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?

We have,

Potential difference, V = 9 V

Total resistance, R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

Now, according to Ohm's law

Current through the series circuit, I =

\frac{V}{R} = \frac{9 V}{13.4 Ω} = 0.67 A

Thus,

Current through 12 Ω resistor = 0.67 A.

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