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128 it would then be apparent that the curve of the rate of change of q (the current) would lead the curve of voltage drop by 908 Thus, in Fig 136 we have that the current LEADS the voltage drop across C by 908 but is, as always, IN PHASE with the voltage drop across R It thus follows that the current wave will lead the applied voltage wave by some intermediate angle , where will be an.

guaranteed to always run, so you don t want behavior that changes depending on whether assertions are enabled.

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code block will never be reached. You can use assert false; for code that should never be reached, so that an assertion error is thrown immediately if the assert statement is executed.

angle between 08 and 908. The actual value of will depend upon the relative values of R, C, and ! in any given case. Next, from inspection of eq. (210), RIp peak voltage drop across R and 1=!C Ip peak voltage drop across C Now recall, from section 5.8, that the peak value of the SUM of two sinusoidal waves of the same frequency is equal to the vector sum of the peak values of the individual sinusoids. Thus, the relationships of the peak values of the voltages in eq. (210) can be stated in the vector polar form, Vp =08 RIp = 8 1=!C Ip = 90 8 211

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The following questions will help you measure your understanding of the material presented in this chapter. You ve heard this before, and this time we really mean it: this chapter s material is crucial for the exam! Regardless of what the exam question is really testing, there s a good chance that flow control code will be part of the question. Expect to see loops and if tests used in questions throughout the entire range of exam objectives.

which shows that, geometrically, the two peak voltages on the right-hand side can be considered to be the adjacent and opposite sides of a right-angled triangle having Vp as the hypotenuse with the angle between RIp and Vp (as will be shown later in connection with Fig. 138). Thus, we have that the steady-state PEAK CURRENT is equal to q 212 Ip Vp = R2 1=!C 2 where also arctan 1=!RC 213

Flow Control (if and switch) (Sun Objective 2.1)

V3 15

1. Given the following,

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Next, in Fig. 136 the voltage drop across R is, as always, in phase with the current; thus, since the peak voltage drop across R is equal to RIp , and since the current LEADS the applied voltage Vp by an angle , we have the vector diagram as in Fig. 137.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. public class Switch2 { final static short x = 2; public static int y = 0; public static void main(String [] args) { for (int z=0; z < 3; z++) { switch (z) { case y: System.out.print("0 "); case x-1: System.out.print("1 "); case x: System.out.print("2 "); } } } }

A. 0 1 2 B. 0 1 2 1 2 2 C. Compilation fails at line 7. D. Compilation fails at line 8. E. Compilation fails at line 9. F.

Next, returning to eq. (210), the quantity 1=!C is called CAPACITIVE REACTANCE and is measured in ohms.* Capacitive reactance is denoted by XC ; thus, XC 1=!C 1=2 fC ohms. Hence, by Ohm s law, XC Ip 1=!C Ip peak voltage drop across the capacitor, which let us next consider as follows. By the basic Kirchho voltage law, the vector sum of the voltage across R, RIp , and the voltage across the capacitor, 1=!C Ip , must be equal to the applied voltage vector Vp . This requirement, and the required relationship between R and 1=!C in eq. (212), will be satis ed only if the voltage drop across the capacitor, 1=!C Ip , lags the current vector Ip by 908, as shown in Fig. 138. Now, in Fig. 138, multiply all the vector magnitudes by 0.7071, thus converting the peak values to rms values. Doing this, and using the usual bar notation to denote vector

V5 15

2. Given the following,

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