A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
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since they are in series total resistance is 0.2+0.3+0.5+0.4+12=13.4 current=V/R=9/13.4= 0.671 A. same current flows through all the resistors in series connection. hence, current in 12 ohm resistor=0.671A