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` respectively. How much current would flow through the `12 Omega` resistor?

Explanation

Since all the given resistors are connected in series, their equivalent resistance
`R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Omega`
The current through the circuit,
`I = V/R_s = 9/13.4 = 0.67 A`
In a series combination, the same current I I flows through all the resistors, so the current flowing through `12Omega`s resistor = 0.67 A.