The inductance of the cable is calculated to be 16.5 mH (approx).
Single-phase testing (ideal) transformer 50 kVA, 0.4/150 kV50 Hz10% leakage reactance 2% resistance on 50 kVA base1% resistance for the additional inductor to be used at connecting leads
The inductance of the cable can be calculated by using the resonant circuit formula.Let;L = inductance of the cableC = Capacitance of the cable
r1 = Resistance of the inductor
r2 = Resistance of the cable
Xm = Magnetizing reactance of the transformer
X1 = Primary reactance of the transformer
X2 = Secondary reactance of the transformer
The resonant frequency formula is; [tex]f = \frac{1}{{2\pi \sqrt{{LC}}}}[/tex]
For the resonant condition, reactance of the capacitor and inductor is equal to each other. Therefore,
[tex]\[XL = \frac{1}{{2\pi fL}}\][/tex]
[tex]\[XC = \frac{1}{{2\pi fC}}\][/tex]
So;
[tex]\[\frac{1}{{2\pi fL}} = \frac{1}{{2\pi fC}}\][/tex] Or [tex]\[LC = \frac{1}{{f^2}}\][/tex] ----(i)
Also;
[tex]Z = r1 + r2 + j(Xm + X1 + X2) + \frac{1}{{j\omega C}} + j\omega L[/tex] ----(ii)
The impedence of the circuit must be purely resistive.
So,
[tex]\text{Im}(Z) = 0 \quad \text{or} \quad Xm + X1 + X2 = \frac{\omega L}{\omega C}[/tex]----(iii)
Substitute the value of impedance in equation (ii)
[tex]Z = r1 + r2 + j(0.1 \times 50 \times 1000) + \frac{1}{j(2\pi \times 50) (1 + L)} + j\omega L = r1 + r2 + j5000 + \frac{j1.59}{1 + L} + j\omega L[/tex]
So, [tex]r1 + r2 + j5000 + \frac{j1.59}{1 + L} + j\omega L = r1 + r2 + j5000 + \frac{j1.59}{1 + L} - j\omega L[/tex]
[tex]j\omega L = j(1 + L) - \frac{1.59}{1 + L}[/tex]
So;
[tex]Xm + X1 + X2 = \frac{\omega L}{\omega C} = \frac{\omega L \cdot C}{1}[/tex]
Substitute the values; [tex]0.1 \times 50 \times 1000 + \omega L (1 + 0.02) = \frac{\omega L C}{1} \quad \omega L C - 0.02 \omega L = \frac{5000 \omega L}{1 + L} \quad \omega L (C - 0.02) = \frac{5000}{1 + L}[/tex] ---(iv)
Substitute the value of L from equation (iv) in equation (i)
[tex]LC = \frac{1}{{f^2}} \quad LC = \left(\frac{1}{{50^2}}\right) \times 10^6 \quad L (C - 0.02) = \frac{1}{2500} \quad L = \frac{{C - 0.02}}{{2500}}[/tex]
Put the value of L in equation (iii)
[tex]0.1 \times 50 \times 1000 + \omega L (1 + 0.02) = \frac{\omega L C}{1} \quad \frac{\omega L C - 0.02 \omega L}{1} = \frac{5000 \omega L}{1 + L} \quad \frac{\omega L C - 0.02 \omega L}{1} = \frac{5000}{1 + \left(\frac{C - 0.02}{2500}\right)} \quad \frac{\omega L C - 0.02 \omega L}{1} = \frac{5000}{1 + \frac{C + 2498}{2500}} \quad \frac{\omega L C - 0.02 \omega L}{1} = \frac{12500000}{C + 2498}[/tex]
Now, substitute the value of ωL in equation (iv);[tex]L = \frac{{C - 0.02}}{{2500}} = \frac{{12500000}}{{C + 2498}} \quad C^2 - 49.98C - 1560.005 = 0[/tex]
Solve for C;[tex]C = 41.28 \mu F \quad \text{or} \quad C = 37.78 \mu F[/tex] (neglect)
Hence, the inductance of the cable is (C-0.02) / 2500 = 16.5 mH (approx).
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Explain the advantages and disadvantages of the 2 ray ground reflection model in the analysis of path loss. (b) In the following cases, tell whether the 2-ray model could be applied, and explain why or why not: h t
=35 m⋅h r
=3 m,d=250 m
h t
=30 m,h r
=1.5 m⋅d=450 m
The two-ray ground reflection model in the analysis of path loss has the following advantages and disadvantages:
Advantages: It provides a quick solution when using hand-held calculators or computers because it is mathematically easy to manipulate. There is no need for the distribution of the building, and the model is applicable to any structure height and terrain. The range is only limited by the radio horizon if the mobile station is located on a slope or at the top of a hill or building.
Disadvantages: It is an idealized model that assumes perfect ground reflection. The model neglects the impact of environmental changes such as soil moisture, surface roughness, and the characteristics of the ground.
The two-ray model does not account for local obstacles, such as building and foliage, in the transmission path.
Therefore, the two-ray model could not be applied in the following cases:
Case 1hₜ = 35 m, hᵣ = 3 m, d = 250 m The distance is too short, and the building is not adequately covered.
Case 2hₜ = 30 m, hᵣ = 1.5 m, d = 450 m The obstacle height is too small, and the distance is too long to justify neglecting other factors.
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QUESTION 1 Which of the followings is true? Narrowband FM is considered to be identical to AM except O A. their bandwidth. O B. a finite and likely large phase deviation. O C. an infinite phase deviation. O D. a finite and likely small phase deviation.
Narrowband FM is considered to be identical to AM except in their bandwidth. In narrowband FM, a finite and likely small phase deviation is present. It is the modulation method in which the frequency of the carrier wave is varied slightly to transmit the information signal.
Narrowband FM is an FM transmission method with a smaller bandwidth than wideband FM, which is a more common approach. Narrowband FM is quite similar to AM, but the key difference lies in the modulation of the carrier wave's amplitude in AM and the modulation of the carrier wave's frequency in Narrowband FM.
The carrier signal in Narrowband FM is modulated by a small frequency deviation, which is inversely proportional to the carrier frequency and directly proportional to the modulation frequency. Therefore, Narrowband FM is identical to AM in every respect except the bandwidth of the modulating signal.
When the modulating signal is a simple sine wave, the carrier wave frequency deviates up and down about its unmodulated frequency. The deviation of the frequency is proportional to the amplitude of the modulating signal, which produces sidebands whose frequency is equal to the carrier frequency plus or minus the modulating signal frequency.
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Prove that a Schmitt oscillator trigger can work as a VCO.
Step 1:
A Schmitt oscillator trigger can work as a VCO (Voltage Controlled Oscillator).
Step 2:
A Schmitt oscillator trigger, also known as a Schmitt trigger, is a circuit that converts an input signal with varying voltage levels into a digital output with well-defined high and low voltage levels. It is commonly used for signal conditioning and noise filtering purposes. On the other hand, a Voltage Controlled Oscillator (VCO) is a circuit that generates an output signal with a frequency that is directly proportional to the input voltage applied to it.
By incorporating a voltage control mechanism into the Schmitt trigger circuit, it can be transformed into a VCO. This can be achieved by introducing a variable voltage input to the reference voltage level of the Schmitt trigger. As the input voltage changes, it will cause the switching thresholds of the Schmitt trigger to vary, resulting in a change in the output frequency.
The VCO functionality of the modified Schmitt trigger circuit allows it to generate a continuous output signal with a frequency that can be controlled by the applied voltage. This makes it suitable for various applications such as frequency modulation, clock generation, and signal synthesis.
Step 3:
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Describe frequency, relative frequency, and cumulative relative frequency.
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Develop a minimum-multiplier realization of a length-7 Type 3 Linear Phase FIR Filter.
A minimum-multiplier realization of a length-7 Type 3 Linear Phase FIR Filter can be developed.
To develop a minimum-multiplier realization of a length-7 Type 3 Linear Phase FIR Filter, we need to understand the key components and design considerations involved. A Type 3 Linear Phase FIR Filter is characterized by its linear phase response, which means that all frequency components of the input signal experience the same constant delay. The minimum-multiplier realization aims to minimize the number of multipliers required in the filter implementation, leading to a more efficient design.
In this case, we have a length-7 filter, which implies that the filter has 7 taps or coefficients. Each tap represents a specific weight or gain applied to a delayed version of the input signal. To achieve a minimum-multiplier realization, we can exploit the symmetry properties of the filter coefficients.
By carefully analyzing the symmetry properties, we can design a structure that reduces the number of required multipliers. For a length-7 Type 3 Linear Phase FIR Filter, the minimum-multiplier realization can be achieved by utilizing symmetric and anti-symmetric coefficients. The symmetric coefficients have the same value at equal distances from the center tap, while the anti-symmetric coefficients have opposite values at equal distances from the center tap.
By taking advantage of these symmetries, we can effectively reduce the number of multipliers needed to implement the filter. This results in a more efficient and resource-friendly design.
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8. Write and execute a query that will delete all countries that are not assigned to an office or a client. You must do this in a single query to receive credit for this question. Write the delete query below and then execute the following statement in SQL Server: Select * from Countries. Take a screenshot of your select query results and paste them below your delete query that you constructed.
The Countries which are not assigned any Office means that the values are Null or Blank:
I created a table:
my sql> select*from Country; + | Country Name | Office | - + | Yes | NULL | Yes | Croatia | Argentina Sweden Brazil Sweden | Au
Here in this table there is Country Name and a Office Column where it is Yes, Null and Blank.
So, we need to delete the Blank and Null values as these means that there are no office assigned to those countries.
The SQL statement:
We will use the delete function,
delete from Country selects the Country table.
where Office is Null or Office = ' ' ,checks for values in Office column which are Null or Blank and deletes it.
Code:
mysql> delete from Country -> where Office is Null or Office = ''; Query OK, 3 rows affected (0.01 sec)
Code Image:
mysql> delete from Country -> where Office is Null or Office Query OK, 3 rows affected (0.01 sec) =
Output:
mysql> select*from Country; + | Country Name | Office | + | Croatia Sweden Sweden | India | Yes | Yes Yes | Yes + 4 rows in s
You can see that all the countries with Null and Blank values are deleted
A cylinder with a movable piston contains 5.00 liters of a gas at 30°C and 5.00 bar. The piston is slowly moved to compress the gas to 8.80bar. (a) Considering the system to be the gas in the cylinder and neglecting ΔEp, write and simplify the closed-system energy balance. Do not assume that the process is isothermal in this part. (b) Suppose now that the process is carried out isothermally, and the compression work done on the gas equals 7.65L bar. If the gas is ideal so that ^ U is a function only of T, how much heat (in joules) is transferred to or from (state which) thes urroundings? (Use the gas-constant table in the back of the book to determine the factor needed to convert Lbar to joules.)(c) Suppose instead that the process is adiabatic and that ^ U increases as T increases. Is the nal system temperature greater than, equal to, or less than 30°C? (Briey state your reasoning.)
A cylinder with a movable piston contains 5.00 liters of a gas at 30°C and 5.00 bar. The piston is slowly moved to compress the gas to 8.80bar.
(a) The closed-system energy balance can be written as follows:ΔU = Q − W, where ΔU is the change in internal energy, Q is the heat transferred to the system, and W is the work done by the system. Neglecting ΔEp, the work done by the system is given by W = PΔV, where P is the pressure and ΔV is the change in volume. Therefore, ΔU = Q − PΔV.
(b) Since the process is carried out isothermally, the temperature remains constant at 30°C. Therefore, ΔU = 0. The work done by the system is
W = −7.65 L bar, since the compression work is done on the gas. Using the gas constant table, we find that 1 L bar = 100 J. Therefore, the work done by the system is
W = −7.65 L bar × 100 J/L bar = −765 J. Since
ΔU = 0, we have Q = W = −765 J. The heat is transferred from the system to the surroundings.
(c) Since the process is adiabatic, Q = 0. Therefore, the closed-system energy balance simplifies to ΔU = −W. Since the gas is ideal and ^ U is a function only of T, the change in internal energy can be written as ΔU = (3/2)nRΔT, where n is the number of moles of gas, R is the gas constant, and ΔT is the change in temperature. Since ^ U increases as T increases, we have ΔU > 0. Therefore, ΔT > 0, and the final system temperature is greater than 30°C.
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To achieve maximum power transfer between a 44 Ω source and a load ZL (ZL > ZG) using a transmission line with a characteristic impedance of 44 Ω, an inductor with a reactance of 82 Ω is connected in series with the source. Determine the distance from the load, ZL, in terms of wavelengths where the inductor should be connected. Length = λ
The inductor should be connected at a distance of 2 wavelengths from the load, ZL, to achieve maximum power transfer.
To determine the distance, we need to consider the conditions for maximum power transfer. When the characteristic impedance of the transmission line matches the complex conjugate of the load impedance, maximum power transfer occurs. In this case, the load impedance is ZL, and we have ZL > ZG, where ZG represents the generator impedance.
Since the transmission line has a characteristic impedance of 44 Ω, we need to match it to the load impedance ZL = 44 Ω + jX. By connecting an inductor with a reactance of 82 Ω in series with the source, we effectively cancel out the reactance of the load impedance.
The electrical length of the transmission line is given by the formula: Length = (2π / λ) * Distance, where λ is the wavelength. Since the inductor cancels the reactance of the load impedance, the transmission line appears purely resistive. Hence, we need to match the resistive components, which are 44 Ω.
For maximum power transfer to occur, the inductor should be connected at a distance of 2 wavelengths from the load, ZL.
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