What is meant by dc drop?
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Redrawn circuit with higher voltages at top.

It should be easier to understand if we redraw in a more conventional manner with higher voltages at the top, etc.


Voltage at A will be equal to Vcc, the supply voltage.
The voltage at B will depend on how much current is able to flow through Q1. This, in turn, will depend on how "turned-on" Q1 is.
The voltage drop is simply \$V_A- V_B\$.


A quick example:


Vcc = 12 V.
Rc = 1 kΩ.
Ic (the current through Rc and Q1) = 5 mA.


From Ohm's Law, \$ V = IR \$ we get the voltage drop across Rc as \$ V = 0.005 \cdot 1000 = 5~V \$.

We can, using this result, calculate \$ V_B = 12 - 5 = 7~V \$.