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120 apart. Each quasi-square wave has a conduction angle of 120 . The 120 quasi-square waveform has the advantage of having no third harmonic content. Each quasi-square wave is ltered by a second-order active low-pass lter. The quasi-square waves are created by averaging two square waves that are phase shifted by 60 . The Fourier results from the output le are shown in Fig. 7.15. Note that as a result of the quasi-square waveform, the rst signi cant harmonic is the fth harmonic. The sine-wave output distortion can be

5. 6. 7.

Now consider the case of a purely PARALLEL connection of n such impedances (see " Fig. 132). In this case the total impedance ZT seen by the generator is given by eq. (207) or " (208), each individual impedance now having the general form Z R j=!C R 1=j!C. (For the special case of two impedances in parallel, eqs. (207) and (208) reduce " to the convenient form of eq. (209).) After ZT is found, the value of the total (generator) " is then, by Ohm s law, equal to current IT " V " IT " ZT Problem 123 In Fig. 140, the reference generator voltage is 60 volts rms, as shown. If the frequency is 100,000 rad/sec, nd magnitude of generator current, " (b) phase angle of current IT . The symbol (capital omega ) denotes ohms. (a)

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In practical work we often encounter networks containing not only resistance but also both inductive and capacitive reactances. To nd the sinusoidal-steady-state response of such a network, we simply replace the inductors and capacitors with the imaginary quantities, jXL j!L and jXC j=!C 1=j!C, where ! 2 f radians per second. If the network is complicated we ll generally use the method of loop currents, applying the Kirchho voltage and current laws and the algebra of complex numbers, as discussed in connection with Fig. 135. Problem 124 In the series circuit of Fig. 141, the R, L, and C values are in ohms, microhenrys, and microfarads. Given that f 28 kHz (kilohertz), nd the vector voltage at point a with respect to ground.* " (Answer (in polar coord.), Va 22:43=143:958 volts)

and the following output,

which two lines of code (inserted independently) will cause this output (Choose two.)

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Problem 125 In Fig. 142, the R, L, and C values are in ohms, microhenrys, and microfarads. As always, the generator voltage is in rms vector volts. Given that ! 106 radians/ second, nd the vector voltage at point y with respect to ground. " (Answer (in polar coord.), Vy 6:19221= 26:5668 volts)

A. for (int x = -1; x < 2; ++x) { B. for (int x = 1; x < 3; ++x ) { C. for (int x = 0; x > 2; ++x ) { D. for (int x = 0; x < 2; x++ ) { E. for (int x = 0; x < 2; ++x ) { 8. Given the following,

1. 2. 3. 4. 5. 6. 7. 8. public class Test { public static void main(String [] args) { int I = 1; do while ( I < 1 ) System.out.print("I is " + I); while ( I > 1 ) ; } }

Problem 126 Making use of the work already done in problem 125, nd, in Fig. 142, the vector voltage at point x with respect to ground.

A. I is 1 B. I is 1 I is 1 C. No output is produced. D. Compilation error E. I is 1 I is 1 I is 1 in an infinite loop. 9. Given the following,

Problem 127 Here we wish to apply the fundamental principle of Thevenin s theorem (section 4.6) to the circuit of Fig. 143, in which we ll assume the internal resistance of the gen" erator is either negligibly small or is included in the resistance R. Find the voltage V 0 " 0 of the equivalent Thevenin generator, as indicated in Fig. and internal impedance Z 144. (Final answers in terms of R, C, and !.)

11. 12. 13. 14. 15. 16. int I = 0; outer: while (true) { I++; inner: for (int j = 0; j < 10; j++) {

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17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

406

I += j; if (j == 3) continue inner; break outer; } continue outer; } System.out.println(I);

A. 1 B. 2 C. 3 D. 4 10. Given the following,

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