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

Given,

R1 = 0.2 Ω R2 = 0.3 Ω R3 = 0.4 Ω R4 = 0.5 Ω R5 = 12 Ω

V = 9V

Therefore, the resultant resistance is given as:

R = R1 + R2 + R3 + R4 + R5

= 0.2 + 0.3 + 0.4 + 0.5 + 12

= 13.4 Ω

The current flow through 12 Ω resistance is given as: V/R Therefore, I = 9/13.4

I = 0.67amp