$(a)$ Three resistors $1 \;\Omega, 2 \;\Omega$ and $3 \;\Omega$ are combined in series. What is the total resistance of the combination?
$(b)$ If the combination is connected to a battery of $emf \; 12 \; V$ and negligible internal resistance,obtain the potential drop across each resistor.

(A) Three resistors of resistances $1 \; \Omega, 2 \; \Omega$ and $3 \; \Omega$ are combined in series. The total resistance of the combination is given by the algebraic sum of individual resistances.
Total resistance $= 1 + 2 + 3 = 6 \; \Omega$
$(b)$ Current flowing through the circuit $= I$. The $emf$ of the battery,$E = 12 \; V$. The total resistance of the circuit,$R = 6 \; \Omega$.
Using Ohm's law,the current is $I = \frac{E}{R} = \frac{12}{6} = 2 \; A$.
Potential drop across $1 \; \Omega$ resistor $= V_1$. From Ohm's law,$V_1 = I \times R_1 = 2 \times 1 = 2 \; V$.
Potential drop across $2 \; \Omega$ resistor $= V_2$. From Ohm's law,$V_2 = I \times R_2 = 2 \times 2 = 4 \; V$.
Potential drop across $3 \; \Omega$ resistor $= V_3$. From Ohm's law,$V_3 = I \times R_3 = 2 \times 3 = 6 \; V$.
Therefore,the potential drops across the $1 \; \Omega, 2 \; \Omega$ and $3 \; \Omega$ resistors are $2 \; V, 4 \; V$ and $6 \; V$ respectively.