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ELECTRICAL WIRING:RESIDENTAL-6 PLANS
18th Edition
ISBN: 9781305098329
Author: MULLIN
Publisher: CENGAGE L
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Textbook Question
Chapter 28, Problem 15R
The utility company has provided a letter to the contractor stating that the available fault current at the line side of the main service-entrance equipment in a residence is 17,000 amperes RMS symmetrical, line to line. In the space following each statement, write in “Meets Code” or “Violation” of 110.9 and 110.10 of the Code.
- a. Main breaker has a 10,000-ampere interrupting rating; branch breakers have a 10,000-ampere interrupting rating. _______
- b. Main current-limiting fuse has a 200,000-ampere interrupting rating; branch breakers have a 10,000-ampere interrupting rating. The panel is marked “Series-Rated.” _______
- c. Main breaker has a 22,000-ampere interrupting rating; branch breakers have a 10,000-ampere interrupting rating. The panel is marked “Series-Connected.” _______
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Please confirm that my solution is correct, especially the block diagram. Please DRAW (not type) what the block diagram would look like if it's incorrect.
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use this code on the bottom to answer the question in the photo
clc; clearvars;
% Read the file [y, Fs] = audioread('106miles.wav'); N = length(y); Nfft = 2^nextpow2(N); dt = 1/Fs; t = (0:dt:(N-1)*dt)'; % Ensure t is a column vector y = y - mean(y); % Remove DC component (if not already zero-mean)
% Carrier signal (25 kHz) fc = 25000; % Carrier frequency in Hz carrier = cos(2 * pi * fc * t);
% DSB-SC Modulation modulated_signal = y .* carrier;
% Plot Time Domain Signal figure; subplot(2,1,1); plot(t, y); title('Original Signal (Time Domain)'); xlabel('Time (s)'); ylabel('Amplitude');
subplot(2,1,2); plot(t, modulated_signal); title('DSB-SC Modulated Signal (Time Domain)'); xlabel('Time (s)'); ylabel('Amplitude');
% Frequency Domain (FFT) Y = fft(y, Nfft) / Nfft; Modulated_Y = fft(modulated_signal, Nfft) / Nfft; f = Fs * (0:(Nfft/2)) / Nfft; % Frequency vector
% Plot Frequency Domain Signal figure; subplot(2,1,1); plot(f, abs(Y(1:Nfft/2+1))); title('Original Signal…
5-9 A 230 V shunt motor has a nominal arma-
ture current of 60 A. If the armature resist-
ance is 0.152, calculate the following:
a. The counter-emf [V]
b. The power supplied to the armature [W]
c. The mechanical power developed by the
motor, [kW] and [hp]
5-10 a. In Problem 5-9 calculate the initial start-
ing current if the motor is directly con-
nected across the 230 V line.
b. Calculate the value of the starting resistor
needed to limit the initial current to 115 A.
Chapter 28 Solutions
ELECTRICAL WIRING:RESIDENTAL-6 PLANS
Ch. 28 - What is the minimum size service for a one-family...Ch. 28 - Prob. 2RCh. 28 - Prob. 3RCh. 28 - Prob. 4RCh. 28 - Prob. 5RCh. 28 - Prob. 6RCh. 28 - Prob. 7RCh. 28 - List the standard sizes of fuses and circuit...Ch. 28 - Prob. 9RCh. 28 - Using the method shown in this unit, what is the...
Ch. 28 - State four possible combinations of service...Ch. 28 - Which Code section states that all overcurrent...Ch. 28 - All electrical components have some sort of...Ch. 28 - Arc-fault damage is closely related to the value...Ch. 28 - The utility company has provided a letter to the...Ch. 28 - While working on the main panelboard that is...Ch. 28 - Repeat problem 16 for a main service panel that...Ch. 28 - NEC ______________________________ generally...Ch. 28 - Prob. 19RCh. 28 - Prob. 20RCh. 28 - Overcurrent devices must be accessible. (True)...
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