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

R=0.2+0.3+0.4+0.5+12 ohm

R= 13.4 ohm

V=IR

9=I×13.4

I=9/13.4

I=0.67 A

Answer 2

Answer:

It is the correct answer.

Step-by-step explanation:

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