Recent content by mathological

Right! Ok so when people tell us that a certain power station provides 2000 MW, then not all of this 2000 MW gets to the consumer. Does anyone know how efficient a typical power distribution system is (in terms of what is being output at a generating station say 2000 MW and what the consumers...

cool! understood! Thanks.

Hi, so I have been having sleepless nights about this concept. From what I understand, say a typical power generating plant that produces 2000MW of energy, for example, will want to transmit as much of 2000 MW of power to the grid as possible and they do so by stepping up the voltages to...

@sophiecentaur Thanks for the clarification! :)

You have assumed a cable resistance of 1Ω which would be in series with a load of 100Ω. The voltage drop across the feeder will be 10kV*1/101. Are you using voltage division here? This gives a voltage across the load of 10kV/1.01 which gives about 98% of the power (proportional to the square...

Hi, ok so everytime I think that I have understood the concept of Energy transmission, losses and voltage drop, I get even more confused about things. I have searched several threads on this forum and the physics forum but failed to find anything that directly answers my query. So I...

Thanks Fredrik. Yeah it was a bit stupid of me to ask what k and m were.

Fredrik, you are absolutely correct. Q1 (b) is supposed to read: (b) Find the modulus and argument of z when |z| takes its least value. What will be the value of k and m (I mean how do you obtain these)?

Thanks Fredrik, I was actually taking 1 <= |z+2| <= 3 as the region between circles centered at (-2,0) and with radii 1 unit and 3 units. I see where I went wrong. But just one thing, why isn't |z+2| a circle centred at (-2,0)?

Complex Locus - PLEASE HELP! Homework Statement Q1: The complex number z satisfies arg(z+3) = pi/3 (a) Sketch the locus of the point P in the Argand diagram which represents z (DONE) (b) Find the modulus and argument of z when z takes its least value. (STUCK)[/color] (c) Hence...