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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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Total resistance of resistors when connected in series is given by
⟹=0.2 Ω+0.3 Ω+0.4 Ω+0.5 Ω+12 Ω=13.4 Ω
According to Ohm’s law, V = IR

There is no current division occurring in a series circuit. So, the current through the 12 Ω resistor will be same as 0.67 A.

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