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A battery of 9v 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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0.5 A

0.75 A

0.67 A

0.45 A

answer is C.

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Detailed Solution

Current flowing in the circuit can be found by using Ohm’s Law,

V = I \times R

Combined resistance,

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

Potential difference ,

V = 9 V

(given)
Hence ,

9 = I \times 13.4

Current ,

I = \frac{9}{13.4}

I = 0.67 A

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