A surge of 100 kv travelling on the line of natural impedance 600 ohm arrives at a junction with two lines of impedances 800 ohm and 200 ohm. Find the surge voltage and the current transmitted into each branch of line.

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