Understanding voltage drop
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Feeding a short-circuit, feeding a reasonable load and feeding an open-circuit.


In the first case you have a short-circuit at the load end so the voltage is zero and no useful work is done. The current through the circuit will be \$ I = \frac V R = \frac 5 {20} = 0.25\ \text A \$.
In the second case you have a reasonable load of 980 &Omega; (just chosen to make the maths easy). The current through the circuit is \$ I = \frac V R = \frac 5 {1000} = 5\ \text {mA} \$. The voltage at the load will be 4.9 V and the total drop caused by the loop will be 0.1 V.
In the third case the circuit is open. No current is flowing so there is no voltage drop across the 10 &Omega; resistors. The voltage measured will be 5 V.