Electricity is supplied to a school along a cable of total resistance \(0.5\,\Omega\) with the maximum current drawn from the mains as \(100\text{ A}\). The...
Question 1 Report
Electricity is supplied to a school along a cable of total resistance \(0.5\,\Omega\) with the maximum current drawn from the mains as \(100\text{ A}\). The maximum energy dissipated as heat for \(1\text{ hr}\) is
Answer Details
Electricity supplied to a school through a cable has a certain resistance. This means that as current flows through the cable, some of the electrical energy is lost as heat due to the resistance of the cable. The amount of energy lost as heat is given by the formula: Energy = Current^2 x Resistance x Time In this question, we are given the resistance of the cable and the maximum current drawn from the mains, which is 100A. We are also told that we need to calculate the maximum energy dissipated as heat for 1 hour (which is 60 minutes). Using the formula above, we can calculate the maximum energy dissipated as heat as follows: Energy = Current^2 x Resistance x Time Energy = 100^2 x 0.5 x 3600 (since 1 hour = 3600 seconds) Energy = 18,000,000 J Energy = 1.8 x 10^7 J Therefore, the maximum energy dissipated as heat for 1 hour is 1.8 x 10^7 J. The correct option is: - 1.8 x 107J
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