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1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. public class If2 { static boolean b1, public static void int x = 0; if ( !b1 ) { if ( !b2 ) { b1 = true; x++; if ( 5 > 6 ) x++; } if ( !b1 ) x else if ( b2 else if ( b1 } } b2; main(String [] args) {

220

= x + 10; = true ) x = x + 100; | b2 ) x = x + 1000;

Equations (217) and (219) can be expressed in the form of the impedance triangle shown in Fig. 139.

17. 18. 19.

(Continued )

System.out.println(x); } }

Comparison of the above with Fig. 138 shows that in the above gure is the phase angle between the voltage and current vectors shown in Fig. 138, where tan 1=!C =R thus arctan 1=!RC the current LEADING the applied voltage in a capacitive circuit. Problem 121 In the series circuit of Fig. 136, let the sinusoidal reference voltage be 95 volts rms, the capacitance be 0.5 F (microfarad), and the resistance 100 ohms. If the frequency is 4000 Hz, nd (a) magnitude of rms current, (b) phase angle of current, (c) (d) (e) reading of voltmeter placed across R, reading of voltmeter placed across C, vector sum of the voltages in (c) and (d). 221

Flow Control (loops) (Sun Objective 2.2)

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6. Given the following,

The foregoing applies to ANY type of series, parallel, or series-parallel connection of R and C components. In doing this, the series circuit of Fig. 136 is the basic building block, " represented by the single symbol , which we ll label Z , as shown below.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(32) DC COMPONENT = 2.475499E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E+02 1.430E+00 1.000E+00 2.003E+01 0.000E+00 2 1.000E+03 1.102E-03 7.704E-04 8.055E+01 4.049E+01 3 1.500E+03 3.476E-04 2.430E-04 1.483E+02 8.822E+01 4 2.000E+03 8.351E-04 5.839E-04 -9.185E+01 -1.720E+02 5 2.500E+03 4.363E-03 3.051E-03 -3.703E+00 -1.038E+02 6 3.000E+03 4.820E-04 3.370E-04 1.227E+02 2.568E+00 7 3.500E+03 4.187E-03 2.928E-03 -7.488E+01 -2.151E+02 8 4.000E+03 7.734E-04 5.408E-04 8.985E+01 -7.038E+01 9 4.500E+03 3.807E-04 2.662E-04 1.590E+02 -2.129E+01 TOTAL HARMONIC DISTORTION = 4.398894E-01 PERCENT

1. public class While { 2. public void loop() { 3. int x= 0; 4. while ( 1 ) { 5. System.out.print("x plus one is " + (x + 1)); 6. } 7. } 8. }

as in eq: 217

A. There is a syntax error on line 1. B. There are syntax errors on lines 1 and 4. C. There are syntax errors on lines 1, 4, and 5. D. There is a syntax error on line 4. E. There are syntax errors on lines 4 and 5. F.

7. Given the following,

" In the above, either expression for Z can be used, but we ll more often use the form " R j=!C R jXC , where XC 1=!C. Z " If, in a given case, ONLY RESISTANCE is present, then Z R, a real number; or, if " j=!C 1=j!C, an imaginary number. ONLY CAPACITANCE is present, then Z The simplest case consists of a series connection of n such impedances (see Fig. 131). In such a case the current is given by eq. (204) in section 8.2, in which each impedance now " has the general form Z R j=!C (or R 1=j!C). Hence, for the case of capacitance (instead of inductance), eq. (205) in section 8.2 would become " " V V " 222 IT " j 1 1 1 ZT R1 R2 Rn ! C1 C2 Cn " " where, to summarize the notation, V and I are vector volts and amperes (being represented as complex numbers), R and C are in ohms and farads, and ! 2 f is frequency in radians per second ( f being frequency in cycles per second). Problem 122 A certain series circuit consists of two resistances, both of 18 ohms, and three capacitors, each of 0.12 F. If the applied sinusoidal reference voltage is 75 volts rms and ! 500,000 rad/sec, nd (a) (b) (c) magnitude of rms current, phase angle of current, magnitude of voltage drop across each capacitor.

1. 2. 3. 4. class For { public void test() { System.out.println("x = "+ x);

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