, ¼ R ¼

factor et , Aet e jð!tþÞ ) Aet cos ð!t þ Þ Ae j eðþj!Þt ¼ Ae j est À1 where s ¼  þ j! The complex frequency s ¼  þ j! has units s , and !, as we know, has units rad/s. Consequently, the units on  must also be sÀ1 . This is the neper frequency with units Np/s. If both  and ! are Fig. 8-10 CHAP. 8] HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY 169 nonzero, the function is a damped cosine. Only negative values of  are considered. If  and ! are zero, the result is a constant. And

same as in Problem 8.2 except the second initial condition, which is now di Q0 di 100 À ð2500=50Þ ¼ 500 A=s ¼V or 0þL þ ¼ dt þ C dt þ 0:1 0 0 The initial values are half those in Problem 8.2, and so, by linearity, i ¼ eÀ250t ð1:35 sin 370:8tÞ 8.4 ðAÞ A parallel RLC network, with R ¼ 50:0 , C ¼ 200 mF, and L ¼ 55:6 mH, has an initial charge Q0 ¼ 5:0 mC on the capacitor. Obtain the expression for the voltage across the network. 1 ¼ 8:99 Â 104 sÀ2 LC qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Since

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