Question

If a line having a total resistance $0.252$ delivers $10kW$ at $250volts$ to a small factory, then the efficiency of transmission is:
A) $97\%$
B) $87\%$
C) $83\%$
D) $76\%$

Answer 1

The efficiency describes the relation between the power given to the system and the output power from the system. Ratio between the power delivered to the power supplied determines the efficiency of the transmission of electricity.

Complete step by step solution:
Let’s define the terms given in the question.
It is given that resistance of the line which is used to transmit the electricity, $R = 0.252\Omega$
Power delivered to the small factory through the line, ${P_D} = 10kW$
The voltage of the electricity which is supplied to the small factory, $v = 250volts$
We know, whenever an electric current passes through a material, the resistance of that particular material causes a loss of power by means of heat. And the power lost is given by the product of resistance of the particular material and the square of the electric current by which the electricity transmitted.
That is, Power lost = ${i^2}R$ ………………………………………. (1)
We know, the electric current is given by the equation,
$i = \dfrac{P}{v}$ …………………………………………….……………….. (2)
Here, $v = 250volts$
$P = 10kW$
By applying simple mathematical conversions, we get,
$P = 10kW = 10000W$
By applying the values of each term in equation (2) we will get,
$i = \dfrac{P}{v}$
$\Rightarrow i = \dfrac{{10000}}{{250}} = 40A$
By applying the values of electric current and resistance in equation (1) we will get,
Power lost = ${i^2}R$
Power lost = ${40^2} \times 0.252$
$\Rightarrow 1600 \times 0.252 = 403W$
$\Rightarrow 403W = 0.403kW$
We know the power supplied to the line will be equal to the sum of power delivered at the small factory and the power lost by the action of resistance while the transmission of the electricity.
Power supplied to the line=power lost + power delivered
That is, ${P_S} = {P_{lost}} + {P_D}$
${P_S} = 10kW + 0.403kW$
$\Rightarrow {P_S} = 10.403kW$
The efficiency is given by the equation,
Efficiency, $\eta = \dfrac{{{P_D}}}{{{P_S}}} \times 100$
$\Rightarrow \eta = \dfrac{{10kW}}{{10.403kW}} \times 100$
$\Rightarrow \eta = 96.12\% \approx 97\%$
So the efficiency of the transmission of electricity to the small factory is $97\%.$

So the final answer is option (A): $97\%.$

Note: Efficient transmission involves reducing the currents by stepping up the voltage prior to transmission, and stepping it down at a substation at the far end. For AC power transmission the stepping up and down is done using transformers.