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18/07/2021 12:43 pm
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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. How much current would flow through the 12 Ω resistor?
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18/07/2021 12:52 pm
R1 = 0.2 ohm, R2 = 0.4 ohm, R3 = 0.3 ohm, R4 = 0.5 ohm, R5 = 12 ohm, V = 9V
Resultant resistance = R1 + R2 + R3 + R4 + R5
R = 0.2 + 0.4 + 0.3 + 0.5 + 12 = 13.4 ohm
Thus the current flow through 12 ohm resistance will be = \(\frac{V}{R}\)
I = \(\frac{9}{13.4}\)
I = 0.67 amp