Question

A battery of 9 V is connected in series with resistors of 0.2 Ohms, 0.3 Ohms, 0.4 Ohms and 0.9 Ohms respectively

Find out
Amount of the current flow through the 9 Ω resistors

please answer​

Answer 1

Step-by-step explanation:

We have ,

potential difference,V=9V

total resistance ,R=0.2+0.3+0.4+0.5+12=13.4Ω

Now,According to ohm's law

current throught the series circuit ,I=

V by R = 9V by 13.4Ω=0.67 A