Anomaly in transmission of electric current?

Question

We step up electricity to reduce heat loss while transmitting over long distances. The heat loss is given by I^2Rt. While stepping up the current is decreased and thus heat loss is reduced. All Right.
But what if we use Ohm's law V = IR here? Now I^2Rt becomes V^2t/R. Then when we step up the voltage the Heat Loss is increased considerably. Can someone please explain this?

Answer 1

The V in V=IR refers to the voltage drop across the resistor, not relative to ground. So, if current I drops, this V drops too and, therefore, V^2t/R drops. It's voltage relative to ground V_0 that increases in a step-up transformer. Transmitted power is V_0*I, so if V_0 increases, the same power is transmitted with less I.

Answer 2

You have got it wrong. The voltage you are saying is a drop along the transmission line not the supply voltage. Let say line Resistance is 10 ohms, if we deliver 1KW power with 100 V supply then the current is 1000/100 = 10 A. Line loss is then 10X10 = 100 W. If we increase supply voltage to 1KV, current will be 1000/1000=1 A.Line loss will be 10X 1=10 W.
Hope you will understand.

Answer 3

Think in terms of budget. We step up the voltage and step down the current so that, for the same allowable losses, we can reduce the copper (size of wire). Insulation goes up, but is still cheaper than copper.

For example, let's say we have a run miles long. The resistance of the copper wire is 1 ohm, the voltage is 10,000 volts, and the current 100 amps. Then the losses in the line would be

I^2R = 10kW

The voltage drop between the ends of the line would be IR = 100(1) = 100V. Note that V^2/R =100^2(1) = 10kW

Now instead step the voltage up and the current down by a factor of 10, to 100kV and 10 amps. Now I^2R losses are

10^2(1)=100W

The voltage drop through the line is now only IR=10V, and note that

V^2/R = 100/1 = 100W

V^2/R is dropping because the voltage drop through the line is rapidly dropping due to the lower current through the same resistancce.

Answer 4

deviation from the regular passing of current

Answer 5

Wow, thanks! I was wondering the same thing yesterday