A 240 kV 2μs rectangular pulse surge on a transmission line has surge impedance of 350 ohms. It approaches a generating station with capacitance of 3000 pF. The transmitted voltage will be ____________

To find the transmitted voltage when a surge encounters a capacitance, we can use the following formula:

[ V_t = V_i times frac{Z_0}{Z_0 + Z_L} ]

Where:

– ( V_t ) = transmitted voltage

– ( V_i ) = incident voltage (240 kV)

– ( Z_0 ) = surge impedance (350 ohms)

– ( Z_L ) = load impedance

The load impedance ( Z_L ) for a capacitive load can be calculated using the formula:

[ Z_L = frac{1}{jomega C} ]

Where:

– ( C = 3000 , text{pF} = 3000 times 10^{-12} , text{F} )

– ( omega = 2pi f ), and for a 2 μs pulse, we can approximate ( f ) using the rise time, but for surge calculations, we can treat it for a high-frequency response effectively causing a near short circuit (ideally Z_L → 0), making it mostly capacitance impedance.

However, for simplification and practical scenarios in surge calculations, we will approximate ( Z_L ) as a very low impedance due to capacitance, and thus the transmitted voltage can be calculated as:

Using the approximation: ( Z_L approx 0 ), the formula simplifies to:

[ V_t = V