Can someone please explain a voltage drop in a circuit please?

When you are given this circuit visualize the current flowing in this circuit . Current will be 2 amps 12 by (2 +4) the resistors are in series right so total resistance is 6 ohms and voltage of power source is 12 so 2 amos current will flow .

Voltage drop is current x resistance

So 2 amps is flowing and it passes through the 2 ohm resistance , it causes a voltage drop of 4 volts . So now at the end of 2 ohm resistor the potential difference will be 12 minus 4 volts wrt to the negetive terminal of the battery . Similarly when 2 amp pass through the 4 ohm resistor it agains drops to 4 x 2 is equal to 8 volts

If you change resitors say 1 ohm and 1 ohm then current becomes 6 amps and the there is a voltage drop of 6 volts each on both the 1 ohm resistors which are in series with a 12 volts battery .

More is the resistance more is the voltage drop

To help u understand u know as per ohms law v is equal to current x resistance so in your example 4 volts is needed to run 2 amps current through the first resistor of 2 ohm similarly 8 volts is needed to run 2 amps through 4 ohm resistor voltage drop is like the distribution of voltage to run equal current

Because in series the current will be same now if you connect 2 ohm and 4 ohm across a 24 volts battery what will happen 4 amps will run in the circuit and voltage drop will be 8 volts across 2 ohm and 16 volts across 4 ohm again 24 volts got distributed to run same current

Total current = 12v / 6 ohm = 2 amps.

Volt drop on 2 ohm = 2 ohm* 2 amp = 4V

Volt drop on 4 ohm = 12V - 4V = 8V