Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Chapter 40, Problem 2SQ
To determine
Explain the purpose of interlock (M) in the DC control circuit.
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A linear electrical load draws 11 A at a 0.72 lagging power factor./1 153. When a capacitor is connected, the line current dropped to 122 A and the power factor
improved to 0.98 lagging. Supply frequency is 50 Hz.
a. Let the current drawn from the source before and after introduction of the capacitor be 11 and 12
respectively. Take the source voltage as the reference and express 11 and 12 as vector
quantities in polar form.
b. Obtain the capacitor current, IC = 12 - 11, graphically as well as using complex number
manipulation. Compare the results.
c. Express the waveforms of the source current before (11(t)) and after (12(t)) introduction of the
capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your
plots.
d. Analytically solve i2(t) – i1(t) using the theories of trigonometry to obtain the capacitor current
in the form, ¡C(t) = ICm sin(2πft + OC). Compare the result with the result in Part b.
Transmission line data:Data:• Active power of the load (P): 50 kW• Power factor of the load (PF): 0.8 (lagging)• Line-to-line voltage at the load (V_LC): 13.8 kV• Line resistance (R): 2 Ω• Line inductance (L): 0.8 H• Line capacitance (C): 0.0003 F• Required series compensation: 60% of the line impedance.• Line length: 250 kmDetermine:1. Characteristic impedance and propagation constant.2. The generalized long line constants A, B, C, D.3. Total voltage, current and power at the generating end.4. Voltage regulation.5. Parameters A, B, C, D of the compensation circuit.6. New generalized constants of the power system afterseries compensation.7. Conclusion of the results obtained.
Chapter 40 Solutions
Electric Motor Control
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- A linear electrical load draws I₁ A at a 0.72 lagging power factor. 11 = 153 When a capacitor is connected, the line current dropped to 122 A and the power factor improved to 0.98 lagging. Supply frequency is 50 Hz. a. Let the current drawn from the source before and after introduction of the capacitor be 1₁ and 12 respectively. Take the source voltage as the reference and express 11 and 12 as vector quantities in polar form. b. Obtain the capacitor current, Ic = I2 − I₁, graphically as well as using complex number manipulation. Compare the results. c. Express the waveforms of the source current before (i₁(t)) and after (i2(t)) introduction of the capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your plots. d. Analytically solve i₂ (t) - i₁ (t) using the theories of trigonometry to obtain the capacitor current in the form, ic (t) = 1cm sin(2´ft + 0c). Compare the result with the result in Part b.arrow_forwardI need a solution from an expert without artificial intelligence. Choose the correct answer 1. The convolution of two signals in time domain is equivalent to (addition, subtraction, multiplication, division). 2. The Fourier transform of non-periodic signal is random). 3. In ASK). in frequency domain signal. (discrete, continuous, digital, the pulse location is proportional to the amplitude of the signal. (PAM, PWM, PPM 4. In TDM the input signals must have (the same maximum frequency, the same sampling frequency, the same bit rate, the same amplitude). 5. The Delta Modulation is equivalent to (1-bit DPCM, 2-bit DPCM, n-bit PCM, PSK). bandwidth when compared to the NRZ 6. In baseband digital transmission, the RZ signal has signal. (less, greater, equal, not related). 7. In Differential Manchester code, the shapes of "1" and "0" are, each other, next bit dependent, not related to each other). (the same, opposite to 8. In multi-level digital carrier system R, is _ R₁. (equal to, less…arrow_forwardI need a solution from an expert without artificial intelligence. Choose the correct answer: 1. In AMI code, the shapes of "1" and "O" are, bit dependent, not related to each other). 2. In FDM the guard band is used to decrease, maintain, not related to). 3. Higher number of levels in PCM produces (the same, opposite to each other, next the overlap between FDM signals. (increase, (higher quantization error, less number of bits per sample, lower quantization error, the same number of bits per sample). 4. If the maximum shift in frequency is 70 kHz and the minimum deviation in frequency of the actual signal is 109.93 MHz, what is the carrier frequency? (110 MHz, 110 kHz, 107 kHz, 102 MHz) Fc 5. TDM of signals requires them to have the same amplitude, sampling frequency, energy). 6. In standard AM, the last step in the transmitter is subtracting, multiplying, dividing). . In digital carrier systems, PSK). (maximum frequency, maximum the carrier signal. (adding, has higher bandwidth. (ASK,…arrow_forward
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