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A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4Ω, 0.5Ω and 12Ω, respectively. How much current would flow through the 12Ωresistor?
Battery voltage: 9V
Resistors (series)=
R1=0.2Ω R2=0.3Ω R3=0.4Ω R4=0.5Ω R5=12Ω
Net Resistance Rnet= R1+R2+R3+R4+R5
Rnet= (0.2+0.3+0.4+0.5+12)Ω
Rnet=13.4Ω
As per ohm's law
V = IR
9V = IX13.4Ω
I= 9V/13.4Ω
I= 0.671A
0.671A current passes through 12Ω.
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