Why would Low Voltage produce HIGH current with the same resistance? Transformers

We always see in Ohm's law that $$v=IR$$. So if $$R$$ is fixed for a wire, how can the electrons with High P.D have low current. Isn't it the electrons bumping into the resistor that creates the heat? Doesn't the rate of electrons bumping into the atoms of the resistor determine how much energy is dissipated from the resistor? If so how can charges store the P.D (High Voltage) in them when they move through the powerlines with a low current.

Assuming that the electrons are "nearly free", the conductance is given by n e^2 tau / m So a material with higher free electron concentration n or a higher mean time between collisions tau will have a higher conductivity.

To answer the question, It is more useful to think of the current as the independent variable. If I is the current then the number of electrons flowing per second is simply I/e where e is the electron charge. But this number is not equal to the electron concentration n mentioned earlier.

Depending on I a potential difference appears across the resistor. This potential difference will be given by IxR.

The electrons do not store the potential difference. In the free electron approximations, reasonable for many situations, the electrons suffer collisions even when there is no current and no potential difference. The power dissipation however occurs only when the electrons flow through the circuit.

Even in a transformer when the voltage is increased, the current decreases but that is an entirely different phenomenon. There the current decreases because of energy conservation. Current x voltage is the power and this should be the same, (even if one assumes no losses) in the primary and secondary coils of the transformer.

Low voltage does not produce high current.

It's easy to see when we arbitrarily set the resistance to 1.

Then $$V = IR = I*1 = I$$

$$V=I$$

The bigger the voltage, the bigger the current.

The power lines have high voltages to minimize power loss that occurs. And the power loss in the transmission cable ($$P_c$$) would then be: $$P_c=I^2R_c$$, where $$R_c$$ is the resistance of the wire. This is for powering a device ($$P$$ is the power consumed) with resistance $$R$$, with current $$I$$ flowing through it, $$P=VI$$.

Combining the two formulas, you get: $$P_c=\frac{P^2R_c}{V^2}$$ which implies that $$V$$ should ideally be very large in cable lines (to minimise $$P_c$$), and the resulting current in them would be smaller (see that $$I=P/V$$ would indicate so).

By the way, it is incorrect to say that "charges" store potential difference. Charges move only because of a potential difference between the two ends of the wire- an electric field exists because of a potential difference.(and so does the charge move due to the electric force acting on it)