**Question:**

(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.

**Solution:**

(a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.

Total resistance = 1 + 2 + 3 = 6 Ω

(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,

$I=\frac{E}{R}$

$=\frac{12}{6}=2 \mathrm{~A}$

otential drop across 1 Ω resistor = *V*1

From Ohm’s law, the value of *V*1 can be obtained as

*V*1 = 2 × 1= 2 V … (i)

Potential drop across 2 Ω resistor = *V*2

Again, from Ohm’s law, the value of *V*2 can be obtained as

*V*2 = 2 × 2= 4 V … (ii)

Potential drop across 3 Ω resistor = *V*3

Again, from Ohm’s law, the value of *V*3 can be obtained as

*V*3 = 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.