Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Textbook Question
Chapter 48, Problem 1SQ
In Figure 48–2, what is the purpose of normally closed contact F?
Expert Solution & Answer
To determine
Explain the scope of normally closed (NC) contact (F) in Figure 48-2.
Explanation of Solution
Refer to Figure 48-2 in the textbook shows that plugging the motor to stop from one direction only. In this figure, the normally closed (NC) contact F is mentioned as an interlock.
In Figure 48-2, while the stop button is pressed, the forward contactor will drop out and the coil (F) is a de-energized normally closed (NC) contact F. This contact in turn, energizes coil (R) to operate the motor in the reverse direction through the closed plugging switch.
Conclusion:
Thus, the scope of normally closed (NC) contact (F) in the given figure is explained.
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Students have asked these similar questions
The MATLAB code is going well but the last part in bandpass, the legend that is supposed to tell the color of both lower and upper-frequency cutoff does not align with each other. As such I need help
My Matlab code:
% Define frequency range for the plot
f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz
w = 2 * pi * f; % Angular frequency
% Parameters for the filters
R = 1e3; % Resistance in ohms (1 kΩ)
C = 1e-6; % Capacitance in farads (1 μF)
L = 0.1; % Inductance in henries (chosen for proper bandpass response)
% Compute cutoff frequencies
f_cutoff_RC = 1 / (2 * pi * R * C); % RC low-pass/high-pass cutoff
f_resonance = 1 / (2 * pi * sqrt(L * C)); % Resonant frequency of RLC
Q_factor = (1/R) * sqrt(L/C); % Quality factor of the circuit
% Band-pass filter cutoff frequencies
f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor));
f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor));
% Define Transfer Functions
H_low =…
1°
⑤
Aa
"Human-written solution required"
2. Using the characteristics of Fig. 6.11, determine ID for the following levels of VGs (with
VDS > VP):
a. VGs = 0V.
b. VGs=-1 V.
c. VGs -1.5 V.
d. VGS
-1.8 V.
e. VGS = -4 V.
f. VGs=-6V.
3. Using the results of problem 2 plot the transfer characteristics of ID vs. VGS-
4. a. Determine Vps for VGs = 0V and Ip = 6 mA using the characteristics of Fig. 6.11.
b. Using the results of part (a), calculate the resistance of the JFET for the region Ip = 0 to
6 mA for VGs =0V.
c. Determine Vps for VGS = -1 V and ID = 3 mA.
d. Using the results of part (c), calculate the resistance of the JFET for the region ID = 0 to
3 mA for VGs -1 V.
e. Determine Vps for VGs = -2 V and ID = 1.5 mA.
f. Using the results of part (e), calculate the resistance of the JFET for the region ID = 0 to
1.5 mA for VGS-2 V.
g. Defining the result of part (b) as ro, determine the resistance for VGs -1 V using
Eq. (6.1) and compare with the results of part (d).
h. Repeat part (g)…
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Chapter 48 Solutions
Electric Motor Control
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- 4. a. Determine VDs for VGS = 0 V and ID = 6 mA using the characteristics of Fig. 6.11. b. Using the results of part (a), calculate the resistance of the JFET for the region ID = 0 to 6 mA for VGS = 0 V. c. Determine VDs for VGS = -1 V and ID = 3 mA. d. Using the results of part (c), calculate the resistance of the JFET for the region ID = 0 to 3 mA for VGS = -1 V. e. Determine VDs for VGS = -2 V and ID = 1.5 mA. f. Using the results of part (e), calculate the resistance of the JFET for the region ID = 0 to 1.5 mA for VGS = -2 V. g. Defining the result of part (b) as ro, determine the resistance for VGS = -1 V using Eq. (6.1) and compare with the results of part (d). h. Repeat part (g) for VGS = -2 V using the same equation, and compare the results with part (f). i. Based on the results of parts (g) and (h), does Eq. (6.1) appear to be a valid approximation?arrow_forwardA. Using D flip-flops, design a logic circuit for the finite-state machine described by the state assigned table in Fig. 1. Present Next State State Output x=0 x=1 Y2Y1 Y2Y1 YY Z 00 00 01 0 01 10 11 888 00 10 0 00 10 1 00 10 1 Fig. 1arrow_forwardAthree phase a.c. distributor AB has: A B C The distance from A to B is 500 m. The distance from A to C is 800 m. The impedance of each section is (6+j 8) /km. The voltage at the far end is maintained at 250 volt. Find: sending voltage, sending current, supply power factor and 80A 60 A total voltage drop. 0.8 lag. P.f 0.6 lead. p.farrow_forward
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