A coal-fired power station produces 600 MW of electrical power. The electricity is transmitted to distant towns and cities through overhead cables. The stat...

Assessment: Physics 0625 | Paper 4 Mock 01 | Theory (Extended) Subject: Physics - 0625

Question 1 Report

A coal-fired power station produces 600 MW of electrical power. The electricity is transmitted to distant towns and cities through overhead cables. The station manager must choose between transmitting at 132 000 V or at 400 000 V. The total resistance of the cable route is 4.0 Ω. Both step-up and step-down transformers are available and are assumed to be 100% efficient. The manager wants to minimise the energy wasted as heat in the cables while keeping costs reasonable.

(a) Calculate the current in the cables if the electricity is transmitted at 132 000 V. [2]

(b) Calculate the power lost in the cables at this current. [2]

(c) Calculate the power lost in the cables if the electricity is transmitted at 400 000 V instead. [2]

(d) Explain, with reference to the equation P = I²R, why the higher voltage gives a much lower power loss. [2]

Answer Details

Marking Scheme

  • (a) \(I = \frac{P}{V} = \frac{600\,000\,000}{132\,000}\) [1]; \(I = 4545\text{ A}\) [1]
  • (b) \(P_{\text{loss}} = I^2 R = 4545^2 \times 4.0\) [1]; \(P_{\text{loss}} = 82\,600\,000\text{ W} \approx 83\text{ MW}\) [1]
  • (c) \(I = \frac{600\,000\,000}{400\,000} = 1500\text{ A}\); \(P_{\text{loss}} = 1500^2 \times 4.0\) [1]; \(P_{\text{loss}} = 9\,000\,000\text{ W} = 9.0\text{ MW}\) [1]
  • (d) For fixed power, increasing voltage reduces current (since \(P = IV\)) [1]; \(P_{\text{loss}} = I^2 R\) means loss depends on the square of current, so a smaller current gives a much larger reduction in power lost [1]

Explanation

(a) The power station produces 600 MW = 600,000,000 W. Using \(P = IV\):

\(I = \frac{P}{V} = \frac{600\,000\,000}{132\,000} = 4545\text{ A}\)

(b) The power dissipated as heat in the cables depends on the current and resistance:

\(P_{\text{loss}} = I^2 R = (4545)^2 \times 4.0 = 20\,657\,025 \times 4.0 = 82\,600\,000\text{ W} \approx 83\text{ MW}\)

This is about 14% of the total power output - a substantial waste.

(c) At the higher transmission voltage:

\(I = \frac{600\,000\,000}{400\,000} = 1500\text{ A}\)

\(P_{\text{loss}} = (1500)^2 \times 4.0 = 2\,250\,000 \times 4.0 = 9\,000\,000\text{ W} = 9.0\text{ MW}\)

This is only about 1.5% of the total power - a dramatic improvement.

(d) The key insight is the squared relationship in \(P_{\text{loss}} = I^2 R\). When the voltage is increased from 132 kV to 400 kV (a factor of about 3), the current decreases by the same factor of 3 (from 4545 A to 1500 A). But because power loss depends on \(I^2\), the loss decreases by a factor of \(3^2 = 9\). This is confirmed by the numbers: 83 MW / 9.0 MW is approximately 9. This squared dependence is the fundamental reason why electricity is transmitted at very high voltages - even a modest increase in transmission voltage produces a large reduction in energy wasted as heat in the cables.

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