A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The half the voltage regulation will be _____________

To find the half voltage regulation, we first need to calculate the voltage regulation and then divide it by 2.

### Step 1: Calculate the receiving end current (I_R)

For an inductive load, the apparent power (S) can be calculated as:

[ S = frac{P}{text{Power Factor}} = frac{5 , text{MW}}{0.8} = 6.25 , text{MVA} ]

The receiving end line current (I_R) can be calculated using:

[ I_R = frac{S}{sqrt{3} cdot V_R} ]

Given ( V_R = 10 , text{kV} = 10,000 , text{V} ):

[ I_R = frac{6,250,000 , text{VA}}{sqrt{3} cdot 10,000 , text{V}} approx frac{6,250,000}{17,320.5} approx 360.6 , text{A} ]

### Step 2: Calculate the voltage drop (V_drop)

Using the transmission line parameters:

– Resistance (R) = 0.39 Ω

– Reactance (X) = 3.96 Ω

– Total impedance (Z) = ( R + jX = 0.39 + j3.96