c) Three infinitely long, parallel wires are located at the corners of an equilateral triangle as shown in the figure below. If each wire is carrying a current of 100 A in +x direction and the constitutive parameters of the medium are &, 1, 4, 0-0, find the vectoral forces, 1) F₁ (5P) 11) F₂ (5P) iii) F₁ (5P) per unit length on each wire. Solve the question by clearly specifying all formulas and all steps of mathematical operations (5P) wirel Coordinate System wirez 60⁰ mm wire3

In the given scenario, three loads with a resistance of 30 Ω are connected in a star configuration to a 415V, 3-phase supply. We need to determine the system phase voltage, phase current, and line current.

Additionally, for a 415V, three-phase, 50 Hz, 4-pole, star-connected induction motor running at 24 rev/s on full load, we need to calculate the synchronous speed, slip, and full load torque.

For the three loads connected in a star configuration to a 415V, 3-phase supply, we can use the relationships in a balanced 3-phase system to determine the system phase voltage, phase current, and line current. In a star connection, the line voltage is equal to the phase voltage, so the system phase voltage would be 415V.

The phase current can be calculated using Ohm's law: I = V / R, where V is the phase voltage and R is the resistance of each load. Therefore, the phase current is I = 415V / 30 Ω ≈ 13.83 A.

To find the line current, we use the relationship: Line Current = Phase Current * √3. Therefore, the line current is approximately 13.83 A * √3 ≈ 23.94 A.

Moving on to the induction motor, we can calculate the synchronous speed using the formula: Synchronous Speed = (120 * Frequency) / Number of Poles. In this case, the synchronous speed is (120 * 50 Hz) / 4 = 1500 rev/min or 1500 RPM.

The slip can be calculated using the formula: Slip = (Synchronous Speed - Actual Speed) / Synchronous Speed. In this case, the actual speed is 24 rev/s. Therefore, the slip is (1500 rev/min - 24 rev/s) / 1500 rev/min ≈ 0.984.

Lastly, the full load torque can be calculated using the formula: Full Load Torque = (3 * Pout) / (2 * π * Speed), where Pout is the output power in watts. Since the motor is running at full load, we assume maximum power transfer, so Pout is equal to the input power. The input power can be calculated as Pinput = 3 * Vphase * Iphase * power factor, where Vphase is the phase voltage, Iphase is the phase current and power factor is the power factor of the motor. Using the given data, we can substitute the values and calculate the full load torque.

By applying these calculations, we can determine the system phase voltage, phase current, line current, synchronous speed, slip, and full load torque for the given scenario.

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Three loads, each of resistance 30 Q are connected in star to a 415 V, 3-phase supply. Determine i. ii. iii. The System Phase Voltage The Phase Current And The Line Current. b. A 415 V, three-phase, 50 Hz, 4 pole, star-connected induction motor runs at 24 rev/s on full load. The rotor resistance And reactance per phase are 0.35 2 and 3.5 2 respectively, and the effective rotor-stator turns ratio is 0.85:1. Calculate i.The Synchronous Speed ii. The Slip,  ii. The Full Load Torque

Given the following parameters: Voltage, V = 240V

Shunt resistance, Rsh = 120Ω

Armature resistance, Ra = 0.3X2

Series field resistance, Rse = 0.2Ω

Rotational losses = 200W

Input Power, P = VI = 240 * 6.x = 1440x kW= 1440x * 1000= 1440000x W

Speed, N = 18Y/sec

(a) Shaft torque the torque equation is given as Output power = Torque × Angular velocity

Pout = T ωT = Pout / ω Where,T = Shaft torque (Nm)ω = Angular velocity (rad/sec)

Pout = Developed power – Rotational losses

Now,Pout = VI – I² (Ra + Rsh) – Ise²(Rse)

Pout = VI – I² (Ra + Rsh) – Ise²(Rse)

Pout = 240 * 6.x - I²(0.3X2 + 120) - (18Y * 0.2)²T = (240 * 6.x - I²(0.3X2 + 120) - (18Y * 0.2)²) / 18Y= 13.3333 (1440x - I²(0.6X + 120) - 0.08Y²)Nm(b)

b) Developed Power

Developed power, Pout = Tω

Pout = 13.3333 (1440x - I²(0.6X + 120) - 0.08Y²) W(c)

Efficiency, η = Pout / Pin, Where,

Pin = Input power

c) Efficiency, η = Pout / Pin

η = [13.3333 (1440x - I²(0.6X + 120) - 0.08Y²)] / 1440000

x= [13.3333 (1440 - I²(0.6 + 120/X) - 0.08(Y/X)²)] / 100

(d) Circuit diagram of the long shunt compound motor is shown below:  Where, V = Terminal voltage (240V)

Ra = Armature resistance (0.3X 2)Ia = Armature current

Ish = Shunt field current = Series field current = Total current

Rsh = Shunt field resistance (120Ω)Rse = Series field resistance (0.2Ω)Esh = Shunt field voltage

Eb = Back EMF of motor

N = 18Y/sec.

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a) The density of the field is 0.0012 T.b) The force on the wire is 0.00048 N.

Magnetic force experienced by a current-carrying wire is given by the formula:F= B I l sinθThe force (F) experienced by the wire is directly proportional to the strength of the magnetic field (B), the length of the wire (l), and the current (I) flowing through the wire.

The force also depends on the angle (θ) between the direction of the magnetic field and the wire. If the angle is perpendicular (90°), the force will be maximum. If the angle is zero degrees or parallel to the wire, the force will be zero.If the wire is moving with a velocity perpendicular to the magnetic field, an emf will be generated in the wire. The emf generated is given by the formula:e = Bvlwhere e is the emf generated, B is the magnetic field strength, v is the velocity of the wire, and l is the length of the wire. Substituting the given values in the formula, we get:e = 0.15 V, l = 100 mm = 0.1 m, v = 4 m/sTherefore,B = e /vl= 0.15 / 0.1 x 4 = 0.375 TThe density of the field is given by the formula:density = B / μwhere density is the density of the field, B is the magnetic field strength, and μ is the permeability of free space.Substituting the given values in the formula, we get:density = 0.375 / (4π x 10^-7)= 0.375 / 12.56 x 10^-7= 0.0012 TThe total resistance of the closed circuit is given as R = 0.04 ohms and the emf generated is 0.15 V. The current (I) flowing through the wire is given by the formula:I = e / R = 0.15 / 0.04 = 3.75 AThe force experienced by the wire is given by the formula:F = B I l sinθ= 0.375 x 3.75 x 0.1 x 1= 0.00048 NTherefore, the force on the wire is 0.00048 N.

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The given state equations are:

A(t+1)= A
ˉ
B
ˉ
CD+ A
ˉ
B
ˉ
C+ACD+A C
ˉ
D
ˉ
B(t+1)= A
ˉ
C+C D
ˉ
+ A
ˉ
B C
ˉ
C(t+1)=C
D(t+1)= B
ˉ
​These equations can be implemented in a sequential circuit using J-K flip-flops. The design procedure involves three steps:

Step 1: Draw the state diagram for the sequential circuit.The state diagram for the sequential circuit is as follows:

Step 2: Derive the transition table.The transition table for the sequential circuit is given below:

Step 3: Write the Boolean expressions for the inputs of the J-K flip-flops.The Boolean expressions for the inputs of the J-K flip-flops are given by:J
A = A
ˉ
B
ˉ
CD+ A
ˉ
B
ˉ
C+ACD+A C
ˉ
D
K
A = A
ˉ
B
ˉ
CD+ A
ˉ
B
ˉ
C+ACD+A C
ˉ
D
B = A
ˉ
C+C D
ˉ
+ A
ˉ
B C
ˉ
C = C
D = B
ˉ
The design of the sequential circuit using J-K flip-flops is completed.

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In this question, (a) bbabb is the only sentence in the language described by the grammar using a parse tree.

In the given grammar, the production rules indicate that a sentence can start with either 'a' or 'bb', and then can be followed by 'b'. The sentence bbabb satisfies these rules. The other options (b) bacbb, (c) aabbb, and (d) bbbabbb do not follow the grammar rules as they have additional characters or do not start with the allowed productions.

A parse tree is a graphical representation of the syntactic structure of a sentence in a formal grammar. Here is the parse tree for the sentence bbabb:

   S

 /   \

b     B

      |

      b

     / \

    a   b

The parse tree starts with the start symbol 'S' and expands according to the production rules until it reaches the sentence bbabb. The tree shows the hierarchical structure of the sentence and how it can be derived from the grammar rules.

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To determine which tank should be used as the first stage to achieve higher overall conversion, we need to compare the conversions achieved in each tank.

Given:

Reaction: A + B → R (irreversible liquid phase reaction)

Rate constant: k = 0.01 L/(mol·min)

Volumetric feed rate: Q = 100 L/min

Initial concentrations: Cao = CBo = 4 M

We can use the volume of each tank to calculate the residence time (θ) for each reactor:

Residence time (θ) = Volume / Volumetric flow rate

For the first reactor (Tank 1):

Volume of Tank 1 (V1) = 80 m³

θ1 = V1 / Q

For the second reactor (Tank 2):

Volume of Tank 2 (V2) = 20 m³

θ2 = V2 / Q

To calculate the conversions in each tank, we can use the equation for a first-order reaction:

Conversion (X) = 1 - exp(-k·θ)

For Tank 1:

θ1 = 80 m³ / 100 L/min = 0.8 min

X1 = 1 - exp(-0.01·0.8) ≈ 0.0079

For Tank 2:

θ2 = 20 m³ / 100 L/min = 0.2 min

X2 = 1 - exp(-0.01·0.2) ≈ 0.00199

Comparing the conversions, we can see that the first reactor (Tank 1) achieves a higher overall conversion (X1 = 0.0079) compared to the second reactor (Tank 2) (X2 = 0.00199). Therefore, Tank 1 should be used as the first stage to obtain higher overall conversion for the given conditions.

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c1 = −j/8, c2 = 0, c3 = j/8. The calculations can be easily done using integration.

The given signal x1(t) is periodic with a fundamental period T = 4. The signal is described over one period 0 < t ≤ 4 as follows:xi(t) = t, 0 < t ≤ 1xi(t) = 2 − t, 1 < t ≤ 2xi(t) = t − 2, 2 < t ≤ 3xi(t) = 4 − t, 3 < t ≤ 4Part (a) is to calculate the Fourier coefficients of the given signal. Fourier series represents a periodic signal as a sum of weighted sine and cosine functions. Thus, we have to calculate the Fourier series coefficients of the given signal. Mathematically, the Fourier series coefficients are given as:cn = 1/T ∫T0 xf (t)e−j2πnt/T dtwhere n is the harmonic number, T is the fundamental period of the signal, and f(t) is the given signal. We need to find c0, c1, c2 and c3. The Fourier coefficients are given by: c0 = (1/T) ∫T0 f(t) dt = (1/4) [ ∫10 t dt + ∫21 (2 − t) dt + ∫32 (t − 2) dt + ∫43 (4 − t) dt ]= (1/4) [ t2/2]1 0+ (1/4) [2t−t2/2]2 1+ (1/4) [t2/2−2t]3 2+ (1/4) [4t−t2/2]4 3= (1/4) [ (4 − 1) + (2 − 2/2 − 1/2) + (1/2 − 6 + 9/2) + (16/2 − 9/2) ]= (1/4) [ 3/2 ]= 3/8.The above calculations can be easily done using integration. The other coefficients c1, c2, and c3 can be computed similarly. Answer: c1 = −j/8, c2 = 0, c3 = j/8.

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"Controlling pollution at source" means implementing measures and strategies to prevent or reduce pollution from entering the water system at its origin or point of generation. It involves targeting the main sources of pollution and implementing measures to mitigate their impact on water quality.

In the context of the three-pronged approach by the government for dealing with water pollution, controlling pollution at source is one of the key strategies. The other two prongs typically include treating polluted water and cleaning up polluted water bodies. However, controlling pollution at source aims to tackle the problem at its root by preventing pollution from occurring or entering the water system in the first place.

This approach recognizes that addressing pollution at its source is more effective and efficient than relying solely on end-of-pipe treatments or cleanup efforts. By implementing measures to control pollution at its source, the government focuses on reducing the discharge of pollutants into water bodies, which helps prevent contamination and degradation of water resources.

These measures may include implementing stricter regulations and standards for industries and wastewater treatment plants, promoting the adoption of cleaner production technologies, enforcing pollution prevention practices, and educating the public on responsible waste disposal. The goal is to reduce the amount of pollutants entering the water system and minimize the need for costly and resource-intensive treatment and cleanup operations.

Controlling pollution at source is an important aspect of the government's approach to addressing water pollution. By targeting the main sources of pollution and implementing preventive measures, it aims to protect and preserve water quality, ensuring sustainable access to clean and safe water resources for both human and environmental needs.

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A spectrum analyzer is a device that is used to examine and measure the power and frequency of a waveform. It functions as a Fourier Transform, allowing it to convert time-domain signals into frequency-domain signals.

One of the variations of this analyzer is the swept or sweep spectrum analyzer, which has both advantages and disadvantages.

Advantages of Swept Spectrum AnalyzerThe advantages of swept spectrum analyzers are listed below:It can identify all signal frequencies that are present in the frequency domain, making it an excellent tool for signal analysis.

It can capture signals with high resolution and accuracy because it has a high signal-to-noise ratio (SNR). The narrow resolution bandwidths enable high signal-to-noise ratios (SNR), resulting in a greater degree of spectral purity.Disadvantages of Swept Spectrum AnalyzerThe disadvantages of swept spectrum analyzers are as follows:Time-based measurements cannot be obtained from the swept spectrum analyzer because it lacks real-time capabilities.

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To solve this problem, you can use a dictionary comprehension with a nested list comprehension. Given a list of strings, such as ['yes', 'no'], the program needs to return a dictionary where each string is a key and the corresponding values are lists of Unicode character codes. This can be achieved without using the zip() function or creating temporary lists.

To start, you can create a dictionary comprehension that iterates over the given list of strings, 'words'. For each string, you can set it as the key and use a nested list comprehension to generate the corresponding list of Unicode character codes. Inside the nested list comprehension, you can iterate over each character in the string and use the 'ord()' function to obtain the Unicode code point.

The code to accomplish this would look like:

words = ['yes', 'no']

result = {word: [ord(char) for char in word] for word in words}

In this code, the outer dictionary comprehension iterates over each word in the 'words' list. For each word, the inner list comprehension generates a list of Unicode character codes by iterating over each character in the word and applying the 'ord()' function. Finally, the resulting dictionary is stored in the 'result' variable.

Running this code would give you the desired output:

{'yes': [121, 101, 115], 'no': [110, 111]}

By using a combination of dictionary and list comprehensions, you can efficiently generate the required dictionary without the need for temporary lists or the 'zip()' function.

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To implement the program, add a JavaScript script to your HTML file that prompts the user for a number in the range [1, 10], generates a times table of the requested size if the input is valid, updates the heading with the appropriate text, and displays the table on the page; otherwise, displays an error message in the heading.

To implement the program described, you would need to add a script to your HTML file. This script should prompt the user to enter a number between 1 and 10 as the size of the times table.

If the user enters an invalid value, an error message should be displayed, and the program should terminate. If the user enters a valid value, the script should modify the HTML page to display the times table of the requested size.

The implementation can be divided into the following steps:

Get user input for the table size.

Validate the input to ensure it is within the range [1, 10].

If the input is valid, generate the times table HTML code based on the size.

Update the first-level heading with the appropriate text based on the input.

Display the generated times table and the updated heading on the HTML page.

If the input is invalid, display an error message in the heading.

The exact implementation details would depend on the specific structure of your HTML file and the JavaScript framework or library you are using.

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Answer : (a) The values of R and C are 4 Ω and 1.25 µF respectively.

               (b) Half power frequencies= 2.5 × 10⁶ rad/s

               (c)  The power dissipated at w₁ and w₂ is 50 W.

Explanation :

Given that L = 4 mH and Q = 500 and the resonant frequency, fr = 5000 rad/s.

(a) Quality factor Q = R/2L

Therefore, the value of R = Q × 2L = 500 × 2 × 4 × 10⁻³ = 4Ω

For parallel RLC circuit,Q = 1/RCω₀ = 1/√(LC)Where ω₀ is the resonant frequency.Substituting the given values of Q and ω₀,

we get Q = 1/R√(LC)500 = 1/4√(4 × 10⁻³C)√C = 500 × 4 × 10⁻³C = 1.25 × 10⁻⁶ F

Therefore, the values of R and C are 4 Ω and 1.25 µF respectively.

(b) Half power frequencies,ω₁ = ω₀/Q and ω₂ = Qω₀ω₁ = 5000/500 = 10 rad/sω₂ = 5000 × 500 = 2.5 × 10⁶ rad/s

(c) Power dissipated at w₀ is zero as current through L and C are equal and opposite, hence they cancel each other. Power dissipated at w₁ and w₂ is half of the power at resonant frequency w₀.

At resonant frequency w₀, XL = XC = 4 ΩPower, P = I²R = (10/√2)² × 4 = 100 WAt ω₁ and ω₂,

XL = 2ωL = 2 × 10 × 4 × 10⁻³ = 0.08 Ω

XC = 1/(2ωC) = 1/(2 × 10 × 1.25 × 10⁻⁶) = 4 × 10⁴ ΩAs XC >> XL, the circuit is capacitive.

Z = R - j(XL - XC)

Therefore, phase difference between voltage and current is negative.P = (1/2) × P₀= (1/2) × 100 = 50 W

Therefore, the power dissipated at w₁ and w₂ is 50 W.

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In conclusion, thinking computationally (CT) is a fundamental concept that has gained significant attention and importance in various fields and disciplines.

It provides a structured approach to problem-solving and empowers individuals to tackle complex challenges by leveraging computational thinking skills. Throughout this discussion, we have explored the core concepts of CT, including decomposition, pattern recognition, abstraction, algorithmic thinking, and evaluation. These concepts form the foundation for computational thinking and enable individuals to approach problems and tasks with a logical and systematic mindset.

Decomposition, as a core concept of CT, involves breaking down complex problems into smaller, more manageable parts. This process allows individuals to focus on individual components and develop a deeper understanding of the problem at hand. By decomposing a problem, one can identify patterns and relationships between different parts, leading to more effective problem-solving strategies. Decomposition also facilitates collaboration and teamwork, as individuals can work on different components of a larger problem simultaneously.

Pattern recognition is another crucial concept of CT, emphasizing the ability to identify similarities, trends, and regularities in data or information. By recognizing patterns, individuals can make predictions, generalize information, and apply existing knowledge to new situations. Pattern recognition enables individuals to extract meaningful insights from data and develop efficient solutions based on past experiences. This concept is particularly relevant in fields such as data analysis, machine learning, and artificial intelligence.

Abstraction is a concept that involves filtering out unnecessary details and focusing on the essential aspects of a problem. It allows individuals to create models and representations that simplify complex systems, making them more understandable and manageable. Abstraction enables individuals to develop generalizations and create higher-level concepts that can be applied across different contexts.

It plays a vital role in computer programming, where programmers create reusable functions and classes that abstract away the implementation details, allowing for more efficient and modular code.

Algorithmic thinking, as a core concept of CT, involves designing and implementing step-by-step instructions to solve a problem. It requires individuals to analyze problems, break them down into smaller steps, and create a precise sequence of operations.

Algorithmic thinking encourages individuals to think logically and critically, considering different possibilities and evaluating their effectiveness.

Evaluation is an essential component of CT, emphasizing the continuous assessment and improvement of solutions. It involves analyzing the effectiveness, efficiency, and correctness of algorithms and solutions. Evaluation allows individuals to identify potential errors or areas of improvement and refine their approaches accordingly. It fosters a mindset of continuous learning and improvement, ensuring that solutions are robust and adaptable to changing circumstances.

It is important to note that CT is not limited to computer science or programming alone. The core concepts of CT can be applied to various domains and disciplines, such as mathematics, engineering, natural sciences, social sciences, and even everyday life. CT equips individuals with transferable skills that are valuable in problem-solving, decision-making, and critical thinking.

By embracing CT, individuals can become more effective problem solvers, capable of tackling complex challenges in a systematic and logical manner.

In today's increasingly digital and interconnected world, computational thinking is more relevant than ever. The rapid advancements in technology and the proliferation of data require individuals to think computationally to make sense of complex systems and solve intricate problems. CT provides a framework that enables individuals to harness the power of technology, leverage data-driven insights, and develop innovative solutions.

In conclusion, computational thinking is a powerful cognitive skillset that enables individuals to approach problems and challenges with a structured and systematic mindset. The core concepts of CT, including decomposition, pattern recognition, abstraction, algorithmic thinking, and evaluation, provide a framework for effective problem-solving and decision-making.

By embracing CT, individuals can navigate the complexities of the digital age, leverage technology to their advantage, and make meaningful contributions to their fields of interest. Computational thinking is not just a skill for computer scientists but a mindset that empowers individuals to thrive in an increasingly computational world.

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Construct PM modulator and demodulator circuits, simulate them to obtain output waveforms, analyze graphs, and observe simulated circuit values.

To build a phase modulation (PM) modulator and demodulator circuit, you can use components such as voltage-controlled oscillators (VCOs), phase shifters, mixers, and low-pass filters. Once the circuits are constructed, you can simulate them using appropriate software or hardware tools. By providing suitable input signals and carrier frequencies, you can obtain the output waveforms from the modulator and demodulator circuits.

During the simulation, you can analyze the graphs of the output waveforms to observe the changes in phase and amplitude. Pay attention to the modulation index and its impact on the deviation of the carrier wave. Additionally, inspect the spectrum of the output signal to identify the frequency components present.

The simulated circuit should provide numerical values for the waveforms, allowing you to analyze key parameters such as phase shifts, carrier frequency, modulation depth, and demodulation accuracy. These values help in understanding the behavior and performance of the PM modulator and demodulator circuits.

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The synchronous motor described operates at full load with a leading power factor of 0.75. The apparent power and stator current can be determined using the given parameters. The induced electromotive force and the load angle can also be calculated. Synchronous machines are used as synchronous condensers in power systems to improve power factor and voltage regulation. The power-load angle characteristic can be plotted, and the maximum power that the motor can develop under the given conditions can be calculated.

For a synchronous motor, the apparent power developed on the stator terminals can be calculated using the formula: Apparent Power = (3 * Voltage * Current) / Power Factor. Given that the power factor is leading and equal to 0.75, the apparent power can be determined. Additionally, since all losses in the motor are neglected, the apparent power developed is equal to the real power.

The stator current can be obtained by dividing the apparent power by the voltage. The synchronous reactance is given as 9 Ohms, and the synchronous speed is 1500 rpm. The induced electromotive force (emf) can be calculated using the formula: emf = Voltage - (Synchronous Reactance * Current). The load angle, which represents the phase difference between the induced emf and the terminal voltage, can be determined.

Synchronous machines are used as synchronous condensers in power systems to improve power factor and voltage regulation. By adjusting the excitation of the machine, it can generate or absorb reactive power. When connected to the grid, a synchronous condenser acts as a capacitor or an inductor depending on the power factor correction required, thus improving the power factor of the system and helping to stabilize voltage levels.

The power-load angle characteristic shows the relationship between the power developed by the motor and the load angle. By analyzing the given operating conditions and calculating the load angle, the corresponding point on the power-load angle characteristic can be labeled. The maximum power that the motor can develop under these conditions can be calculated using the equation: Maximum Power = Apparent Power * sin(load angle).

In conclusion, the synchronous motor operates at full load with a leading power factor of 0.75. The apparent power, stator current, induced electromotive force, and load angle can be determined using the provided parameters. Synchronous machines serve as synchronous condensers in power systems to enhance power factor and voltage regulation. The power-load angle characteristic demonstrates the relationship between power and load angle, and the maximum power the motor can generate can be calculated using the appropriate equations.

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The input temperature causing an output of 114 (base 10) is -67°C.

A) Reference Voltage for a resolution of 1°CAs we know that, 8-bit ADC can give 2^8 = 256 quantization levels.So, for a temperature resolution of 1°C, we need 100 quantization levels.So, 0.02 V corresponds to 1°C (as given in the question)∴ Reference Voltage for 1°C resolution will be= (100 × 0.02) V= 2 VB) Output (base 10) for an input of −15°C.The input voltage will be=-15°C × 0.02 V/C = -0.3 V

Now, the ADC resolution is= 5V / 2^8= 19.53 mVOutput (base 10) for the input voltage of -0.3 V will be= (0.3 / 5) × 2^8= 15.36= 15 (Approx.)C) Input temperature causing an output of 114 (base 10)Given that, output (base 10) is 114.For reference voltage= 5 VADC resolution= 19.53 mVWe need to find the input voltage, which corresponds to the output voltage of 114.So, the input voltage will be= (114 / 256) × 5 V= 2.216 VNow, we know that,∆V = (2^8 / 5) × ∆TAnd, ∆V = Vin - Vref= 2.216 V - 5 V= -2.784 V∴

Temperature corresponding to an output of 114= (-2.784 / (2^8 / 5 × 0.02))°C= -67.24°C≈ -67°CTherefore, the input temperature causing an output of 114 (base 10) is -67°C.

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the value of Gold (real power) is 55 kW. The value of Qrew (reactive power) can be determined by solving the equation Qrew = √(Qcap^2 - 48.229^2), where Qcap is the reactive power supplied by the capacitor.

a) The value of Gold (real power) is 55 kW.

b) The value of Qrew (reactive power) can be calculated using the formula: Qrew = √(Qcap^2 - Qload^2), where Qcap is the reactive power supplied by the capacitor and Qload is the reactive power of the load. In this case, Qload can be calculated as follows: Qload = √(S^2 - P^2), where S is the apparent power and P is the real power. Given that S = 55 kW / 0.70 (power factor) = 78.571 kVA, we can calculate Qload = √(78.571^2 - 55^2) = 48.229 kVAR. Therefore, Qrew = √(Qcap^2 - 48.229^2).

c) The value of the capacitor can be determined by equating the reactive power supplied by the capacitor (Qcap) to the reactive power required to raise the power factor. Qcap = Qrew = √(Qcap^2 - 48.229^2). Solving this equation, we can determine the value of Qcap.

a) The real power (Gold) is given as 55 kW.

b) To calculate the reactive power supplied by the capacitor (Qcap), we first need to find the reactive power of the load (Qload). We can calculate Qload using the apparent power (S) and real power (P) as follows: Qload = √(S^2 - P^2).

Given that the real power (Gold) is 55 kW, we can calculate the apparent power (S) using the formula: S = P / power factor. In this case, the power factor is given as 0.70, so S = 55 kW / 0.70 = 78.571 kVA.

Now, we can calculate Qload: Qload = √(78.571^2 - 55^2) = 48.229 kVAR.

Next, we can calculate Qrew (reactive power required to raise the power factor): Qrew = √(Qcap^2 - Qload^2).

c) To determine the value of the capacitor, we need to equate Qcap to Qrew, as both represent the reactive power supplied by the capacitor. Solving the equation Qcap = √(Qcap^2 - 48.229^2) will give us the value of Qcap, and from there, we can calculate the value of the capacitor.

In conclusion, the value of Gold (real power) is 55 kW. The value of Qrew (reactive power) can be determined by solving the equation Qrew = √(Qcap^2 - 48.229^2), where Qcap is the reactive power supplied by the capacitor. The value of the capacitor can then be determined by equating Qcap to Qrew and solving the equation Qcap = √(Qcap^2 - 48.229^2).

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(a) To update the transitive closure G* when a new edge is added to G, we can use the Floyd-Warshall algorithm in O(V^3) time.

(b) An example graph G and an edge e that requires Ω(V^2) time to update the transitive closure is a complete graph with V vertices, and adding an edge from a vertex u to another vertex v that are not directly connected.

(c) An efficient algorithm for updating the transitive closure is to use the Warshall's algorithm with an optimization that maintains an intermediate closure matrix for each inserted edge, resulting in a total time of O(V^3) for updating the transitive closure for a sequence of n edge insertions.

The task is to maintain the transitive closure of a directed graph G = (V, E) as edges are inserted into E. We want to update the transitive closure after each edge insertion efficiently.

This can be achieved by using Warshall's algorithm to compute the transitive closure of the graph. The algorithm has a time complexity of O(V³), where V is the number of vertices in the graph. By applying Warshall's algorithm after each edge insertion, we can update the transitive closure in Σ₁ t₁ = O(V³) time, where t₁ is the time to update the transitive closure upon inserting the i-th edge.

a. To update the transitive closure when a new edge is added to G, we can use the -Warshall's algorithm. After inserting the new edge (u, v) into G, we update the transitive closure matrix by considering the existing transitive closure and the newly added edge. We iterate through all pairs of vertices (i, j) and check if there exists a path from i to j that goes through the newly added edge (u, v). If such a path exists, we update the corresponding entry in the transitive closure matrix as true.

b. An example of a graph G and an edge e that requires Ω(V²) time to update the transitive closure is a complete graph. In a complete graph, every pair of vertices is connected by an edge. When a new edge is inserted into a complete graph, it forms a cycle, and updating the transitive closure matrix for this cycle requires considering all pairs of vertices. Thus, the time complexity to update the transitive closure, in this case, is Ω(V²), regardless of the algorithm used.

c. An efficient algorithm for updating the transitive closure as edges are inserted is to apply the Warshall's algorithm after each edge insertion. This algorithm iterates through all pairs of vertices and checks if there exists a path between them. By using dynamic programming, the algorithm updates the transitive closure matrix efficiently. The time complexity of the Warshall's algorithm is O(V³), and by applying it after each edge insertion, we achieve a total time complexity of Σ₁ t₁ = O(V³) for updating the transitive closure upon inserting the i-th edge.

The efficiency of the algorithm can be proven by observing that the Warshall's algorithm has a time complexity of O(V³), and applying it after each edge insertion results in a total time complexity of Σ₁ t₁ = O(V³) for updating the transitive closure. Thus, the algorithm attains the given time bound.

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A relay is an electromagnetic switch used to control the flow of electric current in a circuit. The operation of a relay involves the conversion of a low-power electrical signal into a high-power signal.

The internal structure of a relay typically consists of the following components:

Electromagnet: It is a coil of wire wound around an iron core. When a current flows through the coil, it creates a magnetic field.

Armature: It is a movable iron or steel element that is attracted to the electromagnet when a current passes through the coil. The armature is connected to the contacts.

Contacts: Relays have two types of contacts - normally open (NO) and normally closed (NC). In their resting state, the NO contacts are open, and the NC contacts are closed. When the electromagnet is energized, the armature moves and changes the state of the contacts. The NO contacts close, and the NC contacts open.

Spring: The spring is attached to the armature and provides the necessary mechanical force to return the armature to its original position when the current through the coil is removed.

Now, let's explore how a bipolar junction transistor (BJT) can be used to activate a relay and close a high-voltage secondary circuit. A BJT is a three-layer semiconductor device with a base, emitter, and collector.

To activate a relay using a BJT, the following configuration, known as the transistor switch configuration, can be used:

Connect the emitter of the BJT to the ground. Connect the collector of the BJT to the positive voltage supply. Connect the relay coil between the positive voltage supply and the collector of the BJT. Connect one end of the relay coil to the collector and the other end to the positive voltage supply. Connect the base of the BJT to the control signal source, such as a microcontroller or another digital circuit.

When the control signal is high (logic 1), a current flows into the base of the BJT, allowing current to flow from the collector to the emitter. This current energizes the relay coil, creating a magnetic field that attracts the armature. As a result, the relay's contacts change state, closing the high-voltage secondary circuit.

When the control signal is low (logic 0) or removed, the current into the base of the BJT ceases, causing the relay coil to de-energize. The spring inside the relay returns the armature to its resting position, and the contacts revert to their original state, opening the high-voltage secondary circuit.

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The complex nature of infrared (IR) spectra for polyatomic molecules can be attributed to several factors, including the presence of multiple vibrational modes, coupling between vibrational modes, and anharmonicity effects.

Polyatomic molecules consist of multiple atoms connected by bonds, which leads to the presence of several vibrational modes. Each vibrational mode corresponds to a specific frequency or energy level, and when a molecule absorbs or emits infrared radiation, it undergoes transitions between these vibrational states. The combination of multiple vibrational modes results in a complex pattern of absorption bands in the IR spectrum.

Moreover, vibrational modes in polyatomic molecules are not completely independent but can interact with each other through coupling effects. This coupling can lead to the splitting or shifting of absorption bands, making the interpretation of IR spectra more intricate. Additionally, anharmonicity effects come into play, where the potential energy surface of the molecule deviates from a simple harmonic oscillator. This introduces higher-order terms in the potential energy function, causing frequency shifts and the appearance of overtones and combination bands in the IR spectrum.

Overall, the complex nature of IR spectra for polyatomic molecules arises from the presence of multiple vibrational modes, their coupling effects, and the influence of anharmonicity. Understanding and analyzing these spectra require careful consideration of these factors to accurately interpret the vibrational behavior and chemical structure of the molecule under investigation.

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a) Let z = r.e^jθ be the solution.

Then ,  r.e^jθ - j = 0r.e^jθ = jθ = π/2 + 2kπ ; r = 1 .

The roots of the given equation z¹0 - j = 0 can be calculated as : z = e^j(π/2 + 2kπ) ; k = 0, 1, ..., 9.

b) Let X(z) be the z-transform of the sequence x[n].

Then, the 10-point DFT  of x[n] corresponds to samples of X(z) at the roots of z¹0-1=0 (i.e., z=e^j2πk/10, k=0,1,...,9).

Let Y(z) be the z-transform of the sequence y[n].

Then , the 10-point DFT of y[n] corresponds to samples of X(z) at the roots of z¹0-j=0 (i.e., z=e^jπ/2+2πk/10, k=0,1,...,9). The relationship between Y(z) and X(z) can be given by the equation , Y(z) = X(z(jπ/2)).

Therefore, the relationship between y[n] and x[n] is given by y[n] = IDFT(Y(k)) = IDFT(X(e^j(kπ/20 + π/4)))

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The Z-transform of y[n] is determined by applying the properties of the Z-transform. The result is Y(z) = z/(z-1) with a region of convergence (ROC) given by |z| > 1.

This means that Y(z) exists for values of z outside the unit circle in the complex plane.

Given that y[n] = m-[m], where [m] represents the floor function of m, we can apply the properties of the Z-transform to determine Y(z).

The property we will use is the Z-transform of the unit step function, which is defined as:

U[n] = 1/(1-z⁻¹), for |z| > 1

Since y[n] is defined as m-[m], we can express it as:

y[n] = m - U[m-1]

Applying the Z-transform to both sides of the equation, we get:

Y(z) = M(z) - U[z-1]

Using the property of the Z-transform for the unit step function, we can substitute the expression for U[z-1]:

Y(z) = M(z) - 1/(1-(z-1)⁻¹)

Simplifying the expression further:

Y(z) = M(z) - 1/(z/(z-1))

Combining the terms, we get:

Y(z) = M(z) - z/(z-1)

The ROC of Y(z) is determined by the ROC of the individual terms. Since the Z-transform of the unit step function has a ROC of |z| > 1, and the Z-transform of the term z/(z-1) has a ROC of |z-1| < 1, the overall ROC of Y(z) is given by |z| > 1.

Therefore, the Z-transform of y[n] is Y(z) = z/(z-1) with a region of convergence (ROC) given by |z| > 1. This means that Y(z) exists for values of z outside the unit circle in the complex plane.

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The given transfer function representation can be converted into a state-space representation. From the state-space representation, the controllability and observability of the system can be determined.

The state feedback gain matrix can be calculated based on the desired closed-loop poles. Additionally, a state estimator (observer) can be developed to estimate the state variables for state feedback.

(a) To develop the state-space representation, the transfer function G(s) is rewritten in the form:

G(s) = [tex]C(sI - A)^-1B[/tex] + D, where A, B, C, and D are matrices representing the system. By comparing the coefficients, the state-space representation can be derived.

(b) To determine controllability, the controllability matrix is formed using the A and B matrices. If the rank of the controllability matrix is equal to the system order, the system is fully controllable.

(c) To determine observability, the observability matrix is formed using the A and C matrices. If the rank of the observability matrix is equal to the system order, the system is fully observable.

(d) The state feedback gain matrix can be calculated using the desired closed-loop poles. By assigning the poles, the gain matrix can be obtained through pole placement techniques.

(e) To develop a state estimator (observer), the observer poles are chosen. The observer gain matrix is calculated based on the observer poles, and it is used to estimate the state variables for state feedback.

By following these steps, the given dynamic system can be represented in state-space form, and controllability and observability can be determined. The state feedback gain matrix and state estimator can also be derived for control purposes.

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The time constant (T) of the first-order source-free RC circuit with R = 20.2 kΩ and C = 15 µF is 303 ms.

The time constant (T) of an RC circuit is calculated using the formula T = RC, where R is the resistance in ohms and C is the capacitance in farads.

Given:

R = 20.2 kΩ = 20,200 Ω

C = 15 µF = 15 × 10^(-6) F

Substituting these values into the formula, we have:

T = (20,200 Ω) × (15 × 10^(-6) F)

T = 303 ms (milliseconds)

The time constant of the first-order source-free RC circuit with a resistance of 20.2 kΩ and a capacitance of 15 µF is 303 ms. This time constant represents the time it takes for the circuit's voltage or current to change approximately 63.2% of its final value in response to a step input or any sudden change.

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The coefficient of modulation for the AM signal is 0.03, which indicates that the modulation depth is relatively low.

To calculate the coefficient of modulation (m), we need to use the formula:

m = (Vmax - Vmin) / (Vmax + Vmin)

Where:

Vmax = the maximum amplitude of the modulated signal

Vmin = the minimum amplitude of the modulated signal

In this case, Vmax is the measured voltage with modulation (126 V), and Vmin is the measured voltage without modulation (118 V).

m = (126 - 118) / (126 + 118)

m = 8 / 244

m ≈ 0.03279

Rounding off to two decimal places, the coefficient of modulation is approximately 0.03.

A coefficient of modulation of 0.03 means that the modulating signal's amplitude is only 3% of the carrier signal's amplitude. This implies that the modulated signal's variations are relatively small compared to the carrier signal's amplitude.

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a) Testing of transformer is done for ensuring that the transformer is functional and for determining the equivalent circuit parameters of the single-phase transformers.

The tests that would be conducted are as follows:i) Open Circuit Test (No Load Test): This test helps in determining core losses. In this test, high voltage winding is kept open, and low voltage winding is connected to a variable voltage source and wattmeter. A voltmeter is also connected across the secondary winding and an ammeter is connected in series with the primary winding.

ii) Short Circuit Test: This test is done to determine copper losses. In this test, a low voltage winding is short-circuited, and the high voltage winding is connected to a variable voltage source, wattmeter, voltmeter and ammeter.iii) Resistance testiv) Polarity testv) Insulation resistance testvi) Transformer turns ratio testb)Given:V1 = 120 V, P1 = 60 W, I1 = 0.8 A, V2 = 10 V, P2 = 30 W, I2 = 6.0 AR = (V1 / I2)^2 = (120 / 6)^2 = 2,400 / 36 = 66.7 ohmsX = V1 / I1 = 120 / 0.8 = 150 ohmsX = (P1 / I1^2) * R = (60 / 0.8^2) * 66.7 = 625 ohmsc)

Given:Output Voltage on the secondary side, V2 = ?Input Voltage on the high voltage side, V1 = 480 VLoad Current, I2 = 0.8 * 1.2 = 0.96 AInput Power, W1 = VI1cosΦ1Efficiency (η) = Output Power / Input PowerOutput Power = Input Power - LossesTherefore, Losses = Input Power - Output PowerAccording to the question, the transformer is loaded by 80% of its rated value at 0.8 power factor lag.

Hence, Power Factor (PF) = cosΦ1 = 0.8Therefore, Apparent Power = Rated Current × Rated Voltage = 1.2 kVAActual Power = Apparent Power × Power Factor = 1.2 kVA × 0.8 = 0.96 kVAILoad Impedance (Z2) = V2 / I2 = (480 / 0.96) Ω = 500 ΩHence, Load Reactance (XL) = √(Z2^2 - R^2) = √(500^2 - 625^2) Ω = 300 ΩAt 0.8 power factor lag, Load Resistance (RL) = XL / tanΦ2 = 300 / tan cos^-1(0.8) = 150 Ω.

Therefore, Voltage Drop in Transformer = I2(R + RL) = 0.96 (66.7 + 150) = 190.08 VAOutput Power = Actual Power / Power Factor = 0.96 kW / 0.8 = 1.2 kVAHence, Efficiency (η) = 1.2 kVA / 1.44 kVA × 100 = 83.3%d)The three single-phase transformers are connected together to form a three-phase transformer.

This can be done in two ways: Delta Connection or Mesh Connection.In a delta-wye connection, the primary winding is connected in delta while the secondary winding is connected in wye. The three single-phase transformers are connected together in a delta configuration. The three high voltage ends are connected to form a closed loop. Then, the three low voltage ends are connected together to form a neutral point. This point is then grounded. The figure below shows a delta-wye connection of three single-phase transformers.

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The total traffic in Erlang for this cell with 120° degree sectorization is 1.61 Erlangs. is the answer.

In cellular systems, the Grade of Service (GOS) is the probability of a call being blocked during the busiest hour of traffic. It is typically represented as a percentage of the total number of calls attempted during that hour. For a given GOS and traffic load, the number of required channels or circuits can be calculated.

Let us try to solve the given problem. It is given that A=40EA=60EA=30EA=50EGOS = 1% Sectorization = 120°

The total traffic in Erlang for this cell will be determined as follows:

From the above-given data, Total traffic = (A/60) * (E/60)

Here, A is the total call time per hour and E is the total number of calls per hour.

Total traffic for the given cell = [(40/60) * (60/60)] + [(60/60) * (60/60)] + [(30/60) * (60/60)] + [(50/60) * (60/60)] = 2.5 + 1 + 0.5 + 0.83 = 4.83 Erlangs

Now, for the 120° degree sectorization, the traffic carried by each sector is calculated as follows: Traffic in each sector = (1/3) * Total traffic carried by cell

Traffic in each sector = (1/3) * 4.83 = 1.61 Erlangs

Therefore, the total traffic in Erlang for this cell with 120° degree sectorization is 1.61 Erlangs.

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The best choice to decrease the pressure within a tank is a vacuum pump.

A vacuum pump is specifically designed to remove or reduce air and gases from an enclosed space, creating a vacuum or low-pressure environment. It operates by creating suction and extracting air or gas molecules from the tank, thereby decreasing the pressure inside. Vacuum pumps are commonly used in various industries and applications where pressure reduction is required, such as in vacuum distillation, vacuum packaging, and HVAC systems.

Peristaltic pumps, on the other hand, are primarily used for pumping fluids without contaminating or damaging them. They operate by compressing and releasing a flexible tube to push the fluid through. While they are effective for transferring liquids, they are not designed to decrease pressure within a tank.

Gear pumps and centrifugal pumps are both types of positive displacement pumps commonly used for fluid transfer. They are designed to increase pressure and flow rate, rather than decrease pressure. Gear pumps use meshing gears to push the fluid, while centrifugal pumps use an impeller to impart centrifugal force to the fluid. Therefore, neither of these pump types is suitable for reducing pressure within a tank.

In conclusion, if the goal is to decrease the pressure within a tank, the best choice is a vacuum pump, as it is specifically designed for this purpose and can create a vacuum or low-pressure environment by removing air and gases from the tank.

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The given data is 20V, 0.5μF, t=0, 80 mH, 500Ω, 250Ω, 25mA. To find i(0+), V(0*), and dv(0+), we follow the steps below.

Firstly, we find the value of V(0*) and V(0+), which are both 20V, as the switch is initially in its position for a long time. Then, we calculate dv(0+) by dividing V(0+) by the sum of resistances R1 and R2, which is [V(0+)/{250 + 500}] = 20/750 = 0.02667 V/s.

Next, we calculate i(0+) by using KVL at t = 0+ with the equation [L(di/dt) + iR = V]. We obtain i(0+) = V/R2 = 20/500 = 40mA, where R1 and R2 are parallel connected.

Then, we can write the differential equation for the circuit by taking L = 80 mH and R = R1 + R2 = 750Ω. We get [L(di/dt) + iR = V] => [0.08 x (di/dt) + (750)i = 20].

To solve this differential equation and find i(t), we assume i(t) = ke^(st) and differentiate it twice. We get [0.08(di/dt) + 750i = 20] => [0.08(d^2 i/dt^2) + 750(di/dt) = 0].

By putting i(t) = ke^(st), we get s^2 + 9375s + 125000 = 0. The roots of this quadratic equation are s = -125 and -75. Therefore, the solution for i(t) is i(t) = c1e^(-125t) + c2e^(-75t).

In summary, we can find i(0+), V(0*), and dv(0+) by following the above steps and use the obtained values to solve the differential equation and find i(t)..

To find the value of constants c1 and c2, we will use the initial conditions. The initial condition for i(0+) is c1 + c2 = 40 mA, which can be rewritten as c1 + c2 = 0.04A.

Next, we will use the initial condition for dv(0+), which is [V(0+)/{250 + 500}] = [20/750] = [L(di/dt)]0+ + i(0+)R. Substituting the values, we get 0.02667 = [0.08(di/dt)]0+ + (40 x 750).

On integrating, we get the equation i(t) = [c1e^(-125t) + c2e^(-75t)] and dv(t) = L(di/dt) => dv(t) = 0.08c1e^(-125t) + 0.08c2e^(-75t).

To find the values of c1 and c2, we will use the initial condition for dv(0+), which is [V(0+)/{250 + 500}] = [20/750] = [L(di/dt)]0+ + i(0+)R. Substituting the values, we get 0.02667 = [0.08(di/dt)]0+ + (40 x 750).

On solving the equation, we get [c1 + c2 = 0.04]......(1) and [10c1 + 20c2 = -2]......(2).

Solving equation (1) and (2), we get c1 = -0.000444 A and c2 = 0.040444 A. Therefore, the final equations are i(t) = [-0.000444 e^(-125t) + 0.040444 e^(-75t)] and dv(t) = 0.08[-0.000444 e^(-125t) - 0.003033 e^(-75t)].

The required solutions are i(t) and v(t).

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1) The Fourier Series of g(x) has only one non-zero coefficient that is a0. The reason behind it is the function g(x) is an odd function. All the cosine coefficients of odd functions are zero, and only one sine coefficient is non-zero. Thus, b1 = 1 is the only non-zero coefficient, and a0 = 0, a1 = 0, b0 = 0 are zero coefficients.

2) The Fourier series for the function f(x) defined on [-5, 5] where f(x) = 3H(x - 2) can be calculated as follows.

As per the definition of the Heaviside Step Function (H(x)), it is zero when x < 0 and one when x > 0. Therefore, f(x) = 0, x < 2 and f(x) = 3, x > 2.

The Fourier Series equation is given by:
f(x) = a0/2 + Σ[an*cos(nπx/L) + bn*sin(nπx/L)]

Here, L = (b - a)/2, where b = 5, a = -5, and n is an integer.

The function f(x) is an even function because it is symmetrical about the y-axis. Thus, all the sine coefficients will be zero, and only cosine coefficients will be non-zero.

The Fourier coefficients can be calculated as follows:
a0 = (1/L) ∫f(x) dx, where the integral is taken over one period
a0 = (1/10) ∫3 dx, from x = 2 to 5
a0 = 3/10

an = (2/L) ∫f(x)cos(nπx/L) dx, where the integral is taken over one period
an = (2/10) ∫3cos(nπx/10) dx, from x = 2 to 5
an = (6/nπ) sin(nπ/2)

Thus, the Fourier series for f(x) can be written as:
f(x) = 3/10 + Σ[6/(nπ) sin(nπ/2) cos(nπx/10)]

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