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Analysis: First, we de ne the phasors and impedances at the frequency of interest in the problem, = 377 rad/s: Z = R + j L = 4 + j 3 = 5 (0644) V = 10 2 10 V 2 = 2 ( 2215) = I= Z 5 (0644)

The average power can be computed from the phasor quantities: Pav = VI cos( ) = 10 2 cos(0644) = 16 W The instantaneous power is given by the expression: p(t) = v(t) i(t) = 2 10 sin(377t) 2 2 cos(377t 2215) W The instantaneous voltage and current waveforms and the instantaneous and average power are plotted in Figure 76

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1. If necessary, reopen the AutoTester application from Exercise 5-3. Select Form1 in your application. Open the Toolbox window and locate the MenuStrip control. Double-click it to add it to your form. A menu is added to the form (displaying the text Type Here ), and the MenuStrip control is added to the component tray (see Figure 5-6). What s the component tray The component tray shows a list of controls that are attached to the form that either don t display (as you ll see with the Timer control soon) or, in the case of the MenuStrip control, can have more than one in the same location on the form. It allows you to easily locate the control and click on it to view its properties, whereas it might be difficult to find the control otherwise. Up to this point, you have created controls that do not use the component tray; but many controls do get added to this tray, so you ll see more as you create more testware.

Voltage and current waveforms for Example 71 20 10 0 10 20 0 0005 001 0015 t (s) 002 0025 003 0035

v(t) i(t) ) Volts ( ), amps (

When cost is underestimated, for six projects the errors are huge (300 percent to 900 percent), another five exceed 100 percent, and a further seven exceed 50 percent When the six huge errors are excluded, the mean error is an underestimate of 46 percent; the overall median error is an underestimate of 44 percent

Instantaneous and average power for Example 71 40 30 p(t)

important to remember that we have de ned phasors to have rms amplitude in power calculation This is a standard procedure in electrical engineering practice Note that the instantaneous power can be negative for brief periods of time, even though the average power is positive

Known Quantities: Source voltage, internal resistance and frequency, load resistance and inductance values Find: Pav for the RC load

Caution Now that you have two forms, be careful to place the MenuStrip on the correct form. Place it

Next, we compute the load voltage, using the voltage divider rule: VL = ZL 137 ( 0543) 110 (0) = 976 ( 0067) V VS = RS + ZL 2 + 137 ( 0543) 9762 |VL |2 cos( ) = cos( 0543) = 595 W |ZL | 137

or, alternatively, we can compute the load current and calculate average power according to the equation below: VL = 71 (0476) A IL = ZL Pav = |IL |2 |ZL | cos( ) = 712 137 cos( 0543) = 595 W

underestimate of about 45 percent A clear trend can be seen when estimation techniques are considered: FP-based techniques are more accurate more of the time For projects using just work breakdown techniques, 42 percent are accurate to within 10 percent; a further 12 percent are accurate to within 20 percent; nearly half misestimate cost by 20 percent or more; underestimates are twice as common as overestimates, and tend to be worse; and the median underestimate is around 50 percent For projects using functional size based techniques, 61 percent are accurate to within 10 percent; a further 30 percent are accurate to within 20 percent; and only 9 percent misestimate cost by 20 percent or more; the median underestimate is 14 percent

voltage before proceeding to the computation of power; the internal source resistance in this problem causes the source and load voltages to be different

Known Quantities: Source voltage, internal resistance and frequency, load resistance, capacitance and inductance values Find: Pav for the complex load

on the main form (Form1) in your application do not place the menu on your new AboutBox1 form!

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