I don't quite get how stepping up V will reduce the resistance in power cables. V=I*R. So if the transformer increases the V, wouldn't the resistance, R increase as well? and the heat dissipated P=I^2/R also increase and more energy is lost? please help clarify. thanksWhy is the V in transformers stepped up to reduce power loss?NO Power=IV losses=I^2R
Volts is inversely proportional to current
So if we DOUBLE the Volts we can HALVE the current
that means the power lost in the resistance has gone down
by a factor of FOUR (as power lost in resistance is proportional to current squared)
So we can either have 4 times less losses in ALL the wiring
or use thinner wire
The losses in the laminations of the transformer are related to frequency and current. so having larger voltage helps in the design of the transformersWhy is the V in transformers stepped up to reduce power loss?
the two are not proportional,Why is the V in transformers stepped up to reduce power loss?you're not looking at this correctly. you need to focus on the I.
don't just look at the equations but at the transformation process. as the V is stepped up by transformation, the impedance is viewed based on the ratio of the primary and secondary coils (the number of turns determines coil ratios). Based on the ratio, with the increase in voltage (ignoring v=ir), we're able to see a decrease in the current i via the coil transfer
plugging the i into p=i^2/r you see the large reduction in power dissipationWhy is the V in transformers stepped up to reduce power loss?
Let's assume you are trying to transmit power over a fixed length of cable with resistance R. Transmission using a high voltage and low current is always more efficient than high current and low voltage. By keeping the voltage high and the current low the I^2 R losses in the line are kept to a minimum. R for the cable remains constant.Why is the V in transformers stepped up to reduce power loss?yep V=I*R so use this to calculate the voltage drop in a wire of a fixed resistance and you find that the higher the current the higher the voltage drop. power delivered at the load = V*I so for a given power at the load if you increase the V across the terminals you reduce the I carried along the wires (this is all DC). Power dissipated is I^2*R where R is the resistance in the load (electric kettle element say) in series with the resistance in all the wiring between the metering point and the load (call it r) so if you want to dissipate 1kW in your kettle and you double the voltage at the load you need to increase R by 4 to get the same load dissipation - I is reduced to about 1/2 of its original value and the line losses are cut to about 1/4 of what they were (assuming you don't also increase line resistance) - try some simple sums, it really is much better, eventually cranking up the volts becomes dangerous and then you stop.
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