Asked in ETEA MDCAT 2013 2013Moderate

A generator produces 100 kW of power at a potential difference of 10KV. The power is transmitted through cables of total resistance 5Q. How much power is dissipated in the cables?

Correct answer: C. 500 W

  • A. 50 W
  • B. 750 W
  • C. 500 W
  • D. 1000 W

Explanation

To find the power dissipated in the cables, you need to first calculate the current flowing through the circuit. The power produced by the generator is 100 kW, which is equal to 100,000 W, and the potential difference is 10 kV, equal to 10,000 V. Using the formula P = VI, where P is power, V is voltage, and I is current, you can solve for current:I = P / V = 100,000 W / 10,000 V = 10 A.Now, using the formula for power dissipation in the cables, P = I²R, where I is the current (10 A) and R is the resistance of the cables (5 ohms), you calculate:P = 10² x 5 = 100 x 5 = 500 W.Therefore, 500 W of power is dissipated in the cables.Options A (50 W), B (750 W), and D (1000 W) are incorrect as they do not result from the correct application of the power loss formula I²R with the calculated current.

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About Resistance and Resistivity

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