A line has a total resistance of 12 Ω. At the sending end the voltage is 33 kV and the current is…

Step 1. The power sent. $P = VI = 33 \times 10^3\ \text{V} \times 20\ \text{A} = 660\,000\ \text{W} = 660$ kW.

Step 2. The voltage lost. $Ir = 20\ \text{A} \times 12\ \Omega = 240$ V. This is about $0.7\ \%$ of $33$ kV.

Step 3. The heat. $I^2r = (20\ \text{A})^2 \times 12\ \Omega = 4800\ \text{W} = 4.8$ kW. As a check, $240\ \text{V} \times 20\ \text{A} = 4800$ W.

Step 4. The efficiency. The power delivered is $660\ \text{kW} - 4.8\ \text{kW} = 655.2$ kW. So the efficiency is $\dfrac{655.2\ \text{kW}}{660\ \text{kW}} = 0.993$, which is $99.3\ \%$.