Question

A battery of 12V is connected in series with R1(2ohms),
R2(3 ohms) and R3(1

ohm).What is the voltage drop across R2?

Answer 1

equivalent resistance = 3+2+1 = 6 ohms
current i = 12/6 = 2 A
this current is same for all resistor 
hence,voltage drop across 2 ohm = i*R2 = 2*2 = 4 volts

Answer 2

Total resistance in the circuit = 2 + 3 + 1 = 6 ohms - as these are connected in series , we add them
the current flowing in the circuit is EMF voltage of battery / total resistance
           = 12 V / 6 ohms = 2 Amperes
The voltage difference or potential drop across resistance is I R
             = 2 amp * 2 ohms = 4  ohms