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Why can't we use v^2/r to calculate power loss in transmission cables?

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Assuming the cable is being used with a signal that changes with time 1.-It is not only "r" what contributes to the loss. There is also a conductance "g" in parallel that contributes to the loss 2.-A transmission cable cannot be considered as a discrete element (like a resistor, for instance) it is a network whose elements are not discrete but distributed. That is why its parameters are given per unit of length. In the case of "r", it is given in ohms/meter or ohms/Km, etc 3.-There are also other distributed parameters such as inductance (l) and capacitance (c in parallel to conductance) therefore the relation between Voltage and Current is defined by more complex equations than just V=I·r 4.-Normally resistance per unit of length is not used isolatedly to calculate losses. Derived from r,l,g and c (primary parameters) are the secondary parameters such as propagation, attenuation, phase and characteristic impedance that help us to model the line and its behavior. They will help us to calculate the power fed to the cable and the power delivered to the load at the other end. 5.-Have a look at the "Telegrapher´s equations" to see the relation between voltage and current
i don't understand, why is I^2R used to calculate power loss?
In transmission lines? First news to me

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What is the R in a transmission line?
sorry I have no idea ;/ im getting confused. I haven't learnt about the other complex equations yet, do u know of another way to understand this?
Let us clarify one point. Are you talking about transmission lines (telecommunications such as telephone) or electric power transmission lines or distribution lines (the ones that transport the electric power used in our homes and industries)
electric power transmission lines
U can always use.. V^2/R.. you would end up with the same result.. what u need to understand is that usually we know what current flows through the wire.. and so we use i^2R.. so its like this.. |dw:1379223314493:dw| see the power station produces a voltage Va which is send to ur house... since that entire power station is coming in parallel with your load L (load can be anything like bulb, tubelight or a death beam) you would expect the potential difference between your load also to be Va. but here is the problem.. the power line has a resistance,.. and so there will be loss.. because of this loss. the potential difference at your end is less which is Vb now lets calculate how much IS that loss Vb = Va - Vr ( Vr = potential drop across the entire power line) if u know Vr .. then u can use the power loss = Vr^2/R like u said.. instead lets say you know what is the current i ( i = Vr/R) then now i can say power loss = i^2 R... so u see it doesn't matter what you use.. its much easier to calculate current in the wire.. and say i^2R loss.. rather than calculating potential drop across the wire and saying Vr^2/R loss..
@cherrydrops. One thing you must take into account: 1.-Transmission Lines always refer to telecommunications. As I have explained, to determine the power loss is a bit complex. Transmission always refer to sending and propagating an analogue or digital "information signal" over a medium (wired, wireless or optical fiber) 2.-Electric Power Transmission Lines (better call them Electric Power transport or distribution lines) allow you to calculate power loss as v^2/R or i^2R as @Mashy has perfectly explained. Electric Power Lines do not transmit information therefore they are not transmission lines. Sometimes they are used to transmit information about power consumption/distribution but it is not their main purpose.
i think he actually wanted to say power lines.. :)..
Thanks @CarlosGP and @Mashy, i finally got it :)

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