When heat loss is more than energy generated Suppose a electric power station produces 200W of electricity @ 200V. Now instead of stepping it up, it decides to transmit it as 200V to a city 20 km away. The transmission cable has resistance per unit length of  5 ohm/m. As is straightforward, power loss should be 100000W which is way more than what is produced. So final power loss should be 200W only.
Now my question is:
1.Where will this 200W heat be dissipated? (Whether through the whole length of the wire or a part or any other) And why so?
2.What will be the current in the wire? (Since 1A will generate more heat than supplied energy )
Or in an alternative way, suppose a specific power is being generated by a power station (like say 200V, 10A ) which is stepped down to a lower voltage like 20V using a step down transformer. Since power across primary and secondary coil is constant, current in secondary coil should be 100A. But suppose the wire in the secondary coil has a resistance of 10ohm. In that case it cant support a 100A current (since its impossible both by ohms law and energy conservation) and hence the value of current in the secondary will be less than 100A. Hence the power in the secondary will also be less than the primary (less than 2kW). So what happens to the excess power generated? (I.e. the power in the primary coil in excess over the maximum power carried by the secondary coil) 
 A: You cannot have a pd of 200 V at the source and the same pd at the consumer end unless the connecting wires had no resistance.
In your example the resistance of the supply cables is $10^5 \;\Omega$ and so if there was a dead short across the supply at the consumer end the current which would flow in the cables would be $\dfrac {200}{10^5} = 2$ mA and the power loss in the cables would be $200 \times 2 \times 10^{-3} = 0.4 $ W and that would be the power delivered at the power station.
If the consumer connected a $100 \; \Omega$ device across the supply then the voltage across the device would be 100 V the other 100 V being lost across the cables.
The current in the circuit would be be 1 mA.  
The power station would deliver 0.2 W with 0.1 W dissipated in the cables and the consumer using 0.1 W.
A: You MUST be familiar with Ohms law and basic power equations.
There is so much on the internet about them that an explanation here is superfluous.
Ohms law results in the following equation.
The following is the same equation arranged 3 ways.
for V = volts, I = current , R = resistance then. 
R = V/I
I = V/R
V = I x R
Calculate the unknowns from the knowns using this formula.
Energy dissipation in a resistor (derivation on internet) is (3 rearrangements)using Ohms law to convert between variables. 
For P = Power
P = V^2/R
P = I^2 x R
P = V x I 
