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Find the surge voltage and the current transmitted into each branch of line.

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.

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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

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