Question

Three resistances of 1 Ω, 2 Ω and 3 Ω are connected in series combination. Find out equivalent resistance of the combination. If this combination is connected by the battery of 12 V e.m.f. and negligible internal resistance then find out the voltage across each ends of resistance.

Answer 1

Connecting 1 Ω, 2 Ω and 3 Ω in series
R_equi = R₁ + R₂ + R₃ = 1 + 2 + 3 = 6 Ω
Current flowing in circuit

Potential on 1 Ω resistance = V = IR = 2 × 1 = 2V
Potential on 2 Ω resistance V = IR = 2 × 2 = 4V
Potential on 6 Ω resistance (V) = IR = 2 × 3 = 6V