To solve this problem, we need to calculate the reliability index (ẞ) for the power transformer. The reliability index is a measure of how likely the system is to meet its performance requirements, i.e., ensuring that the load current $I_L$ does not exceed the transformer's current capacity $I_C$.

Given Information:

• Load current $I_L$ is normally distributed with:

  • Mean $\mu_L = 1200 \, \text{A}$
  • Standard deviation $\sigma_L = 100 \, \text{A}$

• Current capacity $I_C$ is normally distributed with:

  • Mean $\mu_C = 1500 \, \text{A}$
  • Standard deviation $\sigma_C = 150 \, \text{A}$

Limit State Function:

The limit state function is defined as: $g(I_C, I_L) = I_C - I_L$ The transformer is safe when $g(I_C, I_L) > 0$, meaning the current capacity exceeds the load current.

Step 1: Compute the Reliability Index $ẞ$

The reliability index $ẞ$ is given by the formula: $ẞ = \frac{\mu_C - \mu_L}{\sqrt{\sigma_C^2 + \sigma_L^2}}$

Calculate the combined standard deviation:

$\sigma_g = \sqrt{\sigma_C^2 + \sigma_L^2} = \sqrt{(150)^2 + (100)^2} = \sqrt{22500 + 10000} = \sqrt{32500} \approx 180.28$

Calculate the reliability index $ẞ$:

$ẞ = \frac{\mu_C - \mu_L}{\sigma_g} = \frac{1500 - 1200}{180.28} \approx \frac{300}{180.28} \approx 1.664$

Thus, the reliability index $ẞ$ is approximately 1.664.

Step 2: Determine the Design Point (Most Probable Point, MPP