Answer:
1)b2)b3)d4)c5)c 6). b7).a 8).b 9). b 10)d
Step-by-step explanation:
In AC-DC controlled rectifiers, the average load voltage decreases as the firing angle increases.
The firing angle is the delay between the zero crossing of the input AC voltage and the triggering of the thyristor. As the firing angle is increased, the conduction angle of the thyristor decreases, which reduces the amount of time that the thyristor conducts and the amount of time that the load is connected to the input voltage.
As a result, the average load voltage decreases as the firing angle is increased. Therefore, option b is correct: "The average load voltage decreases as the firing angle increases."
2)
Eliminating the effect of capacitance in transmission lines is not an advantage of conductor bundling.
Conductor bundling is the practice of grouping two or more conductors together in a transmission line to reduce the inductance, increase the capacitance, and improve the overall performance of the line.
The advantages of conductor bundling include reducing the skin effect losses, reducing the series inductance of the transmission lines, and increasing the ratings at less conductor weight of transmission lines.
However, conductor bundling does not eliminate the effect of capacitance in transmission lines. In fact, conductor bundling increases the capacitance of the line, which can be both an advantage and a disadvantage depending on the application.
Therefore, option b is correct: "Eliminate the effect of capacitance in transmission lines" is not an advantage of conductor bundling.
3)
To solve this problem using the Gauss iterative method, we need to follow these steps:
Step 1: Assume an initial value for the voltage magnitude and phase angle at bus 2. Let's assume that the initial voltage at bus 2 is 1.0∠0°.
Step 2: Calculate the complex power injection at bus 2 using the formula:
S₂ = V₂ (Y₂₁ V₁* + Y₂₂ V₂* + Y₂₃ V₃*)
where V₁*, V₂*, and V₃* are the complex conjugates of the voltages at buses 1, 2, and 3, respectively. Using the given values, we get:
S₂ = (1.0∠0°) (40j (1.04∠0°) + (-60j) (1.0∠0°) + 20j (1.04∠0°))
S₂ = -62.4j
Step 3: Calculate the updated value of the voltage at bus 2 using the formula:
V₂,new = (1/S₂*) - Y₂₁ V₁* - Y₂₃ V₃*
where S₂* is the complex conjugate of S₂. Using the given values, we get:
V₂,new = (1/62.4j) - 40j (1.0∠0°) - 20j (1.04∠0°)
V₂,new = 0.016025 - 0.832j
Step 4: Calculate the difference between the updated value and the assumed value of the voltage at bus 2:
ΔV = V₂,new - V₂,old
where V₂,old is the assumed value of the voltage at bus 2. Using the values we assumed and calculated, we get:
ΔV = (0.016025 - 0.832j) - (1.0∠0°)
ΔV = -0.983975 - 0.832j
Step 5: Check if the difference is within the acceptable tolerance. If the difference is greater than the tolerance, go back to step 2 and repeat the process. If the difference is smaller than the tolerance, the solution has converged.
The answer to this problem is option d: 0.9734∠-3.4356°.
4)The most suitable method to solve power flow problems in large power systems is the Newton Raphson method.