Answer 1

 Current flowing through the circuit = I

Emf of the battery, E = 12 V

Total resistance of the circuit, R = 6 Ω

The relation for current using Ohm’s law is,

`I=E/A`

`=12/6=2A`

Potential drop across 1 Ω resistor = V₁

From Ohm’s law, the value of V₁ can be obtained as

V₁ = 2 × 1= 2 V … (i)

Potential drop across 2 Ω resistor = V₂

Again, from Ohm’s law, the value of V₂ can be obtained as

V₂ = 2 × 2= 4 V … (ii)

Potential drop across 3 Ω resistor = V₃

Again, from Ohm’s law, the value of V₃ can be obtained as

V₃ = 2 × 3= 6 V … (iii)

Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.