Test12Solution

Test12Solution - Solution of ECE 300 Test 12 F09 1 An...

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ECE 300 Test 12 F09 1. An industrial load operating at 60 Hz can be modeled as a resistance of 20 Ω in series with an inductance of 300 mH. The impedance is Z = R + j ω L = 20 Ω + j 377 / s ( ) (0.3 H) = 20 + j 113.1 Ω = 114.85 79.97 ° . Impedance is the ratio of voltage to current. Therefore the current lags the voltage by 79.97°. Power factor is PF = cos 79.97 ° ( ) = 0.1742 lagging. (b) Is the power factor leading or lagging? Lagging 2. With reference to the graph below of a periodic voltage with period 6 ms, (a) Find the numerical average voltage. Average value is the area under the curve over one period, divided by the period. In this case that is Average Value = 6V × 4ms + 2V ( ) × 2ms 6ms = 20 / 6 or 3.33V (b) Find the numerical effective (rms) value of this voltage. RMS value is the square root of the mean of the square. The mean of the square is the area under the square over one period, divided by the period. Mean of the Square = 6V ( ) 2 × 4ms + 2V ( ) 2 × 2ms 6ms = 25.33 V 2 Therefore the RMS value is 5.033 V. (c) If this voltage is across a 15 Ω resistor what is the numerical maximum instantaneous power delivered to it? The peak power is the maximum squared voltage divided by the resistance. P peak = 6V ( ) 2 15 = 2.4 W (d) If this voltage is across a 15 Ω resistor what is the numerical average power delivered to it? The average power delivered to a resistor is the effective voltage squared, divided by the resistance. P
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Test12Solution - Solution of ECE 300 Test 12 F09 1 An...

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