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MDCAT
Physics
2013

Physics · Magnetism and Electromagnetic Induction

Work through this past-paper style MCQ, then read the full explanation. Practice more physics questions on mMCQ with adaptive practice and topic analytics.

Question

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? 

Options
  • 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.

This calculation underestimates the power dissipation. Ensure you use the correct formula for power loss.

This is incorrect as it does not result from the correct application of the power loss formula I²R.

Using the formula P = I²R and calculating the current as 10 A, the power dissipated is correctly calculated as 500 W.

This option is incorrect as the calculation exceeds the actual power dissipation calculated by the formula I²R.

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Tagged under Physics · Magnetism and Electromagnetic Induction · 2013