Generator is supplying power to a factory by cables of resistance 20 ohm if the generator is generating 50 kilowatt power at 5000 volt.what is the power receiver factory?

Power = i V

50,000 = i (5000)
i = 10 amps

Power lost = i^2 R = 100(20) = 2000 watts
so
Power received at factory = 50,000 - 2,000
= 48 kilowtts

Given,

Resistance(R)=20 ohm
Volts supplied(V)=5000V
Power(P)=50kW=50,000W
We know,
P=Vi
Or, i=50,000/5000
Or, i=10 A
Power lost= i^2R
=10×10×20
=2000 W
Power recieved at the factory=(50,000-2000)W
=48000 W = 58 kW.
I think it will help.👍👍

Sorry. It will be 48 kW

To determine the power received by the factory, we need to first calculate the current flowing through the cables using Ohm's Law. Ohm's Law states that the current (I) flowing through a conductor is equal to the voltage (V) divided by the resistance (R). Mathematically, it can be expressed as:

I = V / R

Given that the resistance of the cables is 20 ohms and the voltage generated by the generator is 5000 volts, we can substitute these values into the formula to determine the current flowing through the cables:

I = 5000 V / 20 Ω
I = 250 Amperes

Now, we can calculate the power received by the factory using the formula: P = I * V, where P represents power, I represents current, and V represents voltage. Substituting the values into the equation:

P = 250 A * 5000 V
P = 1,250,000 Watts

Therefore, the power received by the factory is 1,250,000 Watts or 1.25 Megawatts.

It is given in the question.

50,000kese aaya