Homework 8 Solutions

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Electrical Engineering

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Jan 9, 2024

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Homework #8 Solutions Q1-Q2 A 1,000 W electric motor is connected to a 120 Vms, 60 Hz source. The power factor seen by the lagging pf of 0.8. To correct the pf to 0.95 lagging, a capacitor is placed in parallel with the motor. Calculate the current drawn from the source with and without the capacitor connected. Determine the value of the capacitor required to make the correction. Solution: Known quantities: Find: The value of capacitor to make the correction. Analysis: Q3-Q4 A center-tapped transformer has the schematic representation shown in Figure P13.1. The primary-side voltage is stepped down to two secondary-side voltages. Assume that each secondary supplies a 7 − kW resistive load and that the primary is connected to 100 V rms . Find: a. The primary power. b. The primary current. Solution: Known quantities: Circuit shown in Figure P13.41, 3ach secondary connected to 7-kW resistive load, the primary connected to 100-V rms. ( )( ) A 9 36 4 10 8 0 120 1000 ° - Ð = = . . . I old ( )( ) A 2 18 77 . 8 95 0 120 1000 ° - Ð = = . . I new VAR 750 9 36 tan 1000 = ° = . Q old VAR 8 . 328 2 . 18 tan 1000 = ° = new Q VAR 2 421 8 328 750 . . Q C = - = W = 34.2 120 = X 2 C C Q ( )( ) “F 6 77 2 34 377 1 . = . C =

Find: a) Primary power. b) Primary current. Analysis: a) b) Q5-Q6 For the circuit shown in Figure P13.43, assume that ࠵? " ࠵? = 80∠0 V rms, ࠵? " = 2 Ω , and ࠵? # = 12 Ω . Assume an ideal transformer. Find: (a) the equivalent resistance seen by the voltage source and (b) the power ࠵? $%&’() supplied by the voltage source. Known quantities: ࠵? " ࠵? = 80∠0 V rms, ࠵? " = 2 Ω , ࠵? # = 12 Ω, ࠵? = 5 Assume ideal transformer. Find: (a) The equivalent resistance seen by the voltage source (b) The power ࠵? $%&’() supplied by the voltage source Analysis: (a) The equivalent resistance seen by the voltage source: To find the equivalent impedance seen by the voltage source, use equation 13.32: ࠵? ࠵? = 1 ࠵? + ࠵? ࠵? In this case, the load impedance, ࠵? ࠵? , is the output resistance ࠵? # , which has the impedance: ࠵? ࠵? = ࠵? # = 12 Ω = 12∠0 kW 14 2 sec 1 sec = + = P P P prim A 140 100 14000 ~ ~ = = = V P I prim prim A 0 140 ~ Ð = prim I

Substitute ࠵? ࠵? into equation 13.32: ࠵? ࠵? = 1 5 + ∗ 12∠0 = 0.48∠0 Ω Final answer: ࠵? ࠵? = ࠵?. ࠵?࠵?∠࠵? ࠵? (b) The power ࠵? ࠵?࠵?࠵?࠵?࠵?࠵? supplied by the voltage source: From equation 13.31, we know that ideal transformers conserve power. Therefore: ࠵? ࠵? = ࠵? ࠵? Using this property, we may solve for the source power by computing the real power at the primary terminal using equation 13.12: ࠵? $%&’() = ࠵? 34" = ࠵? " ࠵? |࠵?| cos(࠵? 5 ) where ࠵? " = ࠵? " ࠵? and ࠵? = ࠵? ࠵? + ࠵? ࠵? . Substitute these values into equation 13.12: ࠵? $%&’() = 80 + (0.48 + 2) ∗ cos(0) = 2.58 ࠵?࠵? Final answer: ࠵? ࠵?࠵?࠵?࠵?࠵?࠵? = ࠵?. ࠵?࠵? ࠵? W Q7 Using a 500 Ω resistance, design an RC low-pass filter that would attenuate a 120 Hz sinusoidal voltage by 20 dB with respect to the DC gain. Solution: Known quantities: The resistance of the RC low-pass filter. Find: Design an RC low-pass filter that would attenuate a 120 Hz sinusoidal voltage by 20 dB with respect to the DC gain. Analysis: The frequency response of the RC low-pass filter is:

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The response of the circuit to the periodic input is: In order to attenuate the sinusoidal input by 20 dB (a factor of 10) with respect to the DC gain, Þ μ F OR… 20 dB attenuation from a First Order filter means that the signal is one decade above the cutoff of the filter. So if a 120 Hz signal is down by 20 dB, and we’re talking about a low pass filter, the cutoff had to be 12 Hz. (1/10 th of 120 Hz = 1 decade) 1 . 12 = 1 2࠵?࠵?࠵? ࠵?࠵? ࠵? = 1 2࠵? × 12 × 500 = 26.5 ࠵?࠵? The result differs from the other approach by less than 0.5%. Q8. Identify type of filter: OK, two ways to look at this. First, at DC the C is an OC and at high frequencies it’s a SC. So we get 0 volts out at infinite frequency and something out at 0 Hz. Sounds low-pass since nothing comes out at very high frequencies Other approach: treat C as a “load” and Thevenize the rest of the circuit. You’ll get 1 NOTE : IF this were a high pass filter, the cutoff is ABOVE 120 Hz by a decade and so the cutoff would be at 1200 Hz. H v j w ( ) = V o ( j w ) V i ( j w ) = 1 1 + j w CR u ( t ) = A sin ˆ w t + j ( ) y ¥ ( t ) = H v j ˆ w ( ) A sin ˆ w t + j + Ð H v j ˆ w ( ) ( ) H v j ˆ w ( ) = 1 1 + ˆ w CR ( ) 2 = 0.1 C = 1 R ˆ w 10 2 - 1 = 99 500 × 2 p ´ 120 @ 26.4 R TH

Here R TH = 2000//2000 + 1000 = 2000 Ohms and V TH = ½ Vin. (Q : Can you see why??) Clearly this is low-pass by inspection – it’s the same circuit we started our discussion of filters with and we derived the transfer function already. J Q9. Using the Thevenin equivalent circuit from Q8, ࠵? = 1 2࠵?࠵? 78 ࠵? = 1 2࠵?(2000)(470 × 10 9:+ ) = 169 ࠵?࠵?࠵? V TH