Question

The long distance transmission of electrical energy is done at
(A) High potential and low current
(B) Low potential and high current
(C) High potential and high current
(D) Low potential and low current

Answer 1

Hint: The higher the voltage used to power a load, the lower the current and the lower the voltage used, the higher the current flowing through the load. The power loss due to a resistive load is directly proportional to the square of the current flowing through it. The choice of voltage and current during transmission is to reduce the power losses in the conducting medium.

Formula used: In this solution we will be using the following formulae;
 $\Rightarrow P = IV $ where $ P $ is the power across (or provided to) a load, $ I $ is the current through the load, and $ V $ is the voltage across the load.
 $\Rightarrow P = {I^2}R $ , $ P $ is the power loss across a resistance $ R $ .

Complete step by step solution:
During transmission of electricity, especially long distance transmission, there are losses of energy in the transmitting conductor. Lots of effort is usually put in, to reduce these losses. One of these efforts is in calculating the optimum voltage and current required to transmit to a particular place.
Let’s investigate which of these combinations in the options will reduce losses.
Say, the power required to power a load is $ P $ . Then
 $\Rightarrow P = IV $ where , $ I $ is the current through the load, and $ V $ is the voltage across the load.
The power loss in the transmitting conductor is given as
 $\Rightarrow {P_l} = {I^2}R $ where $ R $ is the resistance of the conductor. Hence, the higher the current the higher the losses. This establishes that the long distance transmission must be done at low current to minimize losses.
To determine the voltage of transmission, we must calculate the corresponding voltage to this low current. For a given load of power $ P = IV $ , the current through is given by
 $\Rightarrow I = \dfrac{P}{V} $ . This implies that to achieve a low current in the circuit, the voltage must be high.
Hence, a high voltage must be used for long distance transmission.
Therefore, our answer is A, high voltage, low current.

Note:
Alternatively, the power losses in the conductor can also be expressed as
 $\Rightarrow {P_l} = \dfrac{{{V^2}}}{R} $ . The voltage in this case is the voltage drop across the conductor due to its resistance.
But from ohm's law, $ V = IR $ , which thus still implies that a higher current in the conductor causes a higher voltage drop, which then causes a higher power loss. We can derive the initial expression by replacing $ V = IR $ into $ {P_l} = \dfrac{{{V^2}}}{R} $ as in
 $\Rightarrow {P_l} = \dfrac{{{{\left( {IR} \right)}^2}}}{R} = \dfrac{{{I^2}{R^2}}}{R} = {I^2}R $ .