Question

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor?

Answer 1

Given: R₁ = 0.2 Ω

R₂ = 0.3 Ω

R₃ = 0.4 Ω

R₄ = 0.5 Ω

R₅ = 12 Ω

V = 9 V

Therefore, the resultant resistance is given as:

R = R₁ +  R₂ + R₃ + R₄ + R₅

R = 0.2 + 0.3 + 0.4 + 0.5 + 12

R = 13.4 Ω

The current flowing through the 12 Ω resistance is given as:

I = `V/R`

I = `9/13.4`

I = 0.67 ampere