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 ?

Options

A. 0.67 A
B. 2.3 A
C. 4.7 A
D. 0.5 A

Answer

A

Explanation

Since all the resistors are in series, equivalent resistance,
`R_s = 0.2 Omega + 0.3 Omega + 0.4 Omega + 0.5 Omega+ 12 Omega = 13.4 Omega`
Current through the circuit, `I = (V)/(R_s) = (9 V)/(13.4 Omega) = 0.67 A`
In series, same current `(I)` flows through all the resistors. Thus, current flowing through `12 Omega` resistor
=`0.67 A`.