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

Explanation

All the resistors are connected in series. So,
Combined resistance, `R = 0.2 + 0.3 +0.4 +0.5 +12`
`=13.4 Omega`
Potential difference, `V = 9 V`
And, Current, `I = ?` (To be calculated)
Current flowing through the whole circuit can be calculated by using Ohm's law:
Now, `(V)/(I) = R`
So, `(9)/(I) = 13.4`
And, `I = (9)/(13.4)`
Current, `I = 0.67 A`
Since the same current flows through all the resistors connected in series in a circuit, therefore, 0.67 ampere of current would flow through the `12 Omega` resistor.