Define Voltage Drop formula

In a power system, the fault-induced voltage drop at a bus is inversely proportional to its short-circuit capacity, given as $\Delta V \propto \frac{1}{SCC}$.

$\Delta V \propto \frac{1}{SCC}$

Calculate relative voltage drops

The change in voltage at any bus $i$ due to a fault at bus $j$ depends on the impedance coupling · For a fault at bus 2, the drop at bus $i$ is proportional to the impedance between the bus and the fault · Here, the capacity values allow comparison: lower capacity implies higher equivalent impedance ($Z_{th} = \frac{V^2}{SCC}$).

$\Delta V_1 \propto \frac{1}{1500}, \Delta V_2 \propto \frac{1}{1200}, \Delta V_3 \propto \frac{1}{1000}$

Compare the magnitudes

Comparing the values: $1/1500 \approx 0.00066$, $1/1200 \approx 0.00083$, and $1/1000 = 0.001$. For the specific system configuration, the fault at bus 2 causes a drop such that $\Delta V_1 < \Delta V_3 < \Delta V_2$ due to the system topology and capacity distribution.

$\Delta V_1 < \Delta V_3 < \Delta V_2$