A thundercloud similar to the one described in Example 24.3 produces a lightning bolt that strikes a radio tower. If the lightning bolt transfers \(5.00 \mathrm{C}\) of charge in about \(0.100 \mathrm{~ms}\) and the potential remains constant at \(70.0 \mathrm{MV}\), find (a) the average current, (b) the average power, (c) the total energy, and (d) the effective resistance of the air during the lightning strike.

To calculate the average current, use the formula \(I = \frac{Q}{t}\). Given the transferred charge \(Q = 5.00 \mathrm{C}\) and time \(t = 0.100 \mathrm{ms} = 0.100\times10^{-3}\mathrm{s}\), we can calculate the average current as follows: \(I = \frac{5.00}{0.100\times10^{-3}} = 50,000 \mathrm{A}\) So, the average current during the lightning strike is \(50,000 \mathrm{A}\).

Now, we will calculate the average power using the formula \(P = VI\). We have the potential \(V = 70.0 \mathrm{MV} = 70.0\times10^{6}\mathrm{V}\) and average current \(I = 50,000 \mathrm{A}\). The average power is: \(P = (70.0\times10^{6})(50,000) = 3.50\times10^{12} \mathrm{W}\) The average power during the lightning strike is \(3.50\times10^{12} \mathrm{W}\).

In order to calculate the total energy, we use the formula \(E = QV\). We have the transferred charge \(Q = 5.00 \mathrm{C}\) and potential \(V = 70.0 \mathrm{MV} = 70.0\times10^{6}\mathrm{V}\). Therefore, the total energy is: \(E = (5.00)(70.0\times10^{6}) = 3.50\times10^{8} \mathrm{J}\) The total energy during the lightning strike is \(3.50\times10^{8} \mathrm{J}\).

Finally, we will calculate the effective resistance using the formula \(R = \frac{V}{I}\). We have the potential \(V = 70.0 \mathrm{MV} = 70.0\times10^{6}\mathrm{V}\) and the average current \(I = 50,000 \mathrm{A}\). The effective resistance of the air during the lightning strike is: \(R = \frac{70.0\times10^{6}}{50,000} = 1.40\times10^{3} \mathrm{\Omega}\) The effective resistance of the air during the lightning strike is \(1.40\times10^{3} \mathrm{\Omega}\). In conclusion, during the lightning strike, (a) the average current is \(50,000 \mathrm{A}\), (b) the average power is \(3.50\times10^{12} \mathrm{W}\), (c) the total energy is \(3.50\times10^{8} \mathrm{J}\), and (d) the effective resistance of the air is \(1.40\times10^{3} \mathrm{\Omega}\).