Connect 1 Semester Access Card For Electric Motors And Control Systems
2nd Edition
ISBN: 9781259550195
Author: Petruzella, Frank
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 3.2, Problem 10RQ
To determine
The winding having the larger diameter and the reason of larger diameter of the conductor.
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A certain signal f(t) has the following PSD (assume 12 load):
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Chapter 3 Solutions
Connect 1 Semester Access Card For Electric Motors And Control Systems
Ch. 3.1 - Prob. 1RQCh. 3.1 - a. If 1 MW of electric power is to be transmitted...Ch. 3.1 - Compare the type of AC power normally supplied to...Ch. 3.1 - a. Outline the basic function of a unit...Ch. 3.1 - list three factors taken into account in selecting...Ch. 3.1 - When motors and motor controllers are installed,...Ch. 3.1 - a. What types of conduit raceways are commonly...Ch. 3.1 - Compare the function of a switchboard, panelboard,...Ch. 3.2 - Define the terms primary and secondary as they...Ch. 3.2 - On what basis is a transformer classified as being...
Ch. 3.2 - Explain how the transfer of energy takes place in...Ch. 3.2 - In an ideal transformer, what is the relationship...Ch. 3.2 - A step-down transformer with a Wins ratio of 10:1...Ch. 3.2 - A step-up transformer has a primary current of 32...Ch. 3.2 - What is meant by the term transformer magnetizing,...Ch. 3.2 - Prob. 8RQCh. 3.2 - Prob. 9RQCh. 3.2 - Prob. 10RQCh. 3.2 - The primary of a transformer is rated for 480 V...Ch. 3.2 - A single-phase transformer is rated for 0.5 kVA, a...Ch. 3.3 - Explain the way in which the high-voltage and...Ch. 3.3 - Prob. 2RQCh. 3.3 - Prob. 3RQCh. 3.3 - Prob. 4RQCh. 3.3 - Prob. 5RQCh. 3.3 - Prob. 6RQCh. 3.3 - Prob. 7RQCh. 3.3 - Prob. 8RQCh. 3.3 - Explain the basic difference between the primary...Ch. 3.3 - Prob. 10RQCh. 3.3 - Prob. 11RQCh. 3.3 - Prob. 12RQCh. 3.3 - What important safety precaution should be...Ch. 3.3 - Prob. 14RQCh. 3.3 - The control transformer for an across-the-line...Ch. 3.3 - The two primary windings of a dual-primary control...Ch. 3.3 - Prob. 3TCh. 3.3 - A dry-type general-purpose power transformer is...Ch. 3.3 - A current transformer is to be tested in circuit...Ch. 3.3 - Discuss how electric power might be distributed...Ch. 3.3 - A block of several transformers arc fed from...Ch. 3.3 - How would you proceed with a DC resistance check...Ch. 3.3 - Prob. 5DT
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- please show full working. I've included the solutionarrow_forwardcan you please show working and steps. The answer is 8kohms.arrow_forwardPSD A certain signal f(t) has the following PSD (assume 12 load): | Sƒ(w) = π[e¯\w\ + 8(w − 2) + +8(w + 2)] (a) What is the mean power in the bandwidth w≤ 1 rad/sec? (b) What is the mean power in the bandwidth 0.99 to 1.01 rad/sec? (c) What is the mean power in the bandwidth 1.99 to 2.01 rad/sec? (d) What is the total mean power in (t)? Pav= + 2T SfLw) dw - SALW)arrow_forward
- An AM modulation waveform signal:- p(t)=(8+4 cos 1000πt + 4 cos 2000πt) cos 10000nt (a) Sketch the amplitude spectrum of p(t). (b) Find total power, sideband power and power efficiency. (c) Find the average power containing of each sideband.arrow_forwardCan you rewrite the solution because it is unclear? AM (+) = 8(1+ 0.5 cos 1000kt +0.5 ros 2000ks) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. -Jet jooort J11000 t = 4 e jqooort jgoort +4e + e +e j 12000rt. 12000 kt + e +e jooxt igoo t te (w) = 8ES(W- 100007) + 8IS (W-10000) USBarrow_forwardCan you rewrite the solution because it is unclear? AM (+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. J4000 t j11000rt $14+) = 45 jqooort +4e + e + e j 12000rt. 12000 kt + e +e +e Le jsoort -; goon t te +e Dcw> = 885(W- 100007) + 8 IS (W-10000) - USBarrow_forward
- Can you rewrite the solution because it is unclear? Q2 AM ①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt) $4+) = 45 = *cos 10000 πt. 8 cos wat + 4 cosat + 4 cos Wat coswet. j1000016 +4e -j10000πt j11000Rt j gooort -j 9000 πt + e +e j sooort te +e J11000 t + e te j 12000rt. -J12000 kt + с = 8th S(W- 100007) + 8 IS (W-10000) <&(w) = USB -5-5 -4-5-4 b) Pc 2² = 64 PSB = 42 + 4 2 Pt Pc+ PSB = y = Pe c) Puss = PLSB = = 32 4² = 8 w 32+ 8 = × 100% = 140 (1)³×2×2 31 = 20% x 2 = 3w 302 USB 4.5 5 5.6 6 ms Ac = 4 mi = 0.5 mz Ac = 4 ५ M2 = =0.5arrow_forwardA. Draw the waveform for the following binary sequence using Bipolar RZ, Bipolar NRZ, and Manchester code. Data sequence= (00110100) B. In a binary PCM system, the output signal-to-quantization ratio is to be hold to a minimum of 50 dB. If the message is a single tone with fm-5 kHz. Determine: 1) The number of required levels, and the corresponding output signal-to-quantizing noise ratio. 2) Minimum required system bandwidth.arrow_forwardFind Io using Mesh analysisarrow_forward
- FM station of 100 MHz carrier frequency modulated by a 20 kHz sinusoid with an amplitude of 10 volt, so that the peak frequency deviation is 25 kHz determine: 1) The BW of the FM signal. 2) The approximated BW if the modulating signal amplitude is increased to 50 volt. 3) The approximated BW if the modulating signal frequency is increased by 70%. 4) The amplitude of the modulating signal if the BW is 65 kHz.arrow_forwardAn FDM is used to multiplex two groups of signals using AM-SSB, the first group contains 25 speech signals, each has maximum frequency of 4 kHz, the second group contains 15 music signals, each has maximum frequency of 10 kHz. A guard bandwidth of 500 Hz is used bety each two signals and before the first one. 1. Find the BWmultiplexing 2. Find the BWtransmission if the multiplexing signal is modulated using AM-DSB-LC.arrow_forwardAn FM signal with 75 kHz deviation, has an input signal-to-noise ratio of 18 dB, with a modulating frequency of 15 kHz. 1) Find SNRO at demodulator o/p. 2) Find SNRO at demodulator o/p if AM is used with m=0.3. 3) Compare the performance in case 1) and 2).. Hint: for single tone AM-DSB-LC, SNR₁ = (2m²) (4)arrow_forward
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