(a) Three resistors of resistances 1$\mathrm{\Omega }$, 2$\mathrm{\Omega }$ and 3$\mathrm{\Omega }$ are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = 1 + 2 + 3 = 6 $\mathrm{\Omega }$

(b) Current flowing through the circuit = I

Emf of the battery, E = 12 V

Total resistance of the circuit, R = 6 $\mathrm{\Omega }$

The relation for current using Ohm's law is,

$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}$
$=\frac{12}{6}=2 \mathrm{A}$

Potential drop across 1 $\mathrm{\Omega }$ resistor = V₁

From Ohm's law, the value of V₁ can be obtained as V₁ = 2 $\times$ 1 = 2 V ...(1)

Potential drop across 2 $\mathrm{\Omega }$ resistor = V₂

Again, from Ohm's law, the value of V₂ can be obtained as V₂ = 2 $\times$ 2 = 4 V ...(2)

Potential drop across 3 $\mathrm{\Omega }$ resistor = V₃

Again, from Ohm's law, the value of V₃ can be obtained as V₃ = 2 $\times$ 3 = 6 V ...(3)

Therefore, the potential drop across 1 $\mathrm{\Omega }$, 2 $\mathrm{\Omega }$, and 3 $\mathrm{\Omega }$ resistors are 2 V, 4 V, and 6 V respectively.