Question

If 2.2 kW power is transmitted through a $10\;\Omega$ line at 22 kV, power loss in the form of heat will be
A. 0.1 watt
B. 14 watts
C. 100 watts
D. 1000 watts

Answer 1

The above problem can be resolved using the general formula for the power loss, for a given magnitude of current and resistance. The power loss is calculated by taking the product of supplied power and the resistance and then taking the ratio with the square of the voltage supply. After that, the given values are substituted to obtain the required result. Along with this, it is necessary to look towards the fact that whether the units are given in specific terms like either in terms of volts or kilovolts and many more.

Complete step by step answer:
Given:
The power transmitted is $P = 2.2\;{\rm{kW}} = 2.2\;{\rm{kW}} \times \dfrac{{1000\;{\rm{W}}}}{{1\;{\rm{kW}}}} = 2.2 \times {10^3}\;{\rm{W}}$
The magnitude of resistance is, $R = 10\;\Omega$.
The voltage supply of the transmission line is,
$V = 22\;{\rm{kV}} = 22\;{\rm{kV}} \times \dfrac{{1000\;{\rm{V}}}}{{1\;{\rm{kV}}}} = 22 \times {10^3}\;{\rm{V}}$
The expression for the power loss in the form of heat is given as,
${P_L} = \dfrac{{{P^2} \times R}}{{{V^2}}}$
Solve by substituting the values in above equation as,

{P_L} = \dfrac{{{P^2} \times R}}{{{V^2}}}\\\ {P_L} = \dfrac{{{{\left( {2.2 \times {{10}^3}\;{\rm{W}}} \right)}^2} \times 10\;\Omega }}{{{{\left( {22 \times {{10}^3}\;{\rm{V}}} \right)}^2}}}\\\ {P_L} = 0.1\;{\rm{watts}} \end{array}$$ Therefore, the required value of power loss in the form of heat is 0.1 watts **So, the correct answer is “Option A”.** **Note:** To resolve the given problem, it is required to understand the concept of power loss taking place in any electrical instrumentations. The mathematical formula of the power supply can be applied using the terms like the supplied voltage, resistance, through which the current tends to flow, along with the appropriate units. Moreover, the transmission possesses many practical applications involving transformers and mechanical instrumentations. Besides, these applications of power transmission involve severe analysis of voltages and the current flowing through the point.