A line of resistance 5.0 Ω carries 240 kW at 6.0 kV, with power factor 1…

Step 1. The first current. $I_1 = \dfrac{240 \times 10^3\ \text{W}}{6.0 \times 10^3\ \text{V}} = 40$ A. The heat was $I_1^2r = (40\ \text{A})^2 \times 5.0\ \Omega = 8000$ W.

Step 2. The new current. $I_2 = \dfrac{240 \times 10^3\ \text{W}}{30 \times 10^3\ \text{V}} = 8$ A. The voltage is five times larger, so the current is five times smaller.

Step 3. The new heat. $I_2^2r = (8\ \text{A})^2 \times 5.0\ \Omega = 320$ W.

Step 4. The factor. $\dfrac{8000\ \text{W}}{320\ \text{W}} = 25$. This is $5^2$.