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

A battery of 9 V is connected in serieswith resistors of 0.2 Omega, 0.

1.	A battery of 9 V is connected in serieswith resistors of 0.2 Omega, 0.3 Omega, 0.4 Omega, 0.5 Omega and 12 Omegarespectively. How much current would flow through the 12 Omega resistor?
Answer» Solution :Since all the given resistors are connected inseries, 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.