(a) Three resistors 1 Ω, 2 Ω, and 3 Ω 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.

Ans (a).

Three resistors of resistances 1 Ω, 2 Ω, and 3Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = l + 2 + 3 = 6Ω

Ans (b).

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/R

=12/6 =2 A

Potential drop across 1 Ω resistor = V₁

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

V₁ = 2 x 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 x 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 x 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.