Question

i. Three resistors 1 Ω, 2 Ω and 3 Ω are combined in series. What is the total resistance of the combination?

ii. If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer 1

Given: R₁ = 1Ω, R₂ = 2 Ω, R₃ = 3 Ω,

V = 12 V

To Find: i. Total resistance (R)

ii. P.D Across R₁, R₂, R₃ (V₁, V₂, V₃ respectively)

Formulae:

i. R_s = R₁ + R₂ + R₃

ii. V = IR

Calculation:

From formula (i):

Rs = l + 2 + 3 = 6 Ω

From formula (ii),

1 = \(\frac{V}{R}=\frac{12}{6}\) = 2A

∴ V₁ = IR₁ = 2 × 1 = 2 V

∴ V₂ = IR₂ = 2 × 2 = 4 V

∴ V₃ = IR₃ = 2 × 3 = 6 V