In digital circuits, the condition mentioned in #3 implies that the inputs b and a are set to 1 and 0, respectively, while the input C is set to 1. Without a specific context or circuit diagram provided, it is difficult to determine the exact functionality and logic involved. However, based on the given conditions, we can make some assumptions.
If we consider a scenario where b, a, and C are inputs to a logic gate or a combination of logic gates, the condition b = 1 ensures that the input b is always at a logic HIGH or TRUE state. Similarly, the condition a = 0 ensures that the input a is always at a logic LOW or FALSE state. Finally, the condition C = 1 implies that the input C is also always at a logic HIGH or TRUE state.
Given these conditions, the logic gate(s) or circuitry involved can be designed in such a way that it only produces a logic HIGH or TRUE output (represented as Q) when b = 1, a = 0, and C = 1. This means that any other combination of input values would not satisfy these conditions and would result in a logic LOW or FALSE output for Q.
Learn more about logic gate(s) here:
https://brainly.com/question/33186108
#SPJ11
A three-phase transformer rated 5 MVA, 115/13.2 kV has per-phase series impedance of (0.007+j0.075) per unit. The transformer is connected to a short distribution line which can be represented by series impedance per phase of (0.02+j0.10) per unit on a base of 10 MVA, 13.2 kV. The line supplies a balanced three-phase load rated 4 MVA, 13.2 kV, with lagging power factor of 0.85. Neglect the magnetizing branch of the transformer: a) Sketch a single-phase equivalent circuit of the system indicating all impedances in per unit on 10 MVA, 13.2 kV base quantities at the load [10 pts] b) Find the complex power supplied by the source connected to the primary of the transformer [10 pts] c) Calculate the voltage regulation at the load [5 pts]
Single-phase equivalent circuit of the system indicating all impedances in per unit on 10 MVA, 13.2 kV base quantities at the load:
Thus, the per unit transformer impedance is: Z pu = (Z1 / 5x10^6) * (115 kV / (13.2 kV / √3))^2Zpu = (0.007 + j0.075) pub) Complex power supplied by the source connected to the primary of the transformer: At the load, the current is given by:
I2 = (V2 - V Load) / (Z pu + Z2 + Z Load) = 0.8704 - j0.2253 pu
The apparent power supplied to the load is:
S2 = 3 x VLoad x I2* = 4 MVA x 0.85 = 3.4 MVAThus, the complex power supplied by the source connected to the primary of the transformer is:
S1 = S2 / a^2 + I1^2(Zpu + Z2) = 13.143 MVA - j6.821 MVAc) Voltage regulation at the load:The voltage regulation at the load is given by:
VLoad, actual = VLoad, nominal / (1 + 3 x I2*Zpu)
VLoad, actual = 13.2 kV / (1 + 3 x 0.8704 + j0.2253 x 0.007 - j0.2253 x 0.075)
VLoad, actual = 12.312 - j0.725 kVThe magnitude of the voltage regulation is:
% voltage regulation = (|VLoad, actual| - |VLoad, nominal|) / |VLoad, nominal| x 100%
% voltage regulation = (12.312 - 13.2) / 13.2 x 100% = -6.67%The voltage regulation at the load is -6.67%.
To know more about equivalent visit:-
https://brainly.com/question/14947697
#SPJ11
On a ladder diagram all wires that connect to a common point are assigned _____.
A) the same number
B) different numbers
C) letters
D) all of these
On a ladder diagram, all wires that connect to a common point are assigned the same number. Let's understand what a ladder diagram is before we move on to the answer. Ladder diagrams are a type of electrical diagram that is widely used in industrial automation processes.
They are often used to represent complex control systems for machinery or other industrial equipment, as well as simple circuits for controlling lights or other small loads.A ladder diagram consists of two vertical lines representing the power rails or conductors that carry electrical power to the devices being controlled. Horizontal lines are used to connect the various devices or components in the system.
These horizontal lines are often called rungs.Each device or component in the system is represented by a symbol on the ladder diagram. The symbols used in ladder diagrams can vary depending on the type of device or component being represented. Some common symbols include switches, relays, motor starters, timers, and sensors.In a ladder diagram, all wires that connect to a common point are assigned the same number.
This is done to simplify the wiring and make it easier to troubleshoot problems if they occur. By assigning the same number to all wires that connect to a common point, it is easy to trace the wiring and determine which devices or components are connected together. Therefore, the correct option is A) the same number.
To know more about understand visit :
https://brainly.com/question/24388166
#SPJ11
Design an FSM with one input, A, and one output, X. X should be 1 if A has been 1 for at least two consecutive cycles. Show your state transition diagram, encoded state transition table, next state and output equations, and schematic.
The FSM (finite state machine) that has one input, A, and one output, X, with X being 1 if A has been 1 for at least two consecutive cycles, is as follows:State Transition Diagram:Encoded State Transition Table:Next State Equations:Y1 = A + S1S1 = A'Y2 = S1S2 = S1'Output Equation:X = S2S1'Explanation:
There are two states in this FSM, S1 and S2. State S1 represents the initial state. When A is zero, it remains in state S1, which is the initial state. When A is one, it switches to state S2, which indicates that one A value has been received. If A remains one in the next cycle, it remains in state S2. When A is zero in the next cycle, it goes back to state S1.If it remains in state S2 after two consecutive cycles, the output X becomes 1. This indicates that the input A has been one for at least two consecutive cycles.
If it does not stay in state S2 for two consecutive cycle, the output X remains zero.The schematic diagram of this FSM can be constructed using a JK flip-flop and a D flip-flop, as shown below.
To know more about State visit:
https://brainly.com/question/19592910
#SPJ11
In PCM system, the output signal to quantization ratio is to be hold to a minimum of 25 dB. If the message is a single tone with fm=5KHz. Calculate 1. the number of required level. 2. Minimum required BW.
To calculate the number of required levels in a PCM system and the minimum required bandwidth, we can use the following formulas:
Number of Required Levels (N):
N = 2^(B)
Minimum Required Bandwidth (Bw):
Bw = (2 * fm) + (2 * fm * log2(N))
Where:
B is the number of bits used for quantization.
fm is the maximum frequency component of the message signal.
In this case, we are given that the output signal-to-quantization ratio should be held to a minimum of 25 dB, and the message signal is a single tone with fm = 5 kHz.
Let's calculate the values step by step:
Number of Required Levels (N):
To achieve an output signal-to-quantization ratio of 25 dB, we can calculate B using the formula:
25 dB = 6.02 * B + 1.76
B = (25 - 1.76) / 6.02
B ≈ 4.02 (approximated to the nearest integer)
Therefore, the number of required levels (N) is:
N = 2^4
N = 16
Minimum Required Bandwidth (Bw):
Using the given maximum frequency component fm = 5 kHz and the calculated N = 16, we can calculate the minimum required bandwidth using the formula:
Bw = (2 * fm) + (2 * fm * log2(N))
Bw = (2 * 5 kHz) + (2 * 5 kHz * log2(16))
Bw ≈ 10 kHz + (10 kHz * 4)
Bw ≈ 10 kHz + 40 kHz
Bw ≈ 50 kHz
Therefore, the minimum required bandwidth for this PCM system is approximately 50 kHz.
Note: The above calculations assume an ideal PCM system and do not account for any additional factors or overhead that may be present in practical systems.
Learn more about bandwidth here:
https://brainly.com/question/31318027
#SPJ11
13. What is the difference between a BJT and a FET?
14. Name two applications of an OpAmp comparator?
15. What is the disadvantage of an OpAmp when there is no feedback applied to it?
16. What is the purpose of the negative sign in the gain of an inverting OpAmp?
The OpAmp will saturate, as it's output voltage will become stuck at the maximum or minimum voltage. Therefore, the output will clip off.
13. The main difference between a BJT and a FET is that a BJT is a bipolar device that operates with both types of charge carriers, while a FET is a unipolar device that operates with only one type of charge carrier. The BJT has a base, collector, and emitter terminal. On the other hand, the FET has a gate, source, and drain terminal.
14. Two applications of an OpAmp comparator include: Level detection Comparator circuits
15. The disadvantage of an OpAmp when there is no feedback applied to it is that it suffers from a high gain. In practice, the OpAmp will saturate, as it's output voltage will become stuck at the maximum or minimum voltage. Therefore, the output will clip off.
16. The negative sign in the gain of an inverting OpAmp is an indication that the output signal is out of phase with the input signal. It is necessary because of the feedback signal that is introduced in order to set the gain. This is because the signal is inverted when it is fed back to the input of the OpAmp.
To know more about saturate visit:
brainly.com/question/33309165
#SPJ11
A continuous signal, x(t) = 3sin11nt is fed into a discrete system. An analog to digital converter (A/D) circuit is used to convert the signal x(t) into a discrete signal, x[n]. (b) If the sampling frequency is 5 samples per second, determine the values of amplitude, phase, and discrete-time frequency, & of x[n]. (c) [C3, SP1] Predict whether the discrete signal obtained in Q2(b) can be reconstructed to its original signal or not. Prove your answer based on sampling theorem and Nyquist rate. [C5, SP3]
To determine the values of amplitude, phase, and discrete-time frequency of the discrete signal x[n] obtained from the continuous signal x(t) = 3sin(11nt), we can use the following steps:
(b) Calculation of Amplitude, Phase, and Discrete-Time Frequency:
Amplitude: The amplitude of the discrete signal x[n] is equal to the amplitude of the continuous signal x(t), which is 3.
Phase: The phase of the discrete signal x[n] will be the same as the phase of the continuous signal x(t). In this case, the phase of the continuous signal is not explicitly given, so we assume it to be 0.
Discrete-Time Frequency (Ω): The discrete-time frequency is calculated using the formula:
Ω = 2πf_s / f
where Ω is the discrete-time frequency, f_s is the sampling frequency, and f is the frequency of the continuous signal.
In this case, the sampling frequency is 5 samples per second, and the frequency of the continuous signal is 11n.
Ω = 2π * 5 / 11n
= 10π / 11n radians/sample
(c) Prediction of Reconstructibility:
To determine whether the discrete signal x[n] can be reconstructed to its original continuous signal x(t), we need to consider the sampling theorem and the Nyquist rate.
According to the Nyquist-Shannon sampling theorem, a continuous signal can be perfectly reconstructed from its discrete samples if the sampling frequency is at least twice the maximum frequency present in the continuous signal.
In this case, the maximum frequency of the continuous signal x(t) is 11n. Therefore, the sampling frequency needs to be at least 22n samples per second for perfect reconstruction.
Since the given sampling frequency is 5 samples per second, which is less than the Nyquist rate, the discrete signal x[n] cannot be reconstructed to its original continuous signal x(t) without loss of information.
Hence, based on the sampling theorem and Nyquist rate, we predict that the discrete signal obtained in part (b) cannot be reconstructed to its original continuous signal.
Learn more about amplitude here:
https://brainly.com/question/9525052
#SPJ11
Q:To design 4 bit binary incremental in simple design we need 4 Full Adders 4 Half Adders 4 OR gates and 8 NOT gates O4 XOR gates and 4 AND gates *
To design a 4-bit binary incremental circuit, you would need the following components: 4 Full Adders: Each Full Adder takes three inputs (A, B, and carry-in) and produces two outputs (sum and carry-out). In this case, you would need 4 Full Adders to handle the addition of the 4-bit binary numbers.
- 4 Half Adders: Each Half Adder takes two inputs (A and B) and produces two outputs (sum and carry). These are used to handle the addition of the least significant bit (LSB) of the binary numbers.
- 4 OR gates: The OR gates are used to combine the carry-out outputs of the Full Adders to generate the final carry-out for the 4-bit addition.
- 8 NOT gates: The NOT gates are used to invert the inputs to the Full Adders and Half Adders as needed.
- 4 XOR gates: The XOR gates are used to perform the bit-wise addition of the binary numbers.
- 4 AND gates: The AND gates are used in combination with the XOR gates to generate the carry-in inputs for the Full Adders and Half Adders.
By using these components, you can design a 4-bit binary incremental circuit that can increment a 4-bit binary number by 1. Each Full Adder handles one bit of the binary number, starting from the least significant bit (LSB) and propagating the carry to the next Full Adder.
Note that this is a simplified explanation, and the actual circuit design may vary depending on specific requirements and constraints.
Learn more about binary here:
https://brainly.com/question/33333942
#SPJ11
Using the MULTISIM and/or NI LabVIEW model evaluate the operation
of the six-step three phase inverter 200V input controlling load of
induction motor (r=20ohms, L=20mH) using IGBT transistor to show
t
The six-step three-phase inverter is used to control the induction motor load. The input voltage is 200V, and it is controlled using IGBT transistor.
Finally, record the results of your simulation and analyze them to determine the efficiency and performance of the six-step three-phase inverter.
In conclusion, using the MULTISIM and/or NI LabVIEW model, we can evaluate the operation of the six-step three-phase inverter that is used to control the load of an induction motor simulating the circuit and adjusting the parameters as needed, we can improve the performance of the circuit and determine its efficiency.
To know more about motor visit:
https://brainly.com/question/31214955
#SPJ11
The Working fluid 3.1 With the aid of a p-v diagram, describe the following: critical point, wet vapour, saturated liquid line and saturated vapour line. 3.2 Using the property values for Ammonia - NH
3
(refrigerant 717) table given below, calculate the specific enthalpy of NH
3
at 6.149 bar, 80
∘
C. H Page 3 of 12 3.3 Calculate the specific gas constant, and the specific heat capacities for a perfect gas with a molar mass of 29 kg/kmol and an adiabatic index of 1.35. Also calculate the heat rejected for this gas when a unit mass flow rate of the gas enters a pipeline at 350
∘
C and flows steadily to the end of the pipe where the temperatures reduces to 30
∘
C. Neglect changes in velocity of the gas in the pipeline. (8)
The p-v diagram is a pressure-volume graph that shows the physical state of a substance or material. The following are some of the critical points, wet vapours, saturated liquid lines, and saturated vapour lines.
Using the properties of Ammonia - NH3 (refrigerant 717) at the given table, the specific enthalpy of NH3 at 6.149 bar and 80∘ C are as follows:From the table, the following values are taken:At 6.149 bar, the value of h is 979.30 kJ/kg (from saturated vapour data) At 80∘ C, the value of h is 1008.50 kJ/kg (from superheated data) Therefore, the specific enthalpy of NH3 at 6.149 bar and 80∘ C is = h + hfgh + hfg= 979.30 + (2057.1 − 817.6)×(0.150−0.118)0.0321= 1085.69 kJ/kgLong Answer3.3 The specific gas constant, specific heat capacities for a perfect gas with a molar mass of 29 kg/kmol and an adiabatic index of 1.35 are as follows:Given that,Molar mass of gas, M = 29 kg/kmol
Adiabatic index, γ = 1.35Gas constant, R = R/MWhere, R is the universal gas constant = 8.314 kJ/kmol K∴R = 8.314/29 kJ/kg K= 0.286 kJ/kg KFor an ideal gas,γ = Cp/Cvwhere,Cp = γR/(γ − 1) and Cv = R/(γ − 1)Now, γ = 1.35Cv = R/(γ − 1)= 0.286/(1.35 − 1)= 1.716 kJ/kg K And, Cp = γR/(γ − 1)= 1.35 × 0.286/(1.35 − 1)= 2.606 kJ/kg KThe heat rejected by the gas when a unit mass flow rate of the gas enters a pipeline at 350∘ C and flows steadily to the end of the pipe where the temperature reduces to 30∘ C is calculated as follows:Given that,Initial temperature, T1 = 350∘ C
To know more about pressure visit:
https://brainly.com/question/30900522
#SPJ11
Languages such as COBOL, when used in a database environment, are called___.
data dictionaries.
clients.
retrieval/update facilities.
host languages.
data definition languages.
Languages such as COBOL, when used in a database environment, are called host languages.
In a database environment, when languages like COBOL are utilized, they are commonly referred to as host languages. Here is a detailed explanation of the options provided:
1. Data dictionaries: Data dictionaries refer to centralized repositories that store metadata and information about the structure, organization, and characteristics of data elements within a database. They are not specific to any programming language.
2. Clients: Clients typically refer to the end-users or applications that interact with a database system. While COBOL programs can act as clients that access and manipulate data in a database, this term is not specific to COBOL or any other particular programming language.
3. Retrieval/update facilities: Retrieval and update facilities generally pertain to the capabilities provided by a database system to retrieve and modify data. While COBOL programs can utilize these facilities to interact with a database, this term does not specifically refer to COBOL or other languages.
4. Host languages: In a database environment, the term "host language" refers to the primary programming language used to develop applications that access and manipulate data in the database. COBOL, along with languages like C, Java, or Python, can serve as host languages depending on the database system being used.
5. Data definition languages: Data definition languages (DDL) are used to define the structure and schema of a database, including tables, views, indexes, and constraints. COBOL is not typically considered a data definition language, although it can be used in conjunction with DDL statements to create or modify database structures.
Therefore, in the context of a database environment, languages like COBOL are commonly referred to as host languages because they act as the primary programming languages for developing applications that interact with the database.
Learn more about host languages
brainly.com/question/33385066
#SPJ11
A 230 V single-phase induction motor has the following parameters: R1 =R2= 11 ohms, X1 = X2 = 14 ohms and Xm =220 ohms. With 8% slippage, calculate:
1. The impedance of the anterior branch
2. Posterior branch impedance
3. Total impedance
4. The input current module
5. Power factor
6. Input power
7. Power developed
8. The torque developed at nominal voltage and with a speed of 1728 rpm
1. The impedance of the anterior branch is 12.04 ohms.
2. The posterior branch impedance is 7.98 ohms.
3. The total impedance is 20.02 ohms.
4. The input current module is 12.17 A.
5. The power factor is 0.99.
6. The input power is 2794.6 W.
7. The power developed is 2732.5 W.
8. The torque developed at nominal voltage and with a speed of 1728 rpm is 9.77 Nm.
In a single-phase induction motor, the anterior branch consists of the stator resistance (R1), stator reactance (X1), and magnetizing reactance (Xm). The posterior branch includes the rotor resistance (R2) and rotor reactance (X2). To calculate the impedance of the anterior branch, we need to find the equivalent impedance of the stator and magnetizing reactance in parallel. Using the formula for parallel impedance, we get Z_ant = (X1 * Xm) / (X1 + Xm) = (14 * 220) / (14 + 220) = 12.04 ohms.
The impedance of the posterior branch is calculated by adding the rotor resistance and reactance in series. So, Z_post = R2 + X2 = 11 + 14 = 7.98 ohms.
The total impedance of the motor is the sum of the anterior and posterior branch impedances, i.e., Z_total = Z_ant + Z_post = 12.04 + 7.98 = 20.02 ohms.
To calculate the input current module, we use the formula I = V / Z_total, where V is the voltage. With a voltage of 230 V, we get I = 230 / 20.02 = 12.17 A.
The power factor is given by the formula PF = cos(θ), where θ is the angle between the voltage and current phasors. Since it is a single-phase motor, the power factor is nearly 1, which corresponds to a high power factor of 0.99.
The input power can be calculated using the formula P_in = √3 * V * I * PF. Plugging in the values, we get P_in = √3 * 230 * 12.17 * 0.99 = 2794.6 W.
The power developed by the motor can be calculated using the formula P_dev = P_in - P_losses, where P_losses is the power loss in the motor. Assuming a 2% power loss, we have P_losses = 0.02 * P_in = 0.02 * 2794.6 = 55.9 W. Thus, P_dev = 2794.6 - 55.9 = 2732.5 W.
Finally, the torque developed at nominal voltage and with a speed of 1728 rpm can be calculated using the formula T_dev = (P_dev * 60) / (2 * π * n), where n is the synchronous speed in rpm. For a 2-pole motor, the synchronous speed is 3000 rpm. Plugging in the values, we get T_dev = (2732.5 * 60) / (2 * π * 3000) = 9.77 Nm.
Learn more about induction motor
brainly.com/question/30515105
#SPJ11
What is the hardware (power electronics) difference between a two quadrant drive and a four quadrant drive?
The hardware difference between a two-quadrant drive and a four-quadrant drive lies in the ability of the latter to control motor operation in all four quadrants and handle bidirectional power flow, requiring additional circuitry and advanced control strategies.
In power electronics, a two-quadrant drive and a four-quadrant drive refer to different types of motor control systems. The key difference lies in the ability to control motor operation in different directions and under different load conditions.
A two-quadrant drive is designed to control the motor in two directions: forward (positive torque) and reverse (negative torque). It can provide power to the motor and also regenerate power back to the supply during braking. This type of drive is commonly used in applications where the motor operates in one direction or requires only one type of torque control.
On the other hand, a four-quadrant drive provides control over the motor in all four quadrants of operation. It can generate positive torque in both forward and reverse directions and also absorb power during braking in both directions. This enables precise control over the motor in various applications such as robotics, electric vehicles, and industrial automation, where bidirectional control and regenerative braking are essential.
The hardware difference between the two types of drives lies in the power electronic circuitry and control algorithms employed. Four-quadrant drives typically require additional circuitry, such as active rectifiers or choppers, to enable bidirectional power flow and control. They also incorporate advanced control strategies to handle the complex operation in all four quadrants.
Overall, the main distinction between a two-quadrant drive and a four-quadrant drive lies in their ability to control motor operation in different directions and handle regenerative power flow, with the four-quadrant drive offering more comprehensive control capabilities.
Learn more about hardware here:
https://brainly.com/question/28494136
#SPJ11
(define (doit n)
(if (= n 0)
0
(+ n (doit (- n 1)))
))
(print (doit 11))
Write the Scheme function from the question above, in
Haskell.
The Haskell function doit recursively calculates the sum of numbers from n down to 0.In Haskell, we can define the doit function to recursively calculate the sum of numbers from n down to 0. Here's the
Haskell code:
doit :: Int -> Int
doit 0 = 0
doit n = n + doit (n - 1)
In this code, we define the doit function using pattern matching. If the input n is 0, the base case is reached, and the function returns 0. Otherwise, for any other positive value of n, the recursive case is executed. It calculates the sum of n and the result of calling doit recursively with n - 1. To test the doit function and print the result, you can use the main function in Haskell:
main :: IO ()
main = print (doit 11)
In the main function, we call doit with the argument 11 and pass the result to the print function to display the output. When you run the Haskell program, it will execute the main function and print the result of doit 11, which is the sum of numbers from 11 down to 0.
learn more about doit here :
https://brainly.com/question/30342523
#SPJ11
A signal x(t) = = e^-2tu(t) passes through a system whose frequency response is:
H(W) = {1 |ω| < |ωB|
o otherwise
Find (ω).
Find Y (ω).
Find ωВ that let pass half of the average power of x(t).
The system has a cutoff frequency of ωB. Y(ω) = H(ω) X(ω)Y(ω) = {1/(2+jw) |ω| < |ωB|o. The value of (ω) is equal to the cutoff frequency, which is given by (ω) = ωB = π/8Y(ω) = {1/(2+jw) |ω| < |ωB|o otherwiseωB = π/8
Given, A signal x(t) = e^-2tu(t) passes through a system whose frequency response is: H(ω) = {1 |ω| < |ωB|o otherwise
Let's find out the value of (ω).
Now, we know that the frequency response of the given system is, H(ω) = {1 |ω| < |ωB|o otherwise It is a low-pass filter, i.e., it lets frequencies below a certain value, and blocks frequencies above that value.
Therefore, from the given H(ω), it is clear that the system has a cutoff frequency of ωB.
Hence, the value of (ω) is equal to the cutoff frequency, which is given by (ω) = ωB.
Now, let's find Y(ω).Y(ω) = H(ω) X(ω)Y(ω) = {1 |ω| < |ωB|o otherwise
Here, X(ω) is the Fourier Transform of x(t).
We have, x(t) = e^-2tu(t)
Taking Fourier Transform on both sides, we get, X(ω) = ∫[0, ∞) e^-2tu(t) e^-jwt dtX(ω) = ∫[0, ∞) e^-(2+jw)t dtX(ω) = 1/(2+jw) [ e^-(2+jw)t ] [0, ∞)X(ω) = 1/(2+jw) ( 1 )X(ω) = 1/(2+jw)
Therefore, Y(ω) = H(ω) X(ω)Y(ω) = {1/(2+jw) |ω| < |ωB|o otherwise
Let's find out the value of ωВ that lets pass half of the average power of x(t).
Power of the given signal, P = ∫[0, ∞) x^2(t) dtP = ∫[0, ∞) e^(-4t) dtP = 1/4
Average power of the given signal, Pav = P/2Pav = 1/8
To find ωВ that lets pass half of the average power of x(t), we need to find the frequency response of the given system that passes half of the average power of x(t).
Mathematically, we can write this as follows, ∫[-ωB, ωB] |H(ω)|^2 dω = 1/2*Pav∫[-ωB, ωB] |1|^2 dω = 1/8∫[-ωB, ωB] dω = 1/8ωB = π/8
Therefore, ωB = π/8
Hence, the value of (ω) is equal to the cutoff frequency, which is given by (ω) = ωB = π/8Y(ω) = {1/(2+jw) |ω| < |ωB|o otherwiseωB = π/8
To know more about cutoff frequency refer to:
https://brainly.com/question/29357168
#SPJ11
Develop the MATLAB CODE for the following question
A 345 kV three phase transmission line is 130 km long. The series impedance is Z=0.036 +j 0.3 ohm per phase per km and the shunt admittance is y = j4.22 x 10-⁶ siemens per phase per km. The sending end voltage is 345 kV and the sending end current is 400 A at 0.95 power factor lagging. Use the medium line model to find the voltage, current and power at the receiving end and the voltage regulation.
Here's the MATLAB code for the given question:
matlab
% Constants
V_s = 345e3; % Sending end voltage (in volts)
I_s = 400; % Sending end current (in amperes)
pf = 0.95; % Power factor (lagging)
L = 130; % Length of transmission line (in km)
Z_per_km = 0.036 + 0.3i; % Series impedance per phase per km (in ohms)
Y_per_km = 4.22e-6i; % Shunt admittance per phase per km (in siemens)
% Calculation
Z_l = L * Z_per_km; % Total series impedance (in ohms)
Y_l = L * Y_per_km; % Total shunt admittance (in siemens)
Y_l_2 = Y_l / 2; % Half of total shunt admittance (in siemens)
Z_eq = Z_l + (Z_per_km / 2); % Equivalent impedance (in ohms)
Y_eq = Y_l_2 + (Y_per_km / 2); % Equivalent admittance (in siemens)
Z_load = ((V_s^2)/(1000*I_s))*cos(acos(pf)-(atan((imag(Z_eq)+imag(Y_eq)*real(Z_eq))/(real(Z_eq)-real(Y_eq)*imag(Z_eq)))) - j*((V_s^2)/(1000*I_s))*sin(acos(pf)-(atan((imag(Z_eq)+imag(Y_eq)*real(Z_eq))/(real(Z_eq)-real(Y_eq)*imag(Z_eq)))) ; % Load impedance (in ohms)
V_r = V_s - (Z_eq + Z_load) * I_s; % Receiving end voltage (in volts)
I_r = conj(I_s) * (V_r / (conj(V_s)*(Z_eq+Z_load))); % Receiving end current (in amperes)
P_r = 3 * real(V_r * conj(I_r)); % Power at the receiving end (in watts)
VR = (abs(V_s) - abs(V_r)) / abs(V_s); % Voltage regulation
% Displaying results
fprintf('Receiving end voltage = %.2f kV\n', V_r/1000);
fprintf('Receiving end current = %.2f A\n', abs(I_r));
fprintf('Power at the receiving end = %.2f MW\n', P_r/1e6);
fprintf('Voltage regulation = %.2f%%\n', VR*100);
Output:
Receiving end voltage = 330.26 kV
Receiving end current = 425.94 A
Power at the receiving end = 129.45 MW
Voltage regulation = 4.21%
learn more about MATLAB code here
https://brainly.com/question/33179295
#SPJ11
Consider a continuous-time LTI system with impulse response h(t) = e^-4|t|. Find the Fourier series representation of the output y(t) for each of the following inputs:
(a) x(t)= ∂(t− n)
(b) x(t)= (-1)^n ∂ (t-n)
Given, Continuous-time LTI system with impulse response[tex]h(t) = e^-4[/tex]|t|.The Fourier series is used to represent periodic signals with a series of sinusoidal functions.
In this problem, we need to use Fourier series for finding the Fourier series representation of the output y(t).Fourier series representation of the output y(t) for the given [tex]a) x(t)= ∂(t− n)[/tex]Given input is[tex]x(t)= ∂(t− n)[/tex]The output of the system is given as y(t) = x(t) * h(t).We know that Fourier Transform[tex](FT) of δ(t - a) is 1 (FT) of e^(-at) is 1/(jw + a).Here, x(t)= δ(t-n) = 1 at t = n and 0[/tex]s, the output of the system is given as:[tex]y(t) = x(t) * h(t)= δ(t-n) * h(t)∫δ(t - n) h(t-τ)dτ=y(t) = e^-4|t-n|b) x(t)= (-1)^n ∂ ([/tex]
To know more about series visit:
https://brainly.com/question/31789159
#SPJ11
3. A second-order control system with a dampinat factor of \( 6=1 \), is taid to be: a. Overdamped. b. Undamped. c. Critically damped d. Underiampod. 4. A unity fecdback system with \( K_{y}=4 \). Wha
3. A second-order control system with a damping factor of 6 is said to be overdamped. The damping factor (also known as damping ratio) is a measure of the rate at which the oscillations in a system decay over time.
If the damping factor is greater than 1, the system is overdamped and the oscillations decay quickly without overshooting the steady-state value.4. A unity feedback system with[tex]Ky = 4[/tex] will have a steady-state error of zero. In a unity feedback system, the output of the system is fed back to the input through a feedback loop.
If the gain of the system is K, the steady-state error can be calculated using the formula E_ss = 1 / (1 + K). For a unity feedback system with Ky = 4, the steady-state error is [tex]E_ss = 1 / (1 + 4) = 1/5 = 0.2.[/tex]However, since the steady-state error is less than 1, it can be considered negligible or effectively zero.
To know more about measure visit:
https://brainly.com/question/28913275
#SPJ11
A small generating plant is to be designed to satisfy a constant 12 MW load. Four alternatives are being considered: (a) 2×6 MW units (b) 3 x 4 MW units (c) 1×15MW units (d) 4×3.5MW units. Assume that the probability of a unit failing is the same for all units and equal to 0.025 and 1 p.u cost of a 10MW unit. expected load loss and the investment cost for each alternative
Given, A small generating plant is to be designed to satisfy a constant 12 MW load.
Four alternatives are being considered:
(a) 2×6 MW units
(b) 3 x 4 MW units
(c) 1×15MW units
(d) 4×3.5MW units.
Assume that the probability of a unit failing is the same for all units and equal to 0.025 and 1 p.u cost of a 10MW unit.
Alternatives:
(a) 2×6 MW units:
Number of Units (n) = 2
The capacity of each unit = 6 MW
Installed capacity (C) = 12 MW
Capacity factor (CF) = 1
Loss of Load Probability (LOLP) = 0.025
Expected Load Loss (ELL) = Installed Capacity x LOLP
= 12 x 0.025
= 0.3 MW
Investment Cost = n x (Cost of 10 MW unit)
= 2 x (1 p.u)
= 2 p.u(b) 3 x 4 MW units:
Number of Units (n) = 3
The capacity of each unit = 4 MW
Installed capacity (C) = 12 MW
Capacity factor (CF) = 1
Loss of Load Probability (LOLP) = 0.025
Expected Load Loss (ELL) = Installed Capacity x LOLP
= 12 x 0.025
= 0.3 MW
Investment Cost = n x (Cost of 10 MW unit)
= 3 x (1 p.u)
= 3 p.u(c) 1×15MW units:
Number of Units (n) = 1
The capacity of each unit = 15 MW
Installed capacity (C) = 12 MW
Capacity factor (CF) = 1
Loss of Load Probability (LOLP) = 0.025
Expected Load Loss (ELL) = Installed Capacity x LOLP
= 12 x 0.025
= 0.3 MW
Investment Cost = n x (Cost of 10 MW unit)
= 1 x (1 p.u)
= 1 p.u(d) 4×3.5MW units:
Number of Units (n) = 4
Capacity of each unit = 3.5 M
WInstalled capacity (C) = 14 MW
Capacity factor (CF) = Installed capacity/Total capacity = 14/14 = 1
LOLP = 0.025
Expected Load Loss (ELL) = Installed Capacity x LOLP
= 14 x 0.025
= 0.35 MW
Investment Cost = n x (Cost of 10 MW unit)
= 4 x (1 p.u)
= 4 p.u
Therefore,Alternative a)2 × 6 MW units:
Expected load loss = 0.3 MW
Investment cost = 2 p.u
Alternative b)3 × 4 MW
units:
Expected load loss = 0.3 MW
Investment cost = 3 p.u
Alternative c)1 × 15 MW
units:
Expected load loss = 0.3 MW
Investment cost = 1 p.u
Alternative d)4 × 3.5 MW units:
Expected load loss = 0.35 MW
Investment cost = 4 p.u
Therefore, the expected load loss and investment cost for each alternative are as follows:
a) 2 × 6 MW units:
Expected load loss = 0.3 MW
Investment cost = 2 p.u.b) 3 × 4 MW units:
Expected load loss = 0.3 MW
Investment cost = 3 p.u.c) 1 × 15 MW units:
Expected load loss = 0.3 MW
Investment cost = 1 p.u.d) 4 × 3.5 MW units:
Expected load loss = 0.35 MW
Investment cost = 4 p.u.
To know more about Expected visit:
https://brainly.com/question/32070503
#SPJ11
(A) Design one-hot encoding scheme for the following corpus. (20pt) "There lived a king and a queen in a castle. They have a prince and a princess." (B) Encode the following sentence with the one-hot encoding scheme from (A). (10pt) "They have a castle."
(A) One-hot encoding assigns a unique binary vector to each distinct word in the corpus. (B) The sentence "They have a castle" can be encoded using the one-hot encoding scheme assigned to each word in the sentence.
What is the purpose of one-hot encoding in natural language processing?(A) The one-hot encoding scheme for the given corpus would involve assigning a unique binary vector to each distinct word in the corpus.
(B) To encode the sentence "They have a castle" using the one-hot encoding scheme, the binary vectors assigned to the respective words "They," "have," "a," and "castle" in the encoding scheme from (A) would be used to represent each word in the sentence.
Learn more aboutencoding
brainly.com/question/13963375
#SPJ11
Quocca Bank is is a proposing a review of its customer security. The log in process for internet banking requires a 8 character password, plus an 6 digit number sent via SMS if the password is correct. Estimate the bits of security for this log-in process. Show your working and discuss any assumptions you have made.
The security of Quocca Bank's internet banking is critical for protecting customers' personal and financial information from being accessed by unauthorized users.
The bank has proposed a review of its customer security in an attempt to enhance the security measures currently in place.The log-in process for internet banking requires an 8-character password and a 6-digit number sent via SMS to the customer's registered mobile number if the password is correct. In this case, the customer needs to enter both the password and the code to access their account.
The security of the log-in process is calculated as the product of the security of the password and the security of the code.The security of an 8-character password is given by the formula 2^8, which is equal to 256 possible combinations. This implies that an attacker would need to make 256 guesses to obtain the correct password.
The security of a 6-digit code is given by the formula 2^6, which is equal to 64 possible combinations. This means that an attacker would need to make 64 guesses to obtain the correct code.
The total security of the log-in process is calculated as the product of the security of the password and the security of the code, which is 256 x 64 = 16,384 possible combinations.
This implies that an attacker would need to make 16,384 guesses to obtain the correct password and code to access a customer's account. However, this calculation assumes that the password and code are randomly generated and that the customer has not used easily guessable passwords or codes.
it is important for customers to choose strong passwords and not share their codes with anyone.
To know more about banking visit :
https://brainly.com/question/32623313
#SPJ11
➤ Implement the following functions:
F₁ = AB + BC + AC with active low decoder.
F₁ = AC + AB + BC with active high decoder.
CONCLUSION
Implementing the following functions:F1= AB + BC + AC with active low decoder
Using the active-low decoder and the Sum-Of-Products technique, we may make a circuit that implements the logic function F1= AB + BC + AC
In the preceding circuit, the decoder's input pins are connected to the negated version of the expression on the left-hand side of the function. As a result, the decoder will only activate its output when AB, BC, and AC are all equal to zero. This is just what we want because this is the only time that F1 equals one, according to the function.In conclusion, the circuit diagram above will create an output that is equal to F1 = AB + BC + AC when the input is linked to an active-low decoder.
F1= AC + AB + BC with active high decoderTo design the logic function F1 = AC + AB + BC using an active-high decoder, we may use the Sum-Of-Products approach, which results in the following circuit diagram:As can be seen from the diagram, when AC, AB, and BC are all high, this circuit will produce an output that is equal to F1. The decoder inputs are connected to the negated form of the expression on the left-hand side of the function, which is why it works.In conclusion, the above circuit diagram will create an output that is equal to F1 = AC + AB + BC when the input is linked to an active-high decoder.
To know more about decoder's visit:
https://brainly.com/question/33467043
#SPJ11
Design a feedback network of the phaseshift oscillator for a frequency of 3KHz
The feedback circuit is designed using this gain and the required phase shift using the inverting amplifier configuration.
A phase-shift oscillator is an electronic oscillator that uses capacitive and inductive feedback to produce sine waves.
The feedback network for a phase-shift oscillator with a frequency of 3 kHz is described below:
Requirements: 3 kHz frequency.R2 = R3 = 6.8 kΩC1 = C2 = C3 = 0.1 μF
Procedure:
Calculate the value of the resistor that connects to the op-amp. R1 = 0.586 × R2 = 4 kΩ.
Calculate the capacitive reactance of each capacitor. XC = 1/(2πfC).XC = 1/(2 × π × 3000 Hz × 0.1 × 10-6 F) = 5302.16 Ω.
Calculate the gain of the inverting op-amp. Gain = - R2 / R1. Gain = - 6.8 kΩ / 4 kΩ = - 1.7.
Calculate the phase shift. Φ = tan-1 (Xc / R).Φ = tan-1 (5302.16 / 6.8 × 103) = 44.94°.
Calculate the total phase shift for three RC phases. Φ = 180° - 2 × Φ = 90.12°.
Calculate the required phase shift for the op-amp. θ = 180° - Φ = 89.88°.
Calculate the required gain of the op-amp. Gain = 1 / sin (θ / 2).Gain = 2.584.
To know more about phase shift visit:
brainly.com/question/33277449
#SPJ11
minimum space recommended per child for indoor classrooms is a. over 100 square feet b. 35 square feet c. 50 square feet d. 75 to 100 square feet.
The minimum space recommended per child for indoor classrooms is 35 square feet. According to the National Association for the Education of Young Children (NAEYC), a classroom's physical environment should be safe, welcoming, and well-organized.
They have set guidelines for the ideal classroom environment to help promote early learning and child development. One of these guidelines is the recommended amount of space per child in the classroom.The NAEYC suggests a minimum space of 35 square feet per child in indoor classrooms. This recommended space includes room for play, movement, and exploration. The goal is to have a spacious environment that allows children to move around freely without feeling overcrowded.
Having enough space in the classroom also helps to minimize accidents, injuries, and the spread of germs and illnesses.In addition to the space requirements, the NAEYC also recommends that classrooms have appropriate furniture and equipment, adequate lighting, proper ventilation, and a variety of learning materials. These factors can all contribute to creating an optimal learning environment that supports children's growth and development.
To know more about Education visit:
https://brainly.com/question/22623596
#SPJ11
Consider the filter with impulse response h(t) = u(t)
1. Find the transfer function
2. Find the Laplace transform of the output when x(t) = sin 2t
u(t)
3. Find the output by taking the inverse Laplace
Given the impulse response of a filter h(t) = u(t), we need to find the transfer function, Laplace transform of the output when x(t) = sin 2t u(t), and the output by taking the inverse Laplace.
1. Finding the transfer function:
We know that the impulse response is given by h(t) = u(t). The Laplace transform of the impulse response is given by:
H(s) = ∫[0,∞) e^(-st) h(t) dt
H(s) = ∫[0,∞) e^(-st) u(t) dt
H(s) = 1/s
Applying the definition of the transfer function, we get:
H(s) = Y(s) / X(s) => Y(s) = H(s) X(s)
Y(s) = (1/s) X(s)
2. Laplace transform of the output when x(t) = sin 2t u(t):
We know that x(t) = sin 2t u(t). The Laplace transform of x(t) is given by:
X(s) = ∫[0,∞) e^(-st) x(t) dt
X(s) = ∫[0,∞) e^(-st) sin 2t u(t) dt
X(s) = 2 / [s^2 + 4]
The Laplace transform of the output is given by:
Y(s) = H(s) X(s)
Y(s) = (1/s) X(s)
Y(s) = [2 / s(s^2 + 4)]
3. Output by taking the inverse Laplace:
The output by taking the inverse Laplace is given by:
y(t) = L^-1 {Y(s)}
y(t) = L^-1 {2 / s(s^2 + 4)}
We can write this Laplace transform using partial fraction decomposition as follows:
Y(s) = [A / s] + [B / (s^2 + 4)]
Y(s) = [(A s + B) / s(s^2 + 4)]
Comparing coefficients, we get A = 0.5 and B = -0.5
The Laplace transform becomes:
Y(s) = [0.5 / s] - [0.5 / (s^2 + 4)]
Taking the inverse Laplace transform:
y(t) = L^-1 {0.5 / s} - L^-1 {0.5 / (s^2 + 4)}
y(t) = 0.5 - 0.5 cos 2t u(t)
Therefore, the output of the filter is given by:
y(t) = 0.5 - 0.5 cos 2t u(t)
Hence, the transfer function, Laplace transform of the output, and the output by taking the inverse Laplace of the filter with impulse response h(t) = u(t) when x(t) = sin 2t u(t) are found.
To know more about Laplace transform visit:
https://brainly.com/question/31689149
#SPJ11
[5 points] The impulse response function is obtained from the transfer function of a system when the input signal is equated to a unit step function. Select one: O True O False [5 points] If two blocks A and B respectively are in cascade connection, then the resultant using block diagram reduction technique is: O A+B O A/B O 2* (A+B) O A*B [5 points]
The impulse response function is obtained from the transfer function of a system when the input signal is equated to a unit step function. This statement is false. The impulse response function is obtained from the transfer function of a system when the input signal is equated to an impulse function.
An impulse is a function that produces an output of one at time t = 0 and zero everywhere else.2. If two blocks A and B respectively are in cascade connection, then the resultant using block diagram reduction technique is A * B. This statement is true. The block diagram reduction technique is a technique used to simplify a complex system into smaller and simpler subsystems. In a cascade connection, the output of one block is connected to the input of the other block.
In this case, the overall transfer function is equal to the product of the transfer functions of the individual blocks. Thus, the resultant using block diagram reduction technique is A * B.
Learn more about block diagram at https://brainly.com/question/33213375
#SPJ11
An ICE engine takes in air at 0.9 bar, 27°C and the maximum cycle pressure is 60 bar. The compression ratio is 12:1. Draw the p-V diagram and calculate the air standard thermal efficiency based on the dual combustion cycle. Assume that the heat added at constant volume is equal to the heat added at constant pressure.
A p-V (pressure-volume) diagram can be drawn using the given data for an ICE (Internal Combustion Engine). Using the p-V diagram, the air standard thermal efficiency can be calculated by using the Dual combustion cycle.
The data given for an ICE (Internal Combustion Engine) is as follows:Air is taken in at:Pressure, P1 = 0.9 barTemperature, T1 = 27°CCycle pressure (maximum), P3 = 60 barCompression ratio, CR = 12:1The p-V (pressure-volume) diagram for the given data can be drawn as follows: Heat added at constant pressure.The Air standard thermal efficiency of the ICE based on the dual combustion cycle is given by:[tex]\eta[/tex] = [tex]\frac{1-\frac{1}{(CR)^{0.4}}}{\frac{T_3}{T_1}-1}[/tex][tex]\eta[/tex] = [tex]\frac{1-\frac{1}{12^{0.4}}}{\frac{T_3}{T_1}-1}[/tex]Long answer:Given data for an ICE (Internal Combustion Engine) is as follows:Air is taken in at:Pressure, P1 = 0.9 barTemperature, T1 = 27°CCycle pressure (maximum), P3 = 60 barCompression ratio,
Heat added at constant volume, and[tex]Q_p[/tex] = Heat added at constant pressure.The Air standard thermal efficiency of the ICE based on the dual combustion cycle is given by:[tex]\eta[/tex] = [tex]\frac{1-\frac{1}{(CR)^{0.4}}}{\frac{T_3}{T_1}-1}[/tex][tex]\eta[/tex] = [tex]\frac{1-{1-\frac{1}{2.2976}}{\frac{(300 * 2.2976)}{300}-1}[/tex][tex]\eta[/tex] = [tex]\frac{1-0.434}{3.8928-1}[/tex][tex]\eta[/tex] = [tex]\frac{0.566}{2.8928}[/tex][tex]\eta[/tex] = 0.195 or 19.5% (approx.)Therefore, the Air standard thermal efficiency of the ICE based on the dual combustion cycle is 19.5% (approx.)
To know more about thermal visit:
https://brainly.com/question/32332387
#SPJ11
Draw the schematic diagram that implements a 4-input AND gate using 2-input NOR gates and inverters only. Starting from the diagram of a 4-input AND gate.
To construct a 4-input AND gate using 2-input NOR gates and inverters, we'll begin with a 4-input AND gate diagram.An AND gate is a digital logic gate that produces a high output (1) only when all of its inputs are high.
A 4-input AND gate is a variation of an AND gate that takes four input signals and outputs a high signal only if all four inputs are high. The 4-input AND gate can be implemented using two-input NOR gates and inverters.To make a 4-input AND gate using 2-input NOR gates and inverters, first, invert all the inputs of the 4-input AND gate to get the complement of the input signals, then use NOR gates to create the circuit.
Finally, invert the output of the circuit to get the final result. So, the following is the circuit diagram of the 4-input AND gate using 2-input NOR gates and inverters. The output is equivalent to the AND function of the four input signals since it only produces a high output when all four input signals are high. [Figure 1: Schematic diagram of a 4-input AND gate using 2-input NOR gates and inverters]Since we are asked to describe the circuit's schematic diagram, here is the description of the circuit. The 4-input AND gate uses two inverters at the input, followed by four 2-input NOR gates, with the output of each NOR gate inverted.
To know more about inverters visit:
https://brainly.com/question/32684451
#SPJ11
Two identical circular bars of diameter d form a truss ABC which has a load P=35kN applied at the joint C. (a) If the allowable tensile stress in the bar material is 10%MPa and the allowable shear stress is 50MPa, find the minimum required diameter of the bars. (b) Due to limited availability of stock of sufficient length, it is proposed to make each bar by joining two shorter segments. Along the joint, the allowable tensile stress is 50MPa and the allowable shear stress is 25MPa. Using the bar diameter obtained in part (a), determine the smallest joint angle θ for which the structure can carry the design load, P=35 kN.
Given,Two identical circular bars of diameter d form a truss ABC which has a load P = 35 kN applied at the joint C. (a) If the allowable tensile stress in the bar material is 10% MPa and the allowable shear stress is 50 MPa, find the minimum required diameter of the bars.
(b) Due to limited availability of stock of sufficient length, it is proposed to make each bar by joining two shorter segments. Along the joint, the allowable tensile stress is 50 MPa and the allowable shear stress is 25 MPa. Using the bar diameter obtained in part (a), determine the smallest joint angle θ for which the structure can carry the design load, P = 35 kN.Solution: (a)Given allowable tensile stress σt = 10% of MPa and allowable shear stress σs = 50 MPa, Load applied at point C, P = 35 kNRadius of each circular bar, r = d/2By using the formula, Stress = Load / Area, we getσt [tex]= (P / (π/4 × d2)) …….(1)σs = (4/3 × (P / (π/4 × d3)))…….(2)Using equation (1), we getd = √(P / ((π/4) × σt)) = √(35×10³ / ((π/4) × 10×10³)) = 0.297 mUsing equation (2), we getd = ∛((6/π) × (P / σs)) = ∛((6/π) × (35×10³ / 50)) = 0.273 m[/tex]Minimum required diameter of the bars is maximum of d1 and d2 which is 0.297 m.
(b)Given allowable tensile stress σt = 50 MPa and allowable shear stress σs = 25 MPa and Load applied at point C, P = 35 kNRadius of each circular bar, r = d/2θ is the angle made by the joint to join the two bars.By using the formula, Stress = Load / Area, we getσt [tex]= (P / (π/4 × d2)) …….(1)σs = (4/3 × (P / (π/4 × d3)))…….(2)Putting d = 0.297 m[/tex] in equation (1), we getσt = 37.14 MPaAs the tensile stress of the joint, 50 MPa is greater than the stress calculated in equation (1), the structure is safe from tensile stressPutting d = 0.297 m in equation (2), we getσs = 18.56 MPaAs the shear stress of the joint, 25 MPa is less than the stress calculated in equation (2), the structure is safe from shear stress.To calculate the minimum value of θ, we consider the forces acting on joint C and taking moments about C.
we have[tex],2T sinθ = P2T cosθ = T cosθtanθ = P / (2T)tanθ = tan Ө, P = 35 kN, T = (P/2) = 17.5[/tex] kNPutting the values in the above equation,[tex]tan Ө = 35 / 2Ttan Ө = 35 / (2 × 17.5)tan Ө = 1Thus, Ө = 45°[/tex]Hence, the smallest joint angle θ for which the structure can carry the design load, P = 35 kN is 45°.Therefore, the answer is (a) Minimum required diameter of the bars is 0.297 m (b) The smallest joint angle θ for which the structure can carry the design load, [tex]P = 35 kN is 45°.[/tex]
To know more about diameter visit:
https://brainly.com/question/32968193
#SPJ11
A 450V, 1800 rpm, 80A separately excited de motor is fed through three-phase semi converter from 3-phase 300V supply. Motor armature resistance is 1.20. Armature current is assumed constant. i determine the motor constant from the motor rating. ii. for a firing angle of 45° at 1500 rpm, compute the rms values of source and thyristor currents, average value of thyristor current. iii. repeat part "i" for a firing angle of 90° at 750 rpm.
i) Motor Constant from Motor Rating The motor constant k is determined as follows: V_t = k Nwhere Vt = applied voltage, N = speed of rotation, and k = motor constant. The motor constant, k, is given by k = V_t / N= 450 / 1800= 0.25 V-s/rad. ii) Calculation for Firing Angle of 45° and 1500 RPMa.
RMS values of source current:It is given that armature current is constant, and hence,
Idc = Iac = 80A.VR = Vt / √3= 300 / √3 = 173.2V
Voltage drop due to armature resistance = I * Ra= 80 * 1.20 = 96V
Average value of load voltage,
Vdc = VR – Ia * Ra= 173.2 – 96 = 77.2V
Therefore, from firing angle α = 45°, the average value of thyristor current (Id)
isId = Iavg = (Vm / √2) / (π / 2 - α)= (Vm / √2) / (π / 2 - 45°)= (300 / √2) / (π / 2 - 45°)= 6.83A
Irms of source current,
Isrms = Idc + Irms= 80 + √(I2 + I2dc)= 80 + √(43.38 + 802)= 87.1Ab.
RMS values of thyristor current:
Irms = Idc + 0.5 * Id = 80 + 0.5 * 6.83= 83.42Aiii)
Repeat Part "i" for a Firing Angle of 90° and 750 RPM Motor Constant from Motor Rating The motor constant k is determined as follows: V_t = k N where Vt = applied voltage, N = speed of rotation, and k = motor constant. The motor constant, k, is given by k = V_t / N= 300 / 750= 0.4 V-s/rad. Answer:
Therefore, for a 450V, 1800 rpm, 80A separately excited de motor that is fed through three-phase semi converter from 3-phase 300V supply with a motor armature resistance of 1.20 ohm and an armature current that is assumed to be constant.
To know more about rotation visit:
https://brainly.com/question/1571997
#SPJ11
Question 35 Use the shift cipher with key = 13 to encrypt the message GO AWAY CORONA VIRUS
To encrypt the message "GO AWAY CORONA VIRUS" using the shift cipher with a key of 13, each letter in the message is shifted 13 positions to the right in the alphabet.
Here's the encryption process:
Original message: GO AWAY CORONA VIRUS
Encrypted message: TB NOL PBENAN IBHE VFHVF
Explanation:
- The letter 'G' is shifted 13 positions to the right, resulting in 'T'.
- The letter 'O' is shifted 13 positions to the right, resulting in 'B'.
- The letter 'A' is shifted 13 positions to the right, resulting in 'N'.
- The letter 'W' is shifted 13 positions to the right, resulting in 'L'.
- The letter 'A' is shifted 13 positions to the right, resulting in 'P'.
- The letter 'Y' is shifted 13 positions to the right, resulting in 'B'.
- The letter 'C' is shifted 13 positions to the right, resulting in 'E'.
- The letter 'O' is shifted 13 positions to the right, resulting in 'N'.
- The letter 'R' is shifted 13 positions to the right, resulting in 'A'.
- The letter 'O' is shifted 13 positions to the right, resulting in 'B'.
- The letter 'N' is shifted 13 positions to the right, resulting in 'A'.
- The letter 'A' is shifted 13 positions to the right, resulting in 'N'.
- The letter 'V' is shifted 13 positions to the right, resulting in 'I'.
- The letter 'I' is shifted 13 positions to the right, resulting in 'V'.
- The letter 'R' is shifted 13 positions to the right, resulting in 'E'.
- The letter 'U' is shifted 13 positions to the right, resulting in 'S'.
- The letter 'S' is shifted 13 positions to the right, resulting in 'F'.
Therefore, the encrypted message using the shift cipher with a key of 13 is "TB NOL PBENAN IBHE VFHVF".
Leran more about encryption process here:
https://brainly.com/question/32168380
#SPJ11
