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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?

Current Electricity

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Answer

Given,

V = 9V

Resistors in series = 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω,

Rs = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

According to Ohm's Law:

V = IR

∴I=VR⇒I=913.4⇒I=0.67 A\therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{9}{13.4} \\[1em] \Rightarrow \text{I} = 0.67 \text{ A}

Therefore, current flowing = 0.67 A.

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