Identify Relationship

Since the system resistance is negligible, the fault current is inversely proportional to the total reactance: $I_f = \frac{V_{th}}{X_{th}}$. Initially, $I_{f1} = \frac{V_{th}}{X_{th}} = 8.00$ pu.

$X_{th} = \frac{1}{8.00} = 0.125 \text{ pu}$

Calculate New Reactance

To limit the current to $I_{f2} = 5.00$ pu, we add a series reactance $X_s$. The new total reactance is $X_{total} = X_{th} + X_s$. Thus, $I_{f2} = \frac{V_{th}}{X_{th} + X_s} = 5.00$ pu.

$X_{th} + X_s = \frac{1}{5.00} = 0.200 \text{ pu}$

Find Added Reactance

Substituting $X_{th} = 0.125$ pu into the equation: $0.125 + X_s = 0.200$.

$X_s = 0.200 - 0.125 = 0.075 \text{ pu}$