Chapter 18, Problem 39P

To determine

Dissipated power when the electricity is delivered at $50000V$ rather than $12000V$ .

Answer to Problem 39P

  $8008W$

Explanation of Solution

Given info:

Power, $P=620kW=6.20\times {10}^{5}W$

Voltage, $12000V$

Total resistance, $3.0\Omega$

Formula used:

By the definition of the power,

  $P=VI$

Where, $V$ is voltage and $I$ is current.

By the ohm’s law, a new expression for electrical power will be $P=\frac{V^2}{R}=I^{2}R$

Where, $V$ is voltage, $I$ is current and $R$ is resistance.

Calculation:

By using the formula of power,

  $P_{dissipated}=I^{2}R=\frac{P^2}{V^2}R$

  $P_{dissipatedat12,000V}=\frac{{\left(6.20\times {10}^{5}W\right)}^2}{{\left(1.2\times {10}^{4}V\right)}^2}\left(3.0\Omega \right)=8008W$

Conclusion: The required power is $8008W$ .