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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?
The resistance is connected in series
RS = 0.2Ω + 0.3Ω + 0.4Ω + 0.5Ω + 12Ω
RS = 13.4Ω
Current = I = V/RS
I = 9/13.4
I = 0.67A
So, the current flowing through 12 W resistor = 0.67 A.
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