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, respectively. How much current would flow through the 12 Ω resistor?

Answer 1

When equivalent resistances are connected in series,

R = R₁ + R₂ + R₃ + R₄ + R₅

R₁ = 0.2 Ω, R₂ = 0.3 Ω, R₃ = 0.4 Ω, R₄ = 0.5 Ω, R₅ = 12 Ω  ....(Given)

R = 0.2 + 0.3 + 0.4 + 0.5 + 12

R = 13.4 Ω

V = IR  ....(According to Ohm’s law)

I = `"V"/"R"`

= `9/13.4`

= 0.67 A

There will be no division of electric current because the resistors are connected in series.

0.67 A of electric current will flow in a 12 Ω resistor.