**Surge Impedance Loading (SIL) Problem**

*The given data -*

The Surge Magnitude = **E = 100 kv**

Natural Impedance = $Z_1$ **= 600 Ω**

Impedance at Line 1 = $Z_2$ **= 800 Ω**

Impedance at Line 2 = $Z_3$ **= 200 Ω**

*To find -*

a] Surge Voltage = $E^n$ = ?

b] Current Transmitted into each Branch Line $Z_2$ and $Z_3$ = ?

*Formulae -*

$$ Surge\ Voltage = E^n = \frac {\frac{2E}{Z_1}}{\frac {1}{Z_1} + \frac {1}{Z_2} + \frac {1}{Z_3}} $$

$$ Current\ Transmitted\ into\ Branch\ Line\ Z_2 = \frac {E^n \times (Z_2 + Z_3)}{Z_2} $$

$$ Current\ Transmitted\ into\ Branch\ Line\ Z_3 = \frac {E^n \times (Z_2 + Z_3)}{Z_3} $$

*Solution -*

**a] Surge Voltage $E^n$ Calculation -**

$$ E^n = \frac {\frac{2E}{Z_1}}{\frac {1}{Z_1} + \frac {1}{Z_2} + \frac {1}{Z_3}} $$

$$ E^n = \frac {\frac{2 \times 100}{600}}{\frac {1}{600} + \frac {1}{800} + \frac {1}{200}} $$

$$ E^n = \frac {0.3333}{7.9167 \times 10^{-3}} $$

$$ E^n = 42.1\ kV$$

**b] Current Transmitted into each Branch Line $Z_2$ and $Z_3$**

Therefore,

$$ Current\ Transmitted\ into\ Branch\ Line\ Z_2 = \frac {E^n \times (Z_2 + Z_3)}{Z_2} $$

$$ = \frac {42.1 \times (800 + 200)}{800} $$

$$ = \frac {42100}{800} $$

$$ = 52.625\ A $$

and

$$ Current\ Transmitted\ into\ Branch\ Line\ Z_3 = \frac {E^n \times (Z_2 + Z_3)}{Z_3} $$

$$ = \frac {42.1 \times (800 + 200)}{200} $$

$$ = \frac {42100}{200} $$

$$ = 210.5\ A $$

*Answer -*

a] Surge Voltage = $E^n$ **= 42.1 kV**

b] Current Transmitted into Branch Line and $Z_2$ **= 52.625 A**

Current Transmitted into Branch Line $Z_3$ **= 210.5 A**